§ 26. Invariant Subspaces and the Cayley-Hamilton Theorem¶

Invariant Subspaces¶

Definition 26.1 : \(T\)-Invariant Subspace

Let \(T\) be a linear operator on a vector space \(V\). A subspace \(W\) of \(V\) is called a \(T\)-invariant subspace of \(V\) if \(T(W) \subseteq W\), that is, if \(T(v) \in W\) for all \(v \in W\).

Example 26.2 : Examples of \(T\)-invariant subspaces

Suppose that \(T\) is a linear operator on a vector space \(V\). Then the following subspaces of \(V\) are \(T\)-invariant:

\(\{0\}\)
\(V\)
\(R(T)\)
\(N(T)\)
\(E_{\lambda}\), for any eigenvalue \(\lambda\) of \(T\).

Proof

Since \(T(0)=0\), we have \(T(\{0\})=\{0\}\subseteq \{0\}\), so \(\{0\}\) is \(T\)-invariant.
For every \(x\in V\), \(T(x)\in V\), so \(T(V)\subseteq V\), and hence \(V\) is \(T\)-invariant.
Let \(y\in R(T)\). Then \(y=T(x)\) for some \(x\in V\), so

\[ T(y)=T(T(x))=T^{2}(x)\in R(T), \]

which shows \(T(R(T))\subseteq R(T)\).
Let \(x\in N(T)\). Then \(T(x)=0\), so

\[ T(x)\in N(T) \]

because \(0\in N(T)\). Hence \(T(N(T))\subseteq N(T)\).
Let \(x\in E_{\lambda}\), where \(\lambda\) is an eigenvalue of \(T\). Then \(T(x)=\lambda x\). Applying \(T\) again yields

\[ T(T(x))=T(\lambda x)=\lambda T(x)=\lambda(\lambda x)=\lambda(\lambda x), \]

so \(T(x)=\lambda x\in E_{\lambda}\). Therefore \(T(E_{\lambda})\subseteq E_{\lambda}\), and \(E_{\lambda}\) is \(T\)-invariant.

Definition 26.3 : \(T\)-Cyclic Subspace

Let \(T\) be a linear operator on a vector space \(V\), and let \(x\) be a nonzero vector in \(V\). The subspace

\[ W=\operatorname{span}\left(\left\{x, T(x), T^{2}(x), \ldots\right\}\right) \]

is called the \(T\)-cyclic subspace of \(V\) generated by \(x\).

Theorem 26.4 : \(T\)-Cyclic subspace generated by \(x\) is the minimal \(T\)-invariant subspace containing \(x\).

Let \(T\) be a linear operator on a vector space \(V\), \(x\) be a nonzero vector in \(V\), and \(W\) be the \(T\)-cyclic subspace of \(V\) generated by \(x\).

(a) \(W\) is \(T\)-invariant.
(b) \(W\) is the "smallest" \(T\)-invariant subspace of \(V\) containing \(x\). That is, any \(T\)-invariant subspace of \(V\) containing \(x\) must also contain \(W\).

Proof

(a)
Let \(w \in W\). Then there exists an integer \(m \ge 0\) and scalars \(a_{0}, a_{1}, \ldots, a_{m}\) such that

\[ w=\sum_{i=0}^{m} a_{i} T^{i}(x). \]

Using linearity of \(T\),

\[ T(w)=T\left(\sum_{i=0}^{m} a_{i} T^{i}(x)\right)=\sum_{i=0}^{m} a_{i} T^{i+1}(x). \]

But \(T^{i+1}(x)\) is one of the spanning vectors of \(W\), so the right-hand side is a linear combination of vectors in \(\{x, T(x), T^{2}(x), \ldots\}\). Hence \(T(w)\in W\), and therefore \(T(W)\subseteq W\). Thus \(W\) is \(T\)-invariant.
(b)
Let \(U\) be any \(T\)-invariant subspace of \(V\) such that \(x \in U\). Because \(U\) is \(T\)-invariant, \(T(x)\in U\). Applying \(T\) repeatedly, we obtain

\[ T^{i}(x)\in U \quad \text{for all integers } i\ge 0. \]

Since \(U\) is a subspace, it contains the span of any subset of its vectors. Therefore,

\[ \operatorname{span}\left(\left\{x, T(x), T^{2}(x), \ldots\right\}\right)\subseteq U, \]

that is, \(W\subseteq U\). Hence every \(T\)-invariant subspace of \(V\) containing \(x\) must contain \(W\), so \(W\) is the smallest \(T\)-invariant subspace of \(V\) containing \(x\).

