9.3: Invariant Subspaces and Direct Sums (2024)

Eigenvalues

Let \(T : V \to V\) be a linear operator. A one-dimensional subspace \(\mathbb{R}\mathbf{v}\), \(\mathbf{v} \neq \mathbf{0}\), is \(T\)-invariant if and only if \(T(r\mathbf{v}) = rT(\mathbf{v})\) lies in \(\mathbb{R}\mathbf{v}\) for all \(r\) in \(\mathbb{R}\). This holds if and only if \(T(\mathbf{v})\) lies in \(\mathbb{R}\mathbf{v}\); that is, \(T(\mathbf{v}) = \lambda\mathbf{v}\) for some \(\lambda\) in \(\mathbb{R}\). A real number \(\lambda\) is called an eigenvalue of an operator \(T : V \to V\) if

\[T(\mathbf{v}) = \lambda\mathbf{v} \nonumber \]

holds for some nonzero vector \(\mathbf{v}\) in \(V\). In this case, \(\mathbf{v}\) is called an eigenvector of \(T\) corresponding to \(\lambda\). The subspace

\[E_{\lambda}(T) = \{\mathbf{v} \mbox{ in } V \mid T(\mathbf{v}) = \lambda\mathbf{v}\} \nonumber \]

is called the eigenspace of \(T\) corresponding to \(\lambda\). These terms are consistent with those used in Section [sec:5_5] for matrices. If \(A\) is an \(n \times n\) matrix, a real number \(\lambda\) is an eigenvalue of the matrix operator \(T_{A} : \mathbb{R}^n \to \mathbb{R}^n\) if and only if \(\lambda\) is an eigenvalue of the matrix \(A\). Moreover, the eigenspaces agree:

\[E_{\lambda}(T_A) = \{\mathbf{x} \mbox{ in } \mathbb{R}^n \mid A\mathbf{x} = \lambda\mathbf{x} \} = E_{\lambda}(A) \nonumber \]

The following theorem reveals the connection between the eigenspaces of an operator \(T\) and those of the matrices representing \(T\).

029450 Let \(T : V \to V\) be a linear operator where \(dim \;V = n\), let \(B\) denote any ordered basis of \(V\), and let \(C_{B} : V \to \mathbb{R}^n\) denote the coordinate isomorphism. Then:

1. The eigenvalues \(\lambda\) of \(T\) are precisely the eigenvalues of the matrix \(M_{B}(T)\) and thus are the roots of the characteristic polynomial \(c_{T}(x)\).
2. In this case the eigenspaces \(E_{\lambda}(T)\) and \(E_{\lambda}[M_{B}(T)]\) are isomorphic via the restriction \(C_{B} : E_{\lambda}(T) \to E_{\lambda}[M_{B}(T)]\).

Write \(A = M_{B}(T)\) for convenience. If \(T(\mathbf{v}) = \lambda\mathbf{v}\), then \({\lambda}C_{B}(\mathbf{v}) = C_{B}[T(\mathbf{v})] = AC_{B}(\mathbf{v})\) because \(C_{B}\) is linear. Hence \(C_{B}(\mathbf{v})\) lies in \(E_{\lambda}(A)\), so we do have a function \(C_{B} : E_{\lambda}(T) \to E_{\lambda}(A)\). It is clearly linear and one-to-one; we claim it is onto. If \(\mathbf{x}\) is in \(E_{\lambda}(A)\), write \(\mathbf{x} = C_{B}(\mathbf{v})\) for some \(\mathbf{v}\) in \(V\) (\(C_{B}\) is onto). This \(\mathbf{v}\) actually lies in \(E_{\lambda}(T)\). To see why, observe that

\[C_B[T(\mathbf{v})] = AC_B(\mathbf{v}) = A\mathbf{x} = \lambda\mathbf{x} = {\lambda}C_B(\mathbf{v}) = C_B(\lambda\mathbf{v}) \nonumber \]

Hence \(T(\mathbf{v}) = \lambda\mathbf{v}\) because \(C_{B}\) is one-to-one, and this proves (2). As to (1), we have already shown that eigenvalues of \(T\) are eigenvalues of \(A\). The converse follows, as in the foregoing proof that \(C_{B}\) is onto.

Theorem [thm:029450] shows how to pass back and forth between the eigenvectors of an operator \(T\) and the eigenvectors of any matrix \(M_{B}(T)\) of \(T\):

\[\mathbf{v} \mbox{ lies in } E_{\lambda}(T) \quad \mbox{ if and only if } \quad C_B(\mathbf{v}) \mbox{ lies in } E_{\lambda}[M_B(T)] \nonumber \]

029494 Find the eigenvalues and eigenspaces for \(T :\|{P}_{2} \to\|{P}_{2}\) given by

\[T(a + bx + cx^2) = (2a + b + c) + (2a + b - 2c)x - (a + 2c)x^2 \nonumber \]

If \(B = \{1, x, x^{2}\}\), then

\[M_B(T) = \left[ \begin{array}{ccc} C_B[T(1)] & C_B[T(x)] & C_B[T(x^2)] \end{array} \right] = \left[ \begin{array}{rrr} 2 & 1 & 1 \\ 2 & 1 & -2 \\ -1 & 0 & -2 \end{array} \right] \nonumber \]

Hence \(c_{T}(x) = \det [xI - M_{B}(T)] = (x + 1)^{2}(x - 3)\) as the reader can verify.

Moreover, \(E_{-1}[M_B(T)] = \mathbb{R}\left[ \begin{array}{r} -1 \\ 2 \\ 1 \end{array} \right]\) and \(E_3[M_B(T)] = \mathbb{R}\left[ \begin{array}{r} 5 \\ 6 \\ -1 \end{array} \right]\), so Theorem [thm:029450] gives \(E_{-1}(T) = \mathbb{R}(-1 + 2x + x^{2})\) and \(E_{3}(T) = \mathbb{R}(5 + 6x - x^{2})\).

029517 Each eigenspace of a linear operator \(T : V \to V\) is a \(T\)-invariant subspace of \(V\).

If \(\mathbf{v}\) lies in the eigenspace \(E_{\lambda}(T)\), then \(T(\mathbf{v}) = \lambda\mathbf{v}\), so \(T[T(\mathbf{v})] = T(\lambda\mathbf{v}) = {\lambda}T(\mathbf{v})\). This shows that \(T(\mathbf{v})\) lies in \(E_{\lambda}(T)\) too.

