## 7.1 Invariant Subspaces

In this section we let $$V$$ and $$W$$ denote real or complex vector spaces.

Suppose $$T \in \mathcal{L}(V)$$ and $$U$$ a subspace of $$V$$, then we say $$U$$ is an invariant subspace of $$T$$ if $$u\in U$$ implies $$Tu\in U$$.

Lemma 7.1 Suppose $$T \in \mathcal{L}(V)$$ and $$U$$ a subspace of $$V$$. Then all of the following hold

• $$U$$ is invariant under $$T$$ if and only if $$T|_U$$ is an operator on $$U$$,
• $$\text{ker} T$$ is invariant under $$T$$, and
• $$\text{im} T$$ is invariant under $$T$$.

Proof. The proof of each part follows.

• By definition, invariant subspace $$U$$ is invariant under $$T$$ if and only if $$u\in U \implies Tu\in U$$, which, by definition of operator, is the same as $$T|_U$$ being an operator.
• If $u T$ then $$Tu=0$$, and hence $$Tu\in \text{ker} T$$.
• If $u T$, then by the definition of range $$Tu\in\text{im} T$$.

Theorem 7.1 Suppose $$T\in \mathcal{L}(V)$$. Let $$\lambda_1,\ldots,\lambda_m$$ denote distinct eigenvalues of $$T$$. The the following are equivalent:

• $$T$$ has a diagonal matrix with respect to some basis of $$V$$;
• $$V$$ has a basis consisting of eigenvectors of $$T$$;
• there exist one-dimensional $$T$$-invariant subspaces $$U_1, \ldots, U_n$$ of $$V$$ such that $$V=U_1 \oplus \cdots \oplus U_n$$;
• $$V=\null (T-\lambda_1 I) \oplus \cdots \oplus \null (T-\lambda_m I)$$;
• $$\text{dim} V = \text{dim} \null (T-\lambda_1 I) + \cdots + \text{dim} \null (T-\lambda_m I)$$.
1. Proof.
2. $$\Longleftrightarrow$$ (ii): Exercise.

3. $$\Longleftrightarrow$$ (iii): Suppose (ii) holds; thus suppose $$V$$ has a basis consisting of eigenvectors of $$T$$. For each $$j$$, let $$U_j=\text{span}(v_j)$$. Obviously each $$U_j$$ is a one-dimensional subspace of $$V$$ that is invariant under $$T$$ (because each $$v_j$$ is an eigenvector of $$T$$). Because $$(v_1,\ldots,v_n)$$ is a basis of $$V$$, each vector in $$V$$ can be written uniquely as a linear combination of $$(v_1,\ldots,v_n)$$. In other words, each vector in $$V$$ can be written uniquely as a sum $$u_1+\cdots +u_n$$, where each $$u_j\in U_j$$. Thus $$V=U_1 \oplus \cdots \oplus U_n$$. Hence (ii) implies (iii). Conversely, suppose now that (iii) holds; thus there are one-dimensional subspaces $$U_1,\ldots,U_n$$ of $$V$$, each invariant under $$T$$, such that $$V=U_1 \oplus \cdots \oplus U_n$$. For each $$j$$, let $$v_j$$ be a nonzero vector in $$U_j$$. Then each $$v_j$$ is an eigenvector of $$T$$. Because each vector in $$V$$ can be written uniquely as a sum $$u_1+\cdots + u_n$$, where each $$u_j\in U_j$$ ( so each $$u_j$$ is a scalar multiple of $$v_j$$), we see that $$(v_1,\ldots,v_n)$$ is a basis of $$V$$. Thus (iii) implies (ii).

4. $$\Longrightarrow$$ (iv): Suppose (ii) holds; thus thus suppose $$V$$ has a basis consisting of eigenvectors of $$T$$. Thus every vector in $$V$$ is a linear combination of eigenvectors of $$T$$. Hence $$V=\null (T-\lambda_1)I + \cdots + \null (T-\lambda_m)I.$$ To show that the sum above is direct, suppose that $$0=u_1+\cdots + u_m$$, where each $$u_j\in \null (T-\lambda_j I)$$. Because nonzero eigenvectors correspond to distinct eigenvalues are linearly independent, this implies that each $$u_j$$ equals 0. This implies that the sum is a direct sum, completing the proof that

5. implies (iii).

6. $$\Longrightarrow$$ (v): Exercise.

7. $$\Longrightarrow$$ (ii): Suppose (v) holds; thus $$\text{dim} V=\text{dim} \null (T-\lambda_1I)+\cdots + \text{dim} \null (T-\lambda_m I)$$. Choose a basis of each $$\null(T-\lambda_j I)$$; put all these bases together to form a list $$(v_1,\ldots,v_n)$$ of eigenvectors of $$T$$, where $$n=\text{dim} V$$. To show that this list is linearly independent, suppose $$a_1 v_1+\cdots + a_n v_n=0$$, where $$a_1,\ldots,a_n$$ are scalars. For each $$j=1, \ldots., m$$, let $$u_j$$ denote the sum of all the terms $$a_k v_k$$ such that $$v_k\in \null(T-\lambda_j I)$$. Thus each $$u_j$$ is an eigenvector of $$T$$ with eigenvalue $$\lambda_j$$, and $$u_1+\cdots + u_m=0$$. Because nonzero eigenvectors corresponding to distinct eigenvalues are linearly independent, this implies that each $$u_j$$ equals to 0. Because each $$u_j$$ is a sum of terms $$a_k v_k$$ where the $$v_k$$’s where chosen to be a basis of $$\null (T-\lambda_j I)$$, this implies that all the $$a_k$$’s equal to 0. Thus $$(v_1,\ldots,v_n)$$ is linearly independent and hence a basis of $$V$$. Thus (v) implies (ii).