Definition 26.5 : Restriction to a \(T\)-Invariant Subspace

If \(T\) is a linear operator on \(V\) and \(W\) is a \(T\)-invariant subspace of \(V\), then the restriction \(T_{W}\) of \(T\) to \(W\) is a mapping from \(W\) to \(W\), and it follows that \(T_{W}\) is a linear operator on \(W\).

Theorem 26.6 : Characteristic polynomial of a restriction divides the original characteristic polynomial.

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and let \(W\) be a \(T\)-invariant subspace of \(V\). Then the characteristic polynomial of \(T_{W}\) divides the characteristic polynomial of \(T\).

Proof

Choose an ordered basis \(\gamma=\left\{v_{1}, v_{2}, \ldots, v_{k}\right\}\) for \(W\), and extend it to an ordered basis \(\beta=\left\{v_{1}, v_{2}, \ldots, v_{k}, v_{k+1}, \ldots, v_{n}\right\}\) for \(V\). Let \(A=[T]_{\beta}\) and \(B_{1}=\left[T_{W}\right]_{\gamma}\). Then, by Exercise 26.12, \(A\) can be written in the form

\[ A=\left(\begin{array}{cc} B_{1} & B_{2} \\ O & B_{3} \end{array}\right) \]

Let \(f(t)\) be the characteristic polynomial of \(T\) and \(g(t)\) the characteristic polynomial of \(T_{W}\). Then

\[ f(t)=\operatorname{det}\left(A-t I_{n}\right)=\operatorname{det}\left(\begin{array}{cc} B_{1}-t I_{k} & B_{2} \\ O & B_{3}-t I_{n-k} \end{array}\right)=g(t) \cdot \operatorname{det}\left(B_{3}-t I_{n-k}\right) \]

by Exercise 21.21. Thus \(g(t)\) divides \(f(t)\).

Theorem 26.7 : Basis and characteristic polynomial of a cyclic subspace

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and let \(W\) denote the \(T\)-cyclic subspace of \(V\) generated by a nonzero vector \(v \in V\). Let \(k=\operatorname{dim}(W)\). Then

(a) \(\left\{v, T(v), T^{2}(v), \ldots, T^{k-1}(v)\right\}\) is a basis for \(W\).
(b) If \(a_{0} v+a_{1} T(v)+\cdots+a_{k-1} T^{k-1}(v)+T^{k}(v)=0\), then the characteristic polynomial of \(T_{W}\) is \(f(t)=(-1)^{k}\left(a_{0}+a_{1} t+\cdots+a_{k-1} t^{k-1}+t^{k}\right)\).

Proof

(a)
Since \(v \neq 0\), the set \(\left\{v\right\}\) is linearly independent. Let \(j\) be the largest positive integer for which

\[ \beta=\left\{v, T(v), \ldots, T^{j-1}(v)\right\} \]

is linearly independent. Such a \(j\) must exist because \(V\) is finite-dimensional. Let \(Z=\operatorname{span}(\beta)\). Then \(\beta\) is a basis for \(Z\). Furthermore, \(T^{j}(v) \in Z\) by Theorem 5.9. We use this information to show that \(Z\) is a \(T\)-invariant subspace of \(V\). Let \(w \in Z\). Since \(w\) is a linear combination of the vectors of \(\beta\), there exist scalars \(b_{0}, b_{1}, \ldots, b_{j-1}\) such that

\[ w=b_{0} v+b_{1} T(v)+\cdots+b_{j-1} T^{j-1}(v) \]

and hence

\[ T(w)=b_{0} T(v)+b_{1} T^{2}(v)+\cdots+b_{j-1} T^{j}(v) . \]

Thus \(T(w)\) is a linear combination of vectors in \(Z\), and hence belongs to \(Z\). So \(Z\) is \(T\)-invariant.

Furthermore, \(v \in Z\). By Theorem 26.4, \(W\) is the smallest \(T\)-invariant subspace of \(V\) that contains \(v\), so that \(W \subseteq Z\). Clearly, \(Z \subseteq W\), and so we conclude that \(Z=W\). It follows that \(\beta\) is a basis for \(W\), and therefore \(\operatorname{dim}(W)=j\). Thus \(j=k\). This proves (a).
(b)
Now view \(\beta\) (from (a)) as an ordered basis for \(W\). Let \(a_{0}, a_{1}, \ldots, a_{k-1}\) be the scalars such that

\[ a_{0} v+a_{1} T(v)+\cdots+a_{k-1} T^{k-1}(v)+T^{k}(v)=0 . \]