Direct Sums

Sometimes vectors in a space \(V\) can be written naturally as a sum of vectors in two subspaces. For example, in the space \(\mathbf{M}_{nn}\) of all \(n \times n\) matrices, we have subspaces

\[U = \{P \mbox{ in }\|{M}_{nn} \mid P \mbox{ is symmetric } \} \quad \mbox{ and } \quad W = \{Q \mbox{ in }\|{M}_{nn} \mid Q \mbox{ is skew symmetric} \} \nonumber \]

where a matrix \(Q\) is called skew-symmetric if \(Q^{T} = -Q\). Then every matrix \(A\) in \(\mathbf{M}_{nn}\) can be written as the sum of a matrix in \(U\) and a matrix in \(W\); indeed,

\[A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T) \nonumber \]

where \(\frac{1}{2}(A + A^T)\) is symmetric and \(\frac{1}{2}(A - A^T)\) is skew symmetric. Remarkably, this representation is unique: If \(A = P + Q\) where \(P^{T} = P\) and \(Q^{T} = -Q\), then \(A^{T} = P^{T} + Q^{T} = P - Q\); adding this to \(A = P + Q\) gives \(P = \frac{1}{2}(A + A^T)\), and subtracting gives \(Q = \frac{1}{2}(A - A^{T})\). In addition, this uniqueness turns out to be closely related to the fact that the only matrix in both \(U\) and \(W\) is \(0\). This is a useful way to view matrices, and the idea generalizes to the important notion of a direct sum of subspaces.

If \(U\) and \(W\) are subspaces of \(V\), their sum \(U + W\) and their intersection \(U \cap W\) were defined in Section [sec:6_4] as follows:

\[\begin{aligned} U + W & = \{ \mathbf{u} + \mathbf{w} \mid \mathbf{u} \mbox{ in } U \mbox{ and } \mathbf{w} \mbox{ in } W \} \\ U \cap W & = \{ \mathbf{v} \mid \mathbf{v} \mbox{ lies in both } U \mbox{ and } W \}\end{aligned} \nonumber \]

These are subspaces of \(V\), the sum containing both \(U\) and \(W\) and the intersection contained in both \(U\) and \(W\). It turns out that the most interesting pairs \(U\) and \(W\) are those for which \(U \cap W\) is as small as possible and \(U + W\) is as large as possible.

Direct Sum of Subspaces029545 A vector space \(V\) is said to be the direct sum of subspaces \(U\) and \(W\) if

\[U \cap W = \{ \mathbf{0} \} \quad \mbox{ and } \quad U + W = V \nonumber \]

In this case we write \(V = U \oplus W\). Given a subspace \(U\), any subspace \(W\) such that \(V = U \oplus W\) is called a complement of \(U\) in \(V\).

029550 In the space \(\mathbb{R}^5\), consider the subspaces \(U = \{(a, b, c, 0, 0) \mid a, b, \mbox{ and } c \mbox{ in } \mathbb{R}\}\) and \(W = \{(0, 0, 0, d, e) \mid d \mbox{ and } e \mbox{ in } \mathbb{R}\}\). Show that \(\mathbb{R}^5 = U \oplus W\).

If \(\mathbf{x} = (a, b, c, d, e)\) is any vector in \(\mathbb{R}^5\), then \(\mathbf{x} = (a, b, c, 0, 0) + (0, 0, 0, d, e)\), so \(\mathbf{x}\) lies in \(U + W\). Hence \(\mathbb{R}^5 = U + W\). To show that \(U \cap W = \{\mathbf{0}\}\), let \(\mathbf{x} = (a, b, c, d, e)\) lie in \(U \cap W\). Then \(d = e = 0\) because \(\mathbf{x}\) lies in \(U\), and \(a = b = c = 0\) because \(\mathbf{x}\) lies in \(W\). Thus \(\mathbf{x} = (0, 0, 0, 0, 0) = \mathbf{0}\), so \(\mathbf{0}\) is the only vector in \(U \cap W\). Hence \(U \cap W = \{\mathbf{0}\}\).

029560 If \(U\) is a subspace of \(\mathbb{R}^n\), show that \(\mathbb{R}^n = U \oplus U^{\perp}\).

The equation \(\mathbb{R}^n = U + U^{\perp}\) holds because, given \(\mathbf{x}\) in \(\mathbb{R}^n\), the vector \(\proj{U}{\mathbf{x}}\) lies in \(U\) and \(\mathbf{x} - \proj{U}{\mathbf{x}}\) lies in \(U^{\perp}\). To see that \(U \cap U^{\perp} = \{\mathbf{0}\}\), observe that any vector in \(U \cap U^{\perp}\) is orthogonal to itself and hence must be zero.

029575 Let \(\{\mathbf{e}_{1}, \mathbf{e}_{2}, \dots, \mathbf{e}_{n}\}\) be a basis of a vector space \(V\), and partition it into two parts: \(\{\mathbf{e}_{1}, \dots, \mathbf{e}_{k}\}\) and \(\{\mathbf{e}_{k+1}, \dots, \mathbf{e}_{n}\}\). If \(U = span \;\{\mathbf{e}_{1}, \dots, \mathbf{e}_{k}\}\) and \(W = span \;\{\mathbf{e}_{k+1}, \dots, \mathbf{e}_{n}\}\), show that \(V = U \oplus W\).

If \(\mathbf{v}\) lies in \(U \cap W\), then \(\mathbf{v} = a_{1}\mathbf{e}_{1} + \cdots + a_{k}\mathbf{e}_{k}\) and \(\mathbf{v} = b_{k+1}\mathbf{e}_{k+1} + \cdots + b_{n}\mathbf{e}_{n}\) hold for some \(a_{i}\) and \(b_{j}\) in \(\mathbb{R}\). The fact that the \(\mathbf{e}_{i}\) are linearly independent forces all \(a_{i} = b_{j} = 0\), so \(\mathbf{v} = \mathbf{0}\). Hence \(U \cap W = \{\mathbf{0}\}\). Now, given \(\mathbf{v}\) in \(V\), write \(\mathbf{v} = v_{1}\mathbf{e}_{1} + \cdots + v_{n}\mathbf{e}_{n}\) where the \(v_{i}\) are in \(\mathbb{R}\). Then \(\mathbf{v} = \mathbf{u} + \mathbf{w}\), where \(\mathbf{u} = v_{1}\mathbf{e}_{1} + \cdots + v_{k}\mathbf{e}_{k}\) lies in \(U\) and \(\mathbf{w} = v_{k+1}\mathbf{e}_{k+1} + \cdots + v_{n}\mathbf{e}_{n}\) lies in \(W\). This proves that \(V = U + W\).

Example [exa:029575] is typical of all direct sum decompositions.

029631 Let \(U\) and \(W\) be subspaces of a finite dimensional vector space \(V\). The following three conditions are equivalent:

1. \(V = U \oplus W\).
3. If \(\{\mathbf{u}_{1}, \dots, \mathbf{u}_{k}\}\) and \(\{\mathbf{w}_{1}, \dots, \mathbf{w}_{m}\}\) are bases of \(U\) and \(W\), respectively, then \(B = \{\mathbf{u}_{1}, \dots, \mathbf{u}_{k}, \mathbf{w}_{1}, \dots, \mathbf{w}_{m}\}\) is a basis of \(V\).