Observe that

\[ \left[T_{W}\right]_{\beta}=\left(\begin{array}{ccccc} 0 & 0 & \cdots & 0 & -a_{0} \\ 1 & 0 & \cdots & 0 & -a_{1} \\ \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -a_{k-1} \end{array}\right). \]

We claim that the characteristic polynomial of this matrix is

\[ f(t)=(-1)^{k}\left(a_{0}+a_{1} t+\cdots+a_{k-1} t^{k-1}+t^{k}\right). \]

To prove this, compute \(f(t)=\det\!\left(\left[T_W\right]_{\beta}-tI_k\right)\). The matrix \(\left[T_W\right]_{\beta}-tI_k\) has the form

\[ \left(\begin{array}{ccccc} -t & 0 & \cdots & 0 & -a_{0} \\ 1 & -t & \cdots & 0 & -a_{1} \\ 0 & 1 & \ddots & 0 & -a_{2} \\ \vdots & \vdots & \ddots & -t & \vdots \\ 0 & 0 & \cdots & 1 & -t-a_{k-1} \end{array}\right). \]

Expanding the determinant along the first row yields the recursion

\[ f_k(t)=-t\,f_{k-1}(t)+(-1)^{k}a_{0}, \]

and continuing this expansion down the same pattern (equivalently, applying the same cofactor expansion repeatedly to the resulting minors) produces

\[ f_k(t)=(-1)^{k}\left(t^{k}+a_{k-1}t^{k-1}+\cdots+a_{1}t+a_{0}\right) = (-1)^{k}\left(a_{0}+a_{1} t+\cdots+a_{k-1} t^{k-1}+t^{k}\right). \]

Thus \(f(t)\) is the characteristic polynomial of \(T_{W}\), proving (b).

Example 26.8 : Example of a cyclic subspace in \(\mathbb{R}^{3}\)

Let \(T\) be the linear operator on \(\mathbb{R}^{3}\) defined by

\[ T(a, b, c)=(-b+c, a+c, 3 c) \]

We determine the \(T\)-cyclic subspace generated by \(e_{1}=(1,0,0)\). Since

\[ T\left(e_{1}\right)=T(1,0,0)=(0,1,0)=e_{2} \]

and

\[ T^{2}\left(e_{1}\right)=T\left(T\left(e_{1}\right)\right)=T\left(e_{2}\right)=(-1,0,0)=-e_{1} \]

it follows that

\[ \operatorname{span}\left(\left\{e_{1}, T\left(e_{1}\right), T^{2}\left(e_{1}\right), \ldots\right\}\right)=\operatorname{span}\left(\left\{e_{1}, e_{2}\right\}\right)=\left\{(s, t, 0): s, t \in \mathbb{R}\right\} \]

Let \(W=\operatorname{span}\left(\left\{e_{1}, e_{2}\right\}\right)\), the \(T\)-cyclic subspace generated by \(e_{1}\). We compute the characteristic polynomial \(f(t)\) of \(T_{W}\) in two ways: by means of Theorem 26.7 and by means of determinants.

(a) By means of Theorem 26.7.

\[ 1 e_{1}+0 T\left(e_{1}\right)+T^{2}\left(e_{1}\right)=0 \]

Therefore, by Theorem 26.7(b),

\[ f(t)=(-1)^{2}\left(1+0 t+t^{2}\right)=t^{2}+1 \]
(b) By means of determinants.

Let \(\beta=\left\{e_{1}, e_{2}\right\}\), which is an ordered basis for \(W\). Since \(T\left(e_{1}\right)=e_{2}\) and \(T\left(e_{2}\right)=-e_{1}\), we have

\[ \left[T_{W}\right]_{\beta}=\left(\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right) \]

and therefore,

\[ f(t)=\operatorname{det}\left(\begin{array}{rr} -t & -1 \\ 1 & -t \end{array}\right)=t^{2}+1 \]

The Cayley-Hamilton Theorem¶

Theorem 26.9 : Cayley-Hamilton Theorem

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and let \(f(t)\) be the characteristic polynomial of \(T\). Then \(f(T)=T_{0}\), the zero transformation. That is, \(T\) "satisfies" its characteristic equation.