(The uniqueness in (2) means that if \(\mathbf{v} = \mathbf{u}_{1} + \mathbf{w}_{1}\) is another such representation, then \(\mathbf{u}_{1} = \mathbf{u}\) and \(\mathbf{w}_{1} = \mathbf{w}\).)

Example [exa:029575] shows that (3) \(\Rightarrow\) (1).

(1) \(\Rightarrow\) (2). Given \(\mathbf{v}\) in \(V\), we have \(\mathbf{v} = \mathbf{u} + \mathbf{w}\), \(\mathbf{u}\) in \(U\), \(\mathbf{w}\) in \(W\), because \(V = U + W\).

If also \(\mathbf{v} = \mathbf{u}_{1} + \mathbf{w}_{1}\), then \(\mathbf{u} - \mathbf{u}_{1} = \mathbf{w}_{1} - \mathbf{w}\) lies in \(U \cap W = \{\mathbf{0}\}\), so \(\mathbf{u} = \mathbf{u}_{1}\) and \(\mathbf{w} = \mathbf{w}_{1}\).

(2) \(\Rightarrow\) (3). Given \(\mathbf{v}\) in \(V\), we have \(\mathbf{v} = \mathbf{u} + \mathbf{w}\), \(\mathbf{u}\) in \(U\), \(\mathbf{w}\) in \(W\). Hence \(\mathbf{v}\) lies in span \(B\); that is, \(V = span \;B\). To see that \(B\) is independent, let \(a_{1}\mathbf{u}_{1} + \cdots + a_{k}\mathbf{u}_{k} + b_{1}\mathbf{w}_{1} + \cdots + b_{m}\mathbf{w}_{m} = \mathbf{0}\). Write \(\mathbf{u} = a_{1}\mathbf{u}_{1} + \cdots + a_{k}\mathbf{u}_{k}\) and \(\mathbf{w} = b_{1}\mathbf{w}_{1} + \cdots + b_{m}\mathbf{w}_{m}\). Then \(\mathbf{u} + \mathbf{w} = \mathbf{0}\), and so \(\mathbf{u} = \mathbf{0}\) and \(\mathbf{w} = \mathbf{0}\) by the uniqueness in (2). Hence \(a_{i} = 0\) for all \(i\) and \(b_{j} = 0\) for all \(j\).

Condition (3) in Theorem [thm:029631] gives the following useful result.

029686 If a finite dimensional vector space \(V\) is the direct sum \(V = U \oplus W\) of subspaces \(U\) and \(W\), then

\[dim \; V = dim \; U + dim \; W \nonumber \]

These direct sum decompositions of \(V\) play an important role in any discussion of invariant subspaces. If \(T : V \to V\) is a linear operator and if \(U_{1}\) is a \(T\)-invariant subspace, the block upper triangular matrix

\[\label{eqn:9_3_1} M_B(T) = \left[ \begin{array}{cc} M_{B_1}(T) & Y \\ 0 & Z \end{array} \right] \]

in Theorem [thm:029359] is achieved by choosing any basis \(B_{1} = \{\mathbf{b}_{1}, \dots, \mathbf{b}_{k}\}\) of \(U_{1}\) and completing it to a basis \(B = \{\mathbf{b}_{1}, \dots, \mathbf{b}_{k}, \mathbf{b}_{k+1}, \dots, \mathbf{b}_{n}\}\) of \(V\) in any way at all. The fact that \(U_{1}\) is \(T\)-invariant ensures that the first \(k\) columns of \(M_{B}(T)\) have the form in ([eqn:9_3_1]) (that is, the last \(n - k\) entries are zero), and the question arises whether the additional basis vectors \(\mathbf{b}_{k+1}, \dots, \mathbf{b}_{n}\) can be chosen such that

\[U_2 = span \;\{\mathbf{b}_{k+1}, \dots, \mathbf{b}_n\} \nonumber \]

is also \(T\)-invariant. In other words, does each \(T\)-invariant subspace of \(V\) have a \(T\)-invariant complement? Unfortunately the answer in general is no (see Example [exa:029851] below); but when it is possible, the matrix \(M_{B}(T)\) simplifies further. The assumption that the complement \(U_{2} = span \;\{\mathbf{b}_{k+1}, \dots, \mathbf{b}_{n}\}\) is \(T\)-invariant too means that \(Y = 0\) in equation [eqn:9_3_1] above, and that \(Z = M_{B_{2}}(T)\) is the matrix of the restriction of \(T\) to \(U_{2}\) (where \(B_{2} = \{\mathbf{b}_{k+1}, \dots, \mathbf{b}_{n}\}\)). The verification is the same as in the proof of Theorem [thm:029359].

029723 Let \(T : V \to V\) be a linear operator where \(V\) has dimension \(n\). Suppose \(V = U_{1} \oplus U_{2}\) where both \(U_{1}\) and \(U_{2}\) are \(T\)-invariant. If \(B_{1} = \{\mathbf{b}_{1}, \dots, \mathbf{b}_{k}\}\) and \(B_{2} = \{\mathbf{b}_{k+1}, \dots, \mathbf{b}_{n}\}\) are bases of \(U_{1}\) and \(U_{2}\) respectively, then

\[B = \{\mathbf{b}_1, \dots, \mathbf{b}_k, \mathbf{b}_{k+1}, \dots, \mathbf{b}_n\} \nonumber \]

is a basis of \(V\), and \(M_{B}(T)\) has the block diagonal form

\[M_B(T) = \left[ \begin{array}{cc} M_{B_1}(T) & 0 \\ 0 & M_{B_2}(T) \end{array} \right] \nonumber \]

where \(M_{B_{1}}(T)\) and \(M_{B_{2}}(T)\) are the matrices of the restrictions of \(T\) to \(U_{1}\) and to \(U_{2}\) respectively.

Reducible Linear Operator029747 The linear operator \(T : V \to V\) is said to be reducible if nonzero \(T\)-invariant subspaces \(U_{1}\) and \(U_{2}\) can be found such that \(V = U_{1} \oplus U_{2}\).

Then \(T\) has a matrix in block diagonal form as in Theorem [thm:029723], and the study of \(T\) is reduced to studying its restrictions to the lower-dimensional spaces \(U_{1}\) and \(U_{2}\). If these can be determined, so can \(T\). Here is an example in which the action of \(T\) on the invariant subspaces \(U_{1}\) and \(U_{2}\) is very simple indeed. The result for operators is used to derive the corresponding similarity theorem for matrices.

029759 Let \(T : V \to V\) be a linear operator satisfying \(T^{2} = 1_{V}\) (such operators are called involutions). Define

\[U_1 = \{ \mathbf{v} \mid T(\mathbf{v}) = \mathbf{v} \} \quad \mbox{ and } \quad U_2 = \{ \mathbf{v} \mid T(\mathbf{v}) = -\mathbf{v} \} \nonumber \]

1. Show that \(V = U_{1} \oplus U_{2}\).
2. If \(dim \;V = n\), find a basis \(B\) of \(V\) such that \(M_B(T) = \left[ \begin{array}{cc} I_k & 0 \\ 0 & -I_{n-k} \end{array} \right]\) for some \(k\).
3. Conclude that, if \(A\) is an \(n \times n\) matrix such that \(A^{2} = I\), then \(A\) is similar to \(\left[ \begin{array}{cc} I_k & 0 \\ 0 & -I_{n-k} \end{array} \right]\) for some \(k\).

1. Then \(\mathbf{v} + T(\mathbf{v})\) lies in \(U_{1}\), because \(T[\mathbf{v} + T(\mathbf{v})] = T(\mathbf{v}) + T^{2}(\mathbf{v}) = \mathbf{v} + T(\mathbf{v})\). Similarly, \(\mathbf{v} - T(\mathbf{v})\) lies in \(U_{2}\), and it follows that \(V = U_{1} + U_{2}\). This proves part (a).
2. A similar argument shows that \(M_{B_2}(T) = -I_{n-k}\), so part (b) follows with \(B = \{\mathbf{b}_{1}, \mathbf{b}_{2}, \dots, \mathbf{b}_{n}\}\).
3. But Theorem [thm:028841] shows that \(M_{B}(T_{A}) = P^{-1}AP\) for some invertible matrix \(P\), and this proves part (c).

Note that the passage from the result for operators to the analogous result for matrices is routine and can be carried out in any situation, as in the verification of part (c) of Example [exa:029759]. The key is the analysis of the operators. In this case, the involutions are just the operators satisfying \(T^{2} = 1_{V}\), and the simplicity of this condition means that the invariant subspaces \(U_{1}\) and \(U_{2}\) are easy to find.

Unfortunately, not every linear operator \(T : V \to V\) is reducible. In fact, the linear operator in Example [exa:029341] has no invariant subspaces except \(0\) and \(V\). On the other hand, one might expect that this is the only type of nonreducible operator; that is, if the operator has an invariant subspace that is not \(0\) or \(V\), then some invariant complement must exist. The next example shows that even this is not valid.

029851 Consider the operator \(T : \mathbb{R}^2 \to \mathbb{R}^2\) given by \(T\left[ \begin{array}{c} a \\ b \end{array} \right] = \left[ \begin{array}{c} a + b \\ b \end{array} \right]\). Show that \(U_1 = \mathbb{R}\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]\) is \(T\)-invariant but that \(U_1\) has not \(T\)-invariant complement in \(\mathbb{R}^2\).

Because \(U_1 = span \;\left\{\left[ \begin{array}{c} 1 \\ 0 \end{array} \right] \right\}\) and \(T\left[ \begin{array}{c} 1 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 1 \\ 0 \end{array} \right]\), it follows (by Example [exa:029323]) that \(U_{1}\) is \(T\)-invariant. Now assume, if possible, that \(U_{1}\) has a \(T\)-invariant complement \(U_{2}\) in \(\mathbb{R}^2\). Then \(U_{1} \oplus U_{2} = \mathbb{R}^2\) and \(T(U_{2}) \subseteq U_{2}\). Theorem [thm:029686] gives

\[2 = dim \; \mathbb{R}^2 = dim \; U_1 + dim \; U_2 = 1 + dim \; U_2 \nonumber \]

so \(dim \;U_{2} = 1\). Let \(U_{2} = \mathbb{R}\mathbf{u}_{2}\), and write \(\mathbf{u}_2 = \left[ \begin{array}{c} p \\ q \end{array} \right]\). We claim that \(\mathbf{u}_{2}\) is not in \(U_{1}\). For if \(\mathbf{u}_{2} \in U_{1}\), then \(\mathbf{u}_{2} \in U_{1} \cap U_{2} = \{\mathbf{0}\}\), so \(\mathbf{u}_{2} = \mathbf{0}\). But then \(U_{2} = \mathbb{R}\mathbf{u}_{2} = \{\mathbf{0}\}\), a contradiction, as \(dim \;U_{2} = 1\). So \(\mathbf{u}_{2} \notin U_{1}\), from which \(q \neq 0\). On the other hand, \(T(\mathbf{u}_{2}) \in U_{2} = \mathbb{R}\mathbf{u}_{2}\) (because \(U_{2}\) is \(T\)-invariant), say \(T(\mathbf{u}_2) = \lambda\mathbf{u}_2 = \lambda\left[ \begin{array}{c} p \\ q \end{array} \right]\).

Thus

\[\left[ \begin{array}{c} p + q \\ q \end{array} \right] = T\left[ \begin{array}{c} p \\ q \end{array} \right] = \lambda\left[ \begin{array}{c} p \\ q \end{array} \right] \mbox{ where } \lambda \in \mathbb{R} \nonumber \]

Hence \(p + q = \lambda p\) and \(q = \lambda q\). Because \(q \neq 0\), the second of these equations implies that \(\lambda = 1\), so the first equation implies \(q = 0\), a contradiction. So a \(T\)-invariant complement of \(U_{1}\) does not exist.

This is as far as we take the theory here, but in Chapter [chap:11] the techniques introduced in this section will be refined to show that every matrix is similar to a very nice matrix indeed—its Jordan canonical form.

Exercises for 1

solutions

2

[ex:ex9_3_1] If \(T : V \to V\) is any linear operator, show that \(\text{ker }T\) and \(im \;T\) are \(T\)-invariant subspaces.

[ex:ex9_3_2] Let \(T\) be a linear operator on \(V\). If \(U\) and \(W\) are \(T\)-invariant, show that

1. \(U \cap W\) and \(U + W\) are also \(T\)-invariant.
2. \(T(U)\) is \(T\)-invariant.

2. \(T(U) \subseteq U\), so \(T[T(U)] \subseteq T(U)\).

Let \(S\) and \(T\) be linear operators on \(V\) and assume that \(ST = TS\).

1. Show that \(im \;S\) and \(\text{ker }S\) are \(T\)-invariant.
2. If \(U\) is \(T\)-invariant, show that \(S(U)\) is \(T\)-invariant.

2. If \(\mathbf{v}\) is in \(S(U)\), write \(\mathbf{v} = S(\mathbf{u})\), \(\mathbf{u}\) in \(U\). Then \(T(\mathbf{v}) = T[S(\mathbf{u})] = (TS)(\mathbf{u}) = (ST)(\mathbf{u}) = S[T(\mathbf{u})]\) and this lies in \(S(U)\) because \(T(\mathbf{u})\) lies in \(U\) (\(U\) is \(T\)-invariant).

Let \(T : V \to V\) be a linear operator. Given \(\mathbf{v}\) in \(V\), let \(U\) denote the set of vectors in \(V\) that lie in every \(T\)-invariant subspace that contains \(\mathbf{v}\).