Proof

We show that \(f(T)(v)=0\) for all \(v \in V\). This is obvious if \(v=0\) because \(f(T)\) is linear; so suppose that \(v \neq 0\). Let \(W\) be the \(T\)-cyclic subspace generated by \(v\), and suppose that \(\operatorname{dim}(W)=k\). By Theorem 26.7(a), there exist scalars \(a_{0}, a_{1}, \ldots, a_{k-1}\) such that

\[ a_{0} v+a_{1} T(v)+\cdots+a_{k-1} T^{k-1}(v)+T^{k}(v)=0 . \]

Hence Theorem 26.7(b) implies that

\[ g(t)=(-1)^{k}\left(a_{0}+a_{1} t+\cdots+a_{k-1} t^{k-1}+t^{k}\right) \]

is the characteristic polynomial of \(T_{W}\). Combining these two equations yields

\[ g(T)(v)=(-1)^{k}\left(a_{0} I+a_{1} T+\cdots+a_{k-1} T^{k-1}+T^{k}\right)(v)=0 \]

By Theorem 26.6, \(g(t)\) divides \(f(t)\); hence there exists a polynomial \(q(t)\) such that \(f(t)=q(t) g(t)\). So

\[ f(T)(v)=q(T) g(T)(v)=q(T)(g(T)(v))=q(T)(0)=0 \]

Example 26.10 : Cayley-Hamilton computation in \(\mathbb{R}^{2}\)

Let \(T\) be the linear operator on \(\mathbb{R}^{2}\) defined by \(T(a, b)=(a+2 b,-2 a+b)\), and let \(\beta=\left\{e_{1}, e_{2}\right\}\). Then

\[ A=\left(\begin{array}{rr} 1 & 2 \\ -2 & 1 \end{array}\right) \]

where \(A=[T]_{\beta}\). The characteristic polynomial of \(T\) is, therefore,

\[ f(t)=\operatorname{det}(A-t I)=\operatorname{det}\left(\begin{array}{cc} 1-t & 2 \\ -2 & 1-t \end{array}\right)=t^{2}-2 t+5 \]

It is easily verified that \(T_{0}=f(T)=T^{2}-2 T+5 I\). Similarly.

\[ \begin{aligned} f(A) & =A^{2}-2 A+5 I=\left(\begin{array}{rr} -3 & 4 \\ -4 & -3 \end{array}\right)+\left(\begin{array}{rr} -2 & -4 \\ 4 & -2 \end{array}\right)+\left(\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right) \\ & =\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right) . \end{aligned} \]

Corollary 26.11 : Cayley-Hamilton theorem for matrices

Let \(A\) be an \(n \times n\) matrix, and let \(f(t)\) be the characteristic polynomial of \(A\). Then \(f(A)=O\), the \(n \times n\) zero matrix.

Proof

Let \(T=L_{A}:F^{n}\rightarrow F^{n}\) be the linear operator defined by \(T(x)=Ax\). Then, by Theorem 26.9, the characteristic polynomial of \(T\) satisfies

\[ f(T)=T_{0}. \]

We claim that for any polynomial \(p(t)=c_{0}+c_{1}t+\cdots+c_{m}t^{m}\),

\[ p(T)=L_{p(A)}. \]

Indeed, using linearity and the fact that \(T^{k}=L_{A}^{k}=L_{A^{k}}\), we have

\[ p(T) =c_{0}I+c_{1}T+\cdots+c_{m}T^{m} =L_{c_{0}I_{n}}+L_{c_{1}A}+\cdots+L_{c_{m}A^{m}} =L_{c_{0}I_{n}+c_{1}A+\cdots+c_{m}A^{m}} =L_{p(A)}. \]

Applying this to \(p=f\) gives \(f(T)=L_{f(A)}\). Since \(f(T)=T_{0}\), it follows that \(L_{f(A)}\) is the zero transformation on \(F^{n}\), that is,

\[ f(A)x=0 \quad \text{for all } x\in F^{n}. \]

Therefore \(f(A)=O\).

Invariant Subspaces and Direct Sums¶

Theorem 26.12 : Characteristic polynomial of a direct sum of invariant subspaces

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and suppose that \(V=W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k}\), where \(W_{i}\) is a \(T\)-invariant subspace of \(V\) for each \(i\) \((1 \leq i \leq k)\). Suppose that \(f_{i}(t)\) is the characteristic polynomial of \(T_{W_{i}}\) \((1 \leq i \leq k)\). Then \(f_{1}(t) \cdot f_{2}(t) \cdots f_{k}(t)\) is the characteristic polynomial of \(T\).