1. Show that \(U\) is a \(T\)-invariant subspace of \(V\) containing \(\mathbf{v}\).
2. Show that \(U\) is contained in every \(T\)-invariant subspace of \(V\) that contains \(\mathbf{v}\).

1. If \(T\) is a scalar operator (see Example [exa:020771]) show that every subspace is \(T\)-invariant.
2. Conversely, if every subspace is \(T\)-invariant, show that \(T\) is scalar.

Show that the only subspaces of \(V\) that are \(T\)-invariant for every operator \(T : V \to V\) are \(0\) and \(V\). Assume that \(V\) is finite dimensional. [Hint: Theorem [thm:020916].]

Suppose \(U\) is \(T\)-invariant for every \(T\). If \(U \neq 0\), choose \(\mathbf{u} \neq \mathbf{0}\) in \(U\). Choose a basis \(B = \{\mathbf{u}, \mathbf{u}_2, \dots, \mathbf{u}_n\}\) of \(V\) containing \(\mathbf{u}\). Given any \(\mathbf{v}\) in \(V\), there is (by Theorem [thm:020916]) a linear transformation \(T : V \to V\) such that \(T(\mathbf{u}) = \mathbf{v}\), \(T(\mathbf{u}_2) = \cdots = T(\mathbf{u}_n) = \mathbf{0}\). Then \(\mathbf{v} = T(\mathbf{u})\) lies in \(U\) because \(U\) is \(T\)-invariant. This shows that \(V = U\).

Suppose that \(T : V \to V\) is a linear operator and that \(U\) is a \(T\)-invariant subspace of \(V\). If \(S\) is an invertible operator, put \(T^{\prime} = STS^{-1}\). Show that \(S(U)\) is a \(T^{\prime}\)-invariant subspace.

In each case, show that \(U\) is \(T\)-invariant, use it to find a block upper triangular matrix for \(T\), and use that to compute \(c_{T}(x)\).

1. \(T :\|{P}_2 \to\|{P}_2\),
   \(T(a + bx + cx^2) \\ \hspace*{1em}= (-a + 2b + c) + (a + 3b +c)x + (a + 4b)x^2\),
   \(U = span \;\{1, x + x^2\}\)
2. \(T :\|{P}_2 \to\|{P}_2\),
   \(T(a + bx + cx^2) \\ \hspace*{1em} = (5a - 2b + c) + (5a - b + c)x + (a + 2c)x^2\),
   \(U = span \;\{1 - 2x^2, x + x^2\}\)

2. \(T(1 - 2x^{2}) = 3 + 3x - 3x^{2} = 3(1 - 2x^{2}) + 3(x + x^{2})\) and \(T(x + x^{2}) = -(1 - 2x^{2})\), so both are in \(U\). Hence \(U\) is \(T\)-invariant by Example [exa:029323]. If \(B = \{ 1 - 2x^2, x + x^2, x^2 \}\) then \(M_B(T) = \left[ \begin{array}{rrr} 3 & -1 & 1 \\ 3 & 0 & 1 \\ 0 & 0 & 3 \end{array} \right]\), so \(c_T(x) = \det \left[ \begin{array}{ccc} x - 3 & 1 & -1 \\ -3 & x & -1 \\ 0 & 0 & x - 3 \end{array} \right] = (x - 3) \det \left[ \begin{array}{cc} x - 3 & 1 \\ -3 & x \end{array} \right] = (x - 3)(x^2 -3x + 3)\)

In each case, show that \(T_{A} : \mathbb{R}^2 \to \mathbb{R}^2\) has no invariant subspaces except \(0\) and \(\mathbb{R}^2\).

1. \(A = \left[ \begin{array}{rr} 1 & 2 \\ -1 & -1 \end{array} \right]\)
2. \(A = \left[ \begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right]\), \(0 < \theta < \pi\)

2. Suppose \(\mathbb{R}\mathbf{u}\) is \(T_{A}\)-invariant where \(\mathbf{u} \neq \mathbf{0}\). Then \(T_{A}(\mathbf{u}) = r\mathbf{u}\) for some \(r\) in \(\mathbb{R}\), so \((rI - A)\mathbf{u} = \mathbf{0}\). But \(\det (rI - A) = (r - \cos \theta)^{2} + sin^{2} \theta \neq 0\) because \(0 < \theta < \pi\). Hence \(\mathbf{u} = \mathbf{0}\), a contradiction.

In each case, show that \(V = U \oplus W\).

1. \(V = \mathbb{R}^4\), \(U = span \;\{(1, 1, 0, 0), (0, 1, 1, 0)\}\),
   \(W = span \;\{(0, 1, 0, 1), (0, 0, 1, 1)\}\)
2. \(V = \mathbb{R}^4\), \(U = \{(a, a, b, b) \mid a, b \mbox{ in }\mathbb{R}\}\),
   \(W = \{(c, d, c, -d) \mid c, d \mbox{ in } \mathbb{R}\}\)
3. \(V =\|{P}_{3}\), \(U = \{a + bx \mid a, b \mbox{ in }\mathbb{R}\}\),
   \(W = \{ax^{2} + bx^{3} \mid a, b \mbox{ in } \mathbb{R}\}\)
4. \(V =\|{M}_{22}\), \(U = \left\{ \left[ \begin{array}{rr} a & a \\ b & b \end{array} \right] \middle| a, b \mbox{ in } \mathbb{R} \right\}\),
   \(W = \left\{ \left[ \begin{array}{rr} a & b \\ -a & b \end{array} \right] \middle| a, b \mbox{ in } \mathbb{R} \right\}\)

2. \(U = span \;\{(1, 1, 0, 0), (0, 0, 1, 1)\}\) and \(W = span \;\{(1, 0, 1, 0), (0, 1, 0, -1)\}\), and these four vectors form a basis of \(\mathbb{R}^4\). Use Example [exa:029575].
3. \(U = span \; \left\{ \left[ \begin{array}{rr} 1 & 1 \\ 0 & 0 \end{array} \right], \left[ \begin{array}{rr} 0 & 0 \\ 1 & 1 \end{array} \right] \right\}\) and \(W = span \; \left\{ \left[ \begin{array}{rr} 1 & 0 \\ -1 & 0 \end{array} \right], \left[ \begin{array}{rr} 0 & 1 \\ 0 & 1 \end{array} \right] \right\}\) and these vectors are a basis of \(\mathbf{M}_{22}\). Use Example [exa:029575].

Let \(U = span \;\{(1, 0, 0, 0), (0, 1, 0, 0)\}\) in \(\mathbb{R}^4\). Show that \(\mathbb{R}^4 = U \oplus W_{1}\) and \(\mathbb{R}^4 = U \oplus W_{2}\), where \(W_{1} = span \;\{(0, 0, 1, 0), (0, 0, 0, 1)\}\) and \(W_{2} = span \;\{(1, 1, 1, 1), (1, 1, 1, -1)\}\).