Proof

The proof is by mathematical induction on \(k\). In what follows, \(f(t)\) denotes the characteristic polynomial of \(T\). Suppose first that \(k=2\). Let \(\beta_{1}\) be an ordered basis for \(W_{1}\), \(\beta_{2}\) an ordered basis for \(W_{2}\), and \(\beta=\beta_{1} \cup \beta_{2}\). Then \(\beta\) is an ordered basis for \(V\) by Theorem 24.17 (d). Let \(A=[T]_{\beta}\), \(B_{1}=\left[T_{W_{1}}\right]_{\beta_{1}}\), and \(B_{2}=\left[T_{W_{2}}\right]_{\beta_{2}}\). It follows that

\[ A=\left(\begin{array}{cc} B_{1} & O \\ O^{\prime} & B_{2} \end{array}\right) \]

where \(O\) and \(O^{\prime}\) are zero matrices of the appropriate sizes. Then

\[ f(t)=\operatorname{det}(A-t I)=\operatorname{det}\left(B_{1}-t I\right) \cdot \operatorname{det}\left(B_{2}-t I\right)=f_{1}(t) \cdot f_{2}(t) \]

as in the proof of Theorem 26.6, proving the result for \(k=2\). Now assume that the theorem is valid for \(k-1\) summands, where \(k-1 \geq 2\), and suppose that \(V\) is a direct sum of \(k\) subspaces, say,

\[ V=W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k} \]

Let \(W=W_{1}+W_{2}+\cdots+W_{k-1}\). It is easily verified that \(W\) is \(T\)-invariant and that \(V=W \oplus W_{k}\). So by the case for \(k=2\), \(f(t)=g(t) \cdot f_{k}(t)\), where \(g(t)\) is the characteristic polynomial of \(T_{W}\). Clearly \(W=W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k-1}\), and therefore \(g(t)=f_{1}(t) \cdot f_{2}(t) \cdots f_{k-1}(t)\) by the induction hypothesis. We conclude that \(f(t)=g(t) \cdot f_{k}(t)=f_{1}(t) \cdot f_{2}(t) \cdots f_{k}(t)\).

Concept 26.13 : Characteristic polynomial of diagonalizable linear operator as a direct sum of eigenspaces

As an illustration of this result, suppose that \(T\) is a diagonalizable linear operator on a finite-dimensional vector space \(V\) with distinct eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\). By Theorem 24.18, \(V\) is a direct sum of the eigenspaces of \(T\). Since each eigenspace is \(T\)-invariant, we may view this situation in the context of Theorem 26.12. For each eigenvalue \(\lambda_{i}\), the restriction of \(T\) to \(E_{\lambda_{i}}\) has characteristic polynomial \(\left(\lambda_{i}-t\right)^{m_{i}}\), where \(m_{i}\) is the dimension of \(E_{\lambda_{i}}\). By Theorem 26.12, the characteristic polynomial \(f(t)\) of \(T\) is the product

\[ f(t)=\left(\lambda_{1}-t\right)^{m_{1}}\left(\lambda_{2}-t\right)^{m_{2}} \cdots\left(\lambda_{k}-t\right)^{m_{k}} . \]

It follows that the multiplicity of each eigenvalue is equal to the dimension of the corresponding eigenspace, as expected.

Definition 26.14 : Direct Sum of Matrices

Let \(B_{1} \in \mathrm{M}_{m \times m}(F)\), and let \(B_{2} \in \mathrm{M}_{n \times n}(F)\). We define the direct sum of \(B_{1}\) and \(B_{2}\), denoted \(B_{1} \oplus B_{2}\), as the \((m+n) \times(m+n)\) matrix \(A\) such that

\[ A_{i j}= \begin{cases}\left(B_{1}\right)_{i j} & \text { for } 1 \leq i, j \leq m \\ \left(B_{2}\right)_{(i-m),(j-m)} & \text { for } m+1 \leq i, j \leq m+n \\ 0 & \text { otherwise }\end{cases} \]

If \(B_{1}, B_{2}, \ldots, B_{k}\) are square matrices with entries from \(F\), then we define the direct sum of \(B_{1}, B_{2}, \ldots, B_{k}\) recursively by

\[ B_{1} \oplus B_{2} \oplus \cdots \oplus B_{k}=\left(B_{1} \oplus B_{2} \oplus \cdots \oplus B_{k-1}\right) \oplus B_{k} \]

If \(A=B_{1} \oplus B_{2} \oplus \cdots \oplus B_{k}\), then we often write

\[ A=\left(\begin{array}{cccc} B_{1} & O & \cdots & O \\ O & B_{2} & \cdots & O \\ \vdots & \vdots & & \vdots \\ O & O & \cdots & B_{k} \end{array}\right). \]

Theorem 26.15 : Matrix of an operator relative to a direct-sum basis

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and let \(W_{1}, W_{2}, \ldots, W_{k}\) be \(T\)-invariant subspaces of \(V\) such that \(V=W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k}\). For each \(i\), let \(\beta_{i}\) be an ordered basis for \(W_{i}\), and let \(\beta=\beta_{1} \cup \beta_{2} \cup \cdots \cup \beta_{k}\). Let \(A=[T]_{\beta}\) and \(B_{i}=\left[T_{W_{i}}\right]_{\beta_{i}}\) for \(i=1,2, \ldots, k\). Then \(A=B_{1} \oplus B_{2} \oplus \cdots \oplus B_{k}\).