Let \(U\) be a subspace of \(V\), and suppose that \(V = U \oplus W_{1}\) and \(V = U \oplus W_{2}\) hold for subspaces \(W_{1}\) and \(W_{2}\). Show that \(dim \;W_{1} = dim \;W_{2}\).

If \(U\) and \(W\) denote the subspaces of even and odd polynomials in \(\mathbf{P}_{n}\), respectively, show that \(\mathbf{P}_{n} = U \oplus W\). (See Exercise [ex:ex6_3_36].) [Hint: \(f(x) + f(-x)\) is even.]

Let \(E\) be an \(n \times n\) matrix with \(E^{2} = E\). Show that \(\mathbf{M}_{nn} = U \oplus W\), where \(U = \{A \mid AE = A\}\) and \(W = \{B \mid BE = 0\}\). [Hint: \(XE\) lies in \(U\) for every matrix \(X\).]

The fact that \(U\) and \(W\) are subspaces is easily verified using the subspace test. If \(A\) lies in \(U \cap V\), then \(A = AE = 0\); that is, \(U \cap V = 0\). To show that \(\mathbf{M}_{22} = U + V\), choose any \(A\) in \(\mathbf{M}_{22}\). Then \(A = AE + (A - AE)\), and \(AE\) lies in \(U\) [because \((AE)E = AE^{2} = AE\)], and \(A - AE\) lies in \(W\) [because \((A - AE)E = AE - AE^{2} = 0\)].

Let \(U\) and \(W\) be subspaces of \(V\). Show that \(U \cap W = \{\mathbf{0}\}\) if and only if \(\{\mathbf{u}, \mathbf{w}\}\) is independent for all \(\mathbf{u} \neq \mathbf{0}\) in \(U\) and all \(\mathbf{w} \neq \mathbf{0}\) in \(W\).

Let \(V \stackrel{T}{\to} W \stackrel{S}{\to} V\) be linear transformations, and assume that \(dim \;V\) and \(dim \;W\) are finite.

1. If \(ST = 1_{V}\), show that \(W = im \;T \oplus \text{ker }S\).
2. Illustrate with \(\mathbb{R}^2 \stackrel{T}{\to} \mathbb{R}^3 \stackrel{S}{\to} \mathbb{R}^2\) where
   \(T(x, y) = (x, y, 0)\) and \(S(x, y, z) = (x, y)\).

Let \(U\) and \(W\) be subspaces of \(V\), let \(dim \;V = n\), and assume that \(dim \;U + dim \;W = n\).

1. If \(U \cap W = \{\mathbf{0}\}\), show that \(V = U \oplus W\).
2. If \(U + W = V\), show that \(V = U \oplus W\). [Hint: Theorem [thm:019692].]

2. By (a) it remains to show \(U + W = V\); we show that \(dim \;(U + W) = n\) and invoke Theorem [thm:019525]. But \(U + W = U \oplus W\) because \(U \cap W = 0\), so \(dim \;(U + W) = dim \;U + dim \;W = n\).

Let \(A = \left[ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right]\) and consider
\(T_{A} : \mathbb{R}^2 \to \mathbb{R}^2\).

1. Show that the only eigenvalue of \(T_{A}\) is \(\lambda = 0\).
2. Show that \(\text{ker }(T_A) = \mathbb{R}\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]\) is the unique
   \(T_{A}\)-invariant subspace of \(\mathbb{R}^2\) (except for \(0\) and
   \(\mathbb{R}^2\)).

2. First, \(\text{ker}(T_{A})\) is \(T_{A}\)-invariant. Let \(U = \mathbb{R}\mathbf{p}\) be \(T_{A}\)-invariant. Then \(T_{A}(\mathbf{p})\) is in \(U\), say \(T_{A}(\mathbf{p}) = \lambda\mathbf{p}\). Hence \(A\mathbf{p} = \lambda\mathbf{p}\) so \(\lambda\) is an eigenvalue of \(A\). This means that \(\lambda = 0\) by (a), so \(\mathbf{p}\) is in \(\text{ker}(T_{A})\). Thus \(U \subseteq \text{ker}(T_{A})\). But \(dim \;[\text{ker}(T_{A})] \neq 2\) because \(T_{A} \neq 0\), so \(dim \;[\text{ker}(T_{A})] = 1 = dim \;(U)\). Hence \(U = \text{ker}(T_{A})\).

If \(A = \left[ \begin{array}{rrrr} 2 & -5 & 0 & 0 \\ 1 & -2 & 0 & 0 \\ 0 & 0 & -1 & -2 \\ 0 & 0 & 1 & 1 \end{array} \right]\), show that \(T_{A} : \mathbb{R}^4 \to \mathbb{R}^4\) has two-dimensional \(T\)-invariant subspaces \(U\) and \(W\) such that \(\mathbb{R}^4 = U \oplus W\), but \(A\) has no real eigenvalue.

Let \(T : V \to V\) be a linear operator where \(dim \;V = n\). If \(U\) is a \(T\)-invariant subspace of \(V\), let \(T_{1} : U \to U\) denote the restriction of \(T\) to \(U\)
(so \(T_{1}(\mathbf{u}) = T(\mathbf{u})\) for all \(\mathbf{u}\) in \(U\)). Show that \(c_T(x) = c_{T_1}(x) \cdot q(x)\) for some polynomial \(q(x)\). [Hint: Theorem [thm:029359].]

Let \(B_{1}\) be a basis of \(U\) and extend it to a basis \(B\) of \(V\). Then \(M_B(T) = \left[ \begin{array}{cc} M_{B_1}(T) & Y \\ 0 & Z \end{array} \right]\), so \(c_T(x) = \det [xI - M_B(T)] = \det [xI - M_{B_1}(T)]\det [xI - Z] = c_{T1}(x)q(x)\).

Let \(T : V \to V\) be a linear operator where \(dim \;V = n\). Show that \(V\) has a basis of eigenvectors if and only if \(V\) has a basis \(B\) such that \(M_{B}(T)\) is diagonal.

In each case, show that \(T^{2} = 1\) and find (as in Example [exa:029759]) an ordered basis \(B\) such that \(M_{B}(T)\) has the given block form.

1. \(T :\|{M}_{22} \to\|{M}_{22}\) where \(T(A) = A^T\),
   \(M_B(T) = \left[ \begin{array}{rr} I_3 & 0 \\ 0 & -1 \end{array} \right]\)
2. \(T :\|{P}_3 \to\|{P}_3\) where \(T[p(x)] = p(-x)\),
   \(M_B(T) = \left[ \begin{array}{rr} I_2 & 0 \\ 0 & -I_2 \end{array} \right]\)
3. \(T : \mathbb{C} \to \mathbb{C}\) where \(T(a + bi) = a - bi\),
   \(M_B(T) = \left[ \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \right]\)
4. \(T : \mathbb{R}^3 \to \mathbb{R}^3\) where
   \(T(a, b, c) = (-a + 2b + c, b + c, -c)\),
   \(M_B(T) = \left[ \begin{array}{rr} 1 & 0 \\ 0 & -I_2 \end{array} \right]\)
5. \(T : V \to V\) where \(T(\mathbf{v}) = -\mathbf{v}\), \(dim \;V = n\), \(M_{B}(T) = -I_{n}\)

2. \(T^{2}[p(x)] = p[-(-x)] = p(x)\), so \(T^{2} = 1\); \(B = \{1, x^{2};\ x, x^{3}\}\)
3. \(T^{2}(a, b, c) = T(-a + 2b + c, b + c, -c) = (a, b, c)\), so \(T^{2} = 1\); \(B = \{(1, 1, 0);\ (1, 0, 0), (0, -1, 2)\}\)

[ex:ex9_3_23] Let \(U\) and \(W\) denote subspaces of a vector space \(V\).

1. If \(V = U \oplus W\), define \(T : V \to V\) by \(T(\mathbf{v}) = \mathbf{w}\) where \(\mathbf{v}\) is written (uniquely) as \(\mathbf{v} = \mathbf{u} + \mathbf{w}\) with \(\mathbf{u}\) in \(U\) and \(\mathbf{w}\) in \(W\). Show that \(T\) is a linear transformation, \(U = \text{ker }T\), \(W = im \;T\), and \(T^{2} = T\).
2. Conversely, if \(T : V \to V\) is a linear transformation such that \(T^{2} = T\), show that \(V = \text{ker }T \oplus im \;T\). [Hint: \(\mathbf{v} - T(\mathbf{v})\) lies in \(\text{ker }T\) for all \(\mathbf{v}\) in \(V\).]

2. Use the Hint and Exercise [ex:ex9_3_2].

Let \(T : V \to V\) be a linear operator satisfying \(T^{2} = T\) (such operators are called idempotents). Define \(U_{1} = \{\mathbf{v} \mid T(\mathbf{v}) = \mathbf{v}\}\),
\(U_{2} = \text{ker }T = \{\mathbf{v} \mid T(\mathbf{v}) = 0\}\).

1. Show that \(V = U_{1} \oplus U_{2}\).
2. If \(dim \;V = n\), find a basis \(B\) of \(V\) such that \(M_B(T) = \left[ \begin{array}{rr} I_r & 0 \\ 0 & 0 \end{array} \right]\), where \(r = rank \;T\).
3. If \(A\) is an \(n \times n\) matrix such that \(A^{2} = A\), show that \(A\) is similar to \(\left[ \begin{array}{rr} I_r & 0 \\ 0 & 0 \end{array} \right]\), where \(r = rank \;A\).

In each case, show that \(T^{2} = T\) and find (as in the preceding exercise) an ordered basis \(B\) such that \(M_{B}(T)\) has the form given (\(0_{k}\) is the \(k \times k\) zero matrix).

1. \(T :\|{P}_2 \to\|{P}_2\) where
   \(T(a + bx + cx^2) = (a - b + c)(1 + x + x^2)\),
   \(M_B(T) = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0_2 \end{array} \right]\)
2. \(T : \mathbb{R}^3 \to \mathbb{R}^3\) where
   \(T(a, b, c) = (a + 2b, 0, 4b + c)\),
   \(M_B(T) = \left[ \begin{array}{rr} I_2 & 0 \\ 0 & 0 \end{array} \right]\)
3. \(T :\|{M}_{22} \to\|{M}_{22}\) where
   \(T\left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{rr} -5 & -15 \\ 2 & 6 \end{array} \right]\left[ \begin{array}{rr} a & b \\ c & d \end{array} \right]\),
   \(M_B(T) = \left[ \begin{array}{cc} I_2 & 0 \\ 0 & 0_2 \end{array} \right]\)

2. \(T^{2}(a, b, c) = T(a + 2b, 0, 4b + c) = (a + 2b, 0, 4b + c) = T(a, b, c)\), so \(T^{2} = T\); \(B = \{(1, 0, 0), (0, 0, 1);\ (2, -1, 4)\}\)

Let \(T : V \to V\) be an operator satisfying \(T^{2} = cT\), \(c \neq 0\).

1. Show that \(V = U \oplus \text{ker }T\), where
   \(U = \{\mathbf{u} \mid T(\mathbf{u}) = c\mathbf{u}\}\).
2. If \(dim \;V = n\), show that \(V\) has a basis \(B\) such that \(M_B(T) = \left[ \begin{array}{rr} cI_r & 0 \\ 0 & 0 \end{array} \right]\), where \(r = rank \;T\).
3. If \(A\) is any \(n \times n\) matrix of \(rank \;r\) such that
   \(A^{2} = cA\), \(c \neq 0\), show that \(A\) is similar to \(\left[ \begin{array}{rr} cI_r & 0 \\ 0 & 0 \end{array} \right]\).

Let \(T : V \to V\) be an operator such that \(T^{2} = c^{2}\), \(c \neq 0\).

1. Show that \(V = U_{1} \oplus U_{2}\), where
   \(U_{1} = \{\mathbf{v} \mid T(\mathbf{v}) = c\mathbf{v}\}\) and \(U_{2} = \{\mathbf{v} \mid T(\mathbf{v}) = -c\mathbf{v}\}\). [Hint: \(\mathbf{v} = \frac{1}{2c} \{[T(\mathbf{v}) + c\mathbf{v}] - [T(\mathbf{v}) - c\mathbf{v}]\}\).]
2. If \(dim \;V = n\), show that \(V\) has a basis \(B\) such that \(M_B(T) = \left[ \begin{array}{cc} cI_k & 0 \\ 0 & -cI_{n-k} \end{array} \right]\) for some \(k\).
3. If \(A\) is an \(n \times n\) matrix such that \(A^{2} = c^{2}I\), \(c \neq 0\), show that \(A\) is similar to \(\left[ \begin{array}{cc} cI_k & 0 \\ 0 & -cI_{n-k} \end{array} \right]\) for some \(k\).

If \(P\) is a fixed \(n \times n\) matrix, define \(T :\|{M}_{nn} \to\|{M}_{nn}\) by \(T(A) = PA\). Let \(U_{j}\) denote the subspace of \(\mathbf{M}_{nn}\) consisting of all matrices with all columns zero except possibly column \(j\).

1. Show that each \(U_{j}\) is \(T\)-invariant.
2. Show that \(\mathbf{M}_{nn}\) has a basis \(B\) such that \(M_{B}(T)\) is block diagonal with each block on the diagonal equal to \(P\).