Proof

The proof is by mathematical induction on \(k\).

If \(k=1\), then \(V=W_{1}\) and \(\beta=\beta_{1}\). Hence \(A=[T]_{\beta_{1}}=\left[T_{W_{1}}\right]_{\beta_{1}}=B_{1}\), and therefore

\[ A=B_{1}=B_{1}\oplus\cdots\oplus B_{1}. \]

Thus the result holds for \(k=1\).

Assume that the result holds for \(k-1\) summands, where \(k-1\ge 1\). Suppose now that

\[ V=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{k}. \]

Let

\[ W=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{k-1}. \]

Because each \(W_i\) is \(T\)-invariant, \(T(W_i)\subseteq W_i\), and hence

\[ T(W)=T(W_{1}+W_{2}+\cdots+W_{k-1}) \subseteq T(W_{1})+T(W_{2})+\cdots+T(W_{k-1}) \subseteq W. \]

Thus \(W\) is \(T\)-invariant, and we also have \(V=W\oplus W_k\).

Let \(\beta'=\beta_{1}\cup\beta_{2}\cup\cdots\cup\beta_{k-1}\). Then \(\beta'\) is an ordered basis for \(W\), and \(\beta=\beta'\cup\beta_k\).

Let \(C=\left[T_{W}\right]_{\beta'}\). Applying the induction hypothesis to the direct sum decomposition

\[ W=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{k-1}, \]

we obtain

\[ C=B_{1}\oplus B_{2}\oplus\cdots\oplus B_{k-1}. \]

Now consider the decomposition \(V=W\oplus W_k\) with ordered bases \(\beta'\) for \(W\) and \(\beta_k\) for \(W_k\). Because both \(W\) and \(W_k\) are \(T\)-invariant, we have \(T(W)\subseteq W\) and \(T(W_k)\subseteq W_k\). Hence for any basis vector \(u\in\beta'\) the vector \(T(u)\) lies in \(W\), so its coordinates relative to \(\beta=\beta'\cup\beta_k\) have zero components on the \(\beta_k\) part. Similarly, for any basis vector \(v\in\beta_k\), the vector \(T(v)\) lies in \(W_k\), so its coordinates relative to \(\beta\) have zero components on the \(\beta'\) part. Therefore the matrix \(A=[T]_{\beta}\) has the block form

\[ A=\begin{pmatrix} C & O \\ O' & B_k \end{pmatrix} = C\oplus B_k, \]

where \(O\) and \(O'\) are zero matrices of the appropriate sizes.

Substituting \(C=B_{1}\oplus\cdots\oplus B_{k-1}\) gives

\[ A=(B_{1}\oplus B_{2}\oplus\cdots\oplus B_{k-1})\oplus B_k =B_{1}\oplus B_{2}\oplus\cdots\oplus B_k. \]

This completes the induction and proves the theorem for all \(k\).

Exercise¶

Exercise 26.8

Let \(T\) be a linear operator on a vector space with a \(T\)-invariant subspace \(W\). Prove that if \(v\) is an eigenvector of \(T_{W}\) with corresponding eigenvalue \(\lambda\), then the same is true for \(T\).

Exercise 26.13

Let \(T\) be a linear operator on a vector space \(V\), let \(v\) be a nonzero vector in \(V\), and let \(W\) be the \(T\)-cyclic subspace of \(V\) generated by \(v\). For any \(w \in V\), prove that \(w \in W\) if and only if there exists a polynomial \(g(t)\) such that \(w=g(T)(v)\).

Exercise 26.16

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\).

(a) Prove that if the characteristic polynomial of \(T\) splits, then so does the characteristic polynomial of the restriction of \(T\) to any \(T\)-invariant subspace of \(V\).
(b) Deduce that if the characteristic polynomial of \(T\) splits, then any nontrivial \(T\)-invariant subspace of \(V\) contains an eigenvector of \(T\).