Let \(V\) be a vector space. If \(f : V \to \mathbb{R}\) is a linear transformation and \(\mathbf{z}\) is a vector in \(V\), define \(T_{f,\mathbf{z}} : V \to V\) by \(T_{f,\mathbf{z}}(\mathbf{v}) = f(\mathbf{v})\mathbf{z}\) for all \(\mathbf{v}\) in \(V\). Assume that \(f \neq 0\) and \(\mathbf{z} \neq \mathbf{0}\).

1. Show that \(T_{f,\mathbf{z}}\) is a linear operator of \(rank \;1\).
2. If \(f \neq 0\), show that \(T_{f,\mathbf{z}}\) is an idempotent if and only if \(f(\mathbf{z}) = 1\). (Recall that \(T : V \to V\) is called an idempotent if \(T^{2} = T\).)
3. Show that every idempotent \(T : V \to V\) of \(rank \;1\) has the form \(T = T_{f,\mathbf{z}}\) for some \(f : V \to \mathbb{R}\) and some \(\mathbf{z}\) in \(V\) with \(f(\mathbf{z}) = 1\). [Hint: Write
   \(im \;T = \mathbb{R}\mathbf{z}\) and show that \(T(\mathbf{z}) = \mathbf{z}\). Then use Exercise [ex:ex9_3_23].]

2. \(T_{f,\mathbf{z}}[T_{f,\mathbf{z}}(\mathbf{v})] = T_{f,\mathbf{z}}[f(\mathbf{v})\mathbf{z}] = f[f(\mathbf{v})\mathbf{z}]\mathbf{z} = f(\mathbf{v})\{f[\mathbf{z}]\mathbf{z}\} = f(\mathbf{v})f(\mathbf{z})\mathbf{z}\). This equals \(T_{f,\mathbf{z}}(\mathbf{v}) = f(\mathbf{v})\mathbf{z}\) for all \(\mathbf{v}\) if and only if \(f(\mathbf{v})f(\mathbf{z}) = f(\mathbf{v})\) for all \(\mathbf{v}\). Since \(f \neq 0\), this holds if and only if \(f(z) = 1\).

Let \(U\) be a fixed \(n \times n\) matrix, and consider the operator \(T :\|{M}_{nn} \to\|{M}_{nn}\) given by \(T(A) = UA\).

1. Show that \(\lambda\) is an eigenvalue of \(T\) if and only if it is an eigenvalue of \(U\).
2. If \(\lambda\) is an eigenvalue of \(T\), show that \(E_{\lambda}(T)\) consists of all matrices whose columns lie in \(E_{\lambda}(U)\):
   \(E_{\lambda}(T) \\ \hspace*{1em}= \{ \left[ \begin{array}{cccc} P_1 & P_2 & \cdots & P_n \end{array} \right] \mid P_i \mbox{ in } E_{\lambda}(U) \mbox{ for each } i \}\)
3. Show if \(dim \;[E_{\lambda}(U)] = d\), then \(dim \;[E_{\lambda}(T)] = nd\). [Hint: If \(B = \{\mathbf{x}_{1}, \dots, \mathbf{x}_{d}\}\) is a basis of \(E_{\lambda}(U)\), consider the set of all matrices with one column from \(B\) and the other columns zero.]

2. If \(A = \left[ \begin{array}{cccc} \mathbf{p}_{1} & \mathbf{p}_2 & \cdots & \mathbf{p}_{n} \end{array} \right]\) where \(U\mathbf{p}_{i} = \lambda\mathbf{p}_{i}\) for each \(i\), then \(UA = \lambda A\). Conversely, \(UA = \lambda A\) means that \(U\mathbf{p} = \lambda\mathbf{p}\) for every column \(\mathbf{p}\) of \(A\).

Let \(T : V \to V\) be a linear operator where \(V\) is finite dimensional. If \(U \subseteq V\) is a subspace, let \(\overline{U} = \{ \mathbf{u}_0 + T(\mathbf{u}_1) + T^2(\mathbf{u}_2) + \cdots + T^k(\mathbf{u}_k) \mid \mathbf{u}_i \mbox{ in } U, k \geq 0 \}\). Show that \(\overline{U}\) is the smallest \(T\)-invariant subspace containing \(U\) (that is, it is \(T\)-invariant, contains \(U\), and is contained in every such subspace).

Let \(U_{1}, \dots, U_{m}\) be subspaces of \(V\) and assume that \(V = U_{1} + \cdots + U_{m}\); that is, every \(\mathbf{v}\) in \(V\) can be written (in at least one way) in the form \(\mathbf{v} = \mathbf{u}_{1} + \cdots + \mathbf{u}_{m}\), \(\mathbf{u}_{i}\) in \(U_{i}\). Show that the following conditions are equivalent.

1. If \(\mathbf{u}_{1} + \cdots + \mathbf{u}_{m} = \mathbf{0}\), \(\mathbf{u}_{i}\) in \(U_{i}\), then \(\mathbf{u}_{i} = \mathbf{0}\) for each \(i\).
2. If \(\mathbf{u}_{1} + \cdots + \mathbf{u}_{m} = \mathbf{u}^{\prime}_{1} + \cdots + \mathbf{u}^{\prime}_{m}\), \(\mathbf{u}_{i}\) and \(\mathbf{u}^{\prime}_{i}\) in \(U_{i}\), then \(\mathbf{u}_{i} = \mathbf{u}^{\prime}_{i}\) for each \(i\).
3. \(U_{i} \cap (U_{1} + \cdots + U_{i-1} + U_{i+1} + \cdots + U_{m}) = \{\mathbf{0}\}\) for each \(i = 1, 2, \dots, m\).
4. \(U_{i} \cap (U_{i+1} + \cdots + U_{m}) = \{\mathbf{0}\}\) for each
   \(i = 1, 2, \dots, m - 1\).

When these conditions are satisfied, we say that \(V\) is the direct sum of the subspaces \(U_{i}\), and write
\(V = U_{1} \oplus U_{2} \oplus \cdots \oplus U_{m}\).

1. Let \(B\) be a basis of \(V\) and let \(B = B_{1} \cup B_{2} \cup \cdots \cup B_{m}\) where the \(B_{i}\) are pairwise disjoint, nonempty subsets of \(B\). If \(U_{i} = span \;B_{i}\) for each \(i\), show that \(V = U_{1} \oplus U_{2} \oplus \cdots \oplus U_{m}\) (preceding exercise).
2. Conversely if \(V = U_{1} \oplus \cdots \oplus U_{m}\) and \(B_{i}\) is a basis of \(U_{i}\) for each \(i\), show that \(B = B_{1} \cup \cdots \cup B_{m}\) is a basis of \(V\) as in (a).

Let \(T: V \to V\) be an operator where \(T^3=0\). If \(\mathbf{u} \in V\) and \(U = span \;\{\mathbf{u},T(\mathbf{u}), T^2(\mathbf{u})\}\), show that \(U\) is \(T\)-invariant and has dimension \(3\).