Exercise 26.17

Let \(A\) be an \(n \times n\) matrix. Prove that

\[ \operatorname{dim}\left(\operatorname{span}\left(\left\{I_{n}, A, A^{2}, \ldots\right\}\right)\right) \leq n \]

Exercise 26.18

Let \(A\) be an \(n \times n\) matrix with characteristic polynomial

\[ f(t)=(-1)^{n} t^{n}+a_{n-1} t^{n-1}+\cdots+a_{1} t+a_{0} \]

(a) Prove that \(A\) is invertible if and only if \(a_{0} \neq 0\).
(b) Prove that if \(A\) is invertible, then

\[ A^{-1}=\left(-1 / a_{0}\right)\left[(-1)^{n} A^{n-1}+a_{n-1} A^{n-2}+\cdots+a_{1} I_{n}\right] . \]
(c) Use (b) to compute \(A^{-1}\) for

\[ A=\left(\begin{array}{rrr} 1 & 2 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & -1 \end{array}\right) \]

Exercise 26.20

Let \(T\) be a linear operator on a vector space \(V\), and suppose that \(V\) is a \(T\)-cyclic subspace of itself. Prove that if \(U\) is a linear operator on \(V\), then \(UT=TU\) if and only if \(U=g(T)\) for some polynomial \(g(t)\). Hint: Suppose that \(V\) is generated by \(v\). Choose \(g(t)\) according to Exercise 26.13 so that \(g(T)(v)=U(v)\).

Exercise 26.21

Let \(T\) be a linear operator on a two-dimensional vector space \(V\). Prove that either \(V\) is a \(T\)-cyclic subspace of itself or \(T=c I\) for some scalar \(c\).

Exercise 26.22

Let \(T\) be a linear operator on a two-dimensional vector space \(V\) and suppose that \(T \neq c I\) for any scalar \(c\). Show that if \(U\) is any linear operator on \(V\) such that \(UT=TU\), then \(U=g(T)\) for some polynomial \(g(t)\).

Exercise 26.23

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and let \(W\) be a \(T\)-invariant subspace of \(V\). Suppose that \(v_{1}, v_{2}, \ldots, v_{k}\) are eigenvectors of \(T\) corresponding to distinct eigenvalues. Prove that if \(v_{1}+v_{2}+\cdots+v_{k}\) is in \(W\), then \(v_{i} \in W\) for all \(i\). Hint: Use mathematical induction on \(k\).

Exercise 26.24

Prove that the restriction of a diagonalizable linear operator \(T\) to any nontrivial \(T\)-invariant subspace is also diagonalizable. Hint: Use the result of Exercise 26.23.

Exercise 26.25

(a) Prove the converse to Exercise 24.18(a): If \(T\) and \(U\) are diagonalizable linear operators on a finite-dimensional vector space \(V\) such that \(UT=TU\), then \(T\) and \(U\) are simultaneously diagonalizable. Hint: For any eigenvalue \(\lambda\) of \(T\), show that \(E_{\lambda}\) is \(U\)-invariant, and apply Exercise 26.24 to obtain a basis for \(E_{\lambda}\) of eigenvectors of \(U\).
(b) State and prove a matrix version of (a).

Exercise 26.26

Let \(T\) be a linear operator on an \(n\)-dimensional vector space \(V\) such that \(T\) has \(n\) distinct eigenvalues. Prove that \(V\) is a \(T\)-cyclic subspace of itself. Hint: Use Exercise 26.23 to find a vector \(v\) such that \(\left\{v, T(v), \ldots, T^{n-1}(v)\right\}\) is linearly independent.

Definition 26.16 : Induced Operator on a Quotient Space

Let \(T\) be a linear operator on a vector space \(V\), and let \(W\) be a \(T\)-invariant subspace of \(V\). Define \(\overline{T}: V / W \rightarrow V / W\) by

\[ \overline{T}(v+W)=T(v)+W \quad \text { for any } v+W \in V / W . \]

Exercise 26.27

(a) Prove that \(\overline{T}\) is well defined. That is, show that \(\overline{T}(v+W)=\overline{T}\left(v^{\prime}+W\right)\) whenever \(v+W=v^{\prime}+W\).
(b) Prove that \(\overline{T}\) is a linear operator on \(V / W\).
(c) Let \(\eta: V \rightarrow V / W\) be the linear transformation defined in Exercise 40 of Section 2.1 by \(\eta(v)=v+W\). Show that the diagram of Figure 26.1 commutes; that is, prove that \(\eta T=\overline{T} \eta\). This exercise does not require the assumption that \(V\) is finite-dimensional.

Exercise 26.28

Let \(f(t), g(t)\), and \(h(t)\) be the characteristic polynomials of \(T\), \(T_{W}\), and \(\overline{T}\), respectively. Prove that \(f(t)=g(t) h(t)\). Hint: Extend an ordered basis \(\gamma=\left\{v_{1}, v_{2}, \ldots, v_{k}\right\}\) for \(W\) to an ordered basis \(\beta=\left\{v_{1}, v_{2}, \ldots, v_{k}, v_{k+1}, \ldots, v_{n}\right\}\) for \(V\). Then show that the collection of cosets \(\alpha=\left\{v_{k+1}+W, v_{k+2}+W, \ldots, v_{n}+W\right\}\) is an ordered basis for \(V / W\), and prove that

\[ [T]_{\beta}=\left(\begin{array}{cc} B_{1} & B_{2} \\ O & B_{3} \end{array}\right) \]

where \(B_{1}=[T]_{\gamma}\) and \(B_{3}=[\overline{T}]_{\alpha}\).

Exercise 26.29

Use the hint in Exercise 26.28 to prove that if \(T\) is diagonalizable, then so is \(\overline{T}\).

Exercise 26.30

Prove that if both \(T_{W}\) and \(\overline{T}\) are diagonalizable and have no common eigenvalues, then \(T\) is diagonalizable.

Exercise 26.31

Let

\[ A=\left(\begin{array}{rrr} 1 & 1 & -3 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right), \]

let \(T=L_{A}\), and let \(W\) be the cyclic subspace of \(\mathbb{R}^{3}\) generated by \(e_{1}\).

(a) Use Theorem 26.7 to compute the characteristic polynomial of \(T_{W}\).
(b) Show that \(\left\{e_{2}+W\right\}\) is a basis for \(\mathbb{R}^{3} / W\), and use this fact to compute the characteristic polynomial of \(\overline{T}\).
(c) Use the results of (a) and (b) to find the characteristic polynomial of \(A\).

Exercise 26.32

Prove the converse to Exercise 24.9(a) of Section 5.2: If the characteristic polynomial of \(T\) splits, then there is an ordered basis \(\beta\) for \(V\) such that \([T]_{\beta}\) is an upper triangular matrix. Hints: Apply mathematical induction to \(\operatorname{dim}(V)\). First prove that \(T\) has an eigenvector \(v\), let \(W=\operatorname{span}\left(\left\{v\right\}\right)\), and apply the induction hypothesis to \(\overline{T}: V / W \rightarrow V / W\). Exercise 6.35(b) is helpful here.

Exercise 26.39

Let \(\mathcal{C}\) be a collection of diagonalizable linear operators on a finite-dimensional vector space \(V\). Prove that there is an ordered basis \(\beta\) such that \([T]_{\beta}\) is a diagonal matrix for all \(T \in \mathcal{C}\) if and only if the operators of \(\mathcal{C}\) commute under composition. (This is an extension of Exercise 26.25.) Hints for the case that the operators commute: The result is trivial if each operator has only one eigenvalue. Otherwise, establish the general result by mathematical induction on \(\operatorname{dim}(V)\), using the fact that \(V\) is the direct sum of the eigenspaces of some operator in \(\mathcal{C}\) that has more than one eigenvalue.

Exercise 26.40

Let \(B_{1}, B_{2}, \ldots, B_{k}\) be square matrices with entries in the same field, and let \(A=B_{1} \oplus B_{2} \oplus \cdots \oplus B_{k}\). Prove that the characteristic polynomial of \(A\) is the product of the characteristic polynomials of the \(B_{i}\)'s.

Exercise 26.41

Let

\[ A=\left(\begin{array}{cccc} 1 & 2 & \cdots & n \\ n+1 & n+2 & \cdots & 2 n \\ \vdots & \vdots & & \vdots \\ n^{2}-n+1 & n^{2}-n+2 & \cdots & n^{2} \end{array}\right) . \]

Find the characteristic polynomial of \(A\). Hint: First prove that \(A\) has rank \(2\) and that \(\operatorname{span}\left(\left\{(1,1, \ldots, 1),(1,2, \ldots, n)\right\}\right)\) is \(L_{A}\)-invariant.

Exercise 26.42

Let \(A \in \mathrm{M}_{n \times n}(\mathbb{R})\) be the matrix defined by \(A_{i j}=1\) for all \(i\) and \(j\). Find the characteristic polynomial of \(A\).