A vector $$v$$ is called a generalized eigenvector of $$T$$ corresponding to $$\lambda$$, where $$\lambda$$ is an eigenvalue of $$T$$, if $$(T-\lambda I)^j v=0$$ for some positive integer $$j$$.

Lemma 7.2 If $$T\in \mathcal{L}(V)$$. Then, if $$m$$ is a nonnegative integer such that $$\text{ker} T^m=\text{ker} T^{m+1}$$, then

• $$\{0\}=\text{ker} T^0 \subseteq \text{ker} T^1 \subseteq \cdots \subseteq \text{ker} T^m = \text{ker} T^{m+1} = \text{ker} T^{m+2} = \cdots$$
• $$\text{ker} T^{\text{dim} V}=\text{ker} T^{\text{dim} V+1} =\text{ker} T^{\text{dim} V+2}\cdots$$
• $$V=\text{im} T^0 \supseteq \text{im} T^1 \supseteq \cdots \supseteq \text{im} T^k \supseteq \text{im} T^{k+1} \supseteq \cdots$$
• $$\text{im} T^{\text{dim} V}= \text{im} T^{\text{dim} V+1} = \text{im} T^{\text{dim} V+2}\cdots$$

Lemma 7.3 Suppose $$T\in \mathcal{L}(V)$$ and $$\lambda$$ is an eigenvalue of $$T$$. Then the set of generalized eigenvalues of $$T$$ corresponding to $$\lambda$$ equals $$\text{ker}(T-\lambda I)^{\text{dim} V}$$.

Proof. The proof is left for the reader.

An operator is called nilpotent if some power of it equal 0.

Lemma 7.4 Suppose $$N\in \mathcal{L}(V)$$ is nilpotent, then $$N^{\text{dim} V}=0$$.

Proof. The proof is left for the reader.

Theorem 7.2 Let $$T\in \mathcal{L}(V)$$ and $$\lambda\in\mathbb{F}$$. Then for every basis of $$V$$ with respect to which $$T$$ has an upper-triangular matrix, $$\lambda$$ appears on the diagonal of the matrix of $$T$$ precisely $$\text{dim} (T-\lambda I)^{\text{dim} V}$$ times.

Proof. The proof is left for the reader.

The multiplicity of an eigenvalue $$\lambda$$ of $$T$$ is defined to be the dimension of the subspace of generalized eigenvectors corresponding to $$\lambda$$, that is the multiplicity of $$\lambda$$ is equal to $$\text{dim} \text{ker}(T-\lambda I)^{\text{dim} V}$$.

Theorem 7.3 If $$V$$ is a complex vector space and $$T\in \mathcal{L}(V)$$, then the sum of the multiplicities of all the eigenvalues of $$T$$ equals $$\text{dim} V$$.

Proof. The proof is left for the reader.

Let $$d_j$$ denote the multiplicity of $$\lambda_j$$ as an eigenvalue of $$T$$, the polynomial $(z-\lambda_1)^{d_1} \cdots (z-\lambda_m)^{d_m}$ is called the characteristic polynomial of $$T$$.

Theorem 7.4 Suppose that $$V$$ is a complex vector space and $$T\in \mathcal{L}(V)$$. Let $$q$$ denote the characteristic polynomial of $$T$$. Then $$q(T)=0$$.

Proof. The proof is left for the reader.

Theorem 7.5 If $$T\in \mathcal{L}(V)$$ and $$p\in \mathcal{P}(\mathbb{F})$$, then $$\text{ker} p(T)$$ is invariant under $$T$$.

Proof. The proof is left for the reader.

Theorem 7.6 Suppose $$V$$ is a complex vector space and $$T\in \mathcal{L}(V)$$. Let $$\lambda_1,\ldots,\lambda_m$$ be the distinct eigenvalues of $$T$$, and let $$U_1,\ldots,U_m$$ be the corresponding subspaces of generalized eigenvectors. Then

• $$V=U_1\oplus \cdots \oplus U_m$$;
• each $$U_J$$ is invariant under $$T$$;
• each $$\left.(T-\lambda_j I) \right|_{U_j}$$ is nilpotent.

Proof. The proof is left for the reader.

Theorem 7.7 Suppose $$V$$ is a complex vector space and $$T\in \mathcal{L}(V)$$. Then there is a basis of $$V$$ consisting of generalized eigenvectors of $$T$$.

Proof. The proof is left for the reader.

Theorem 7.8 Suppose $$N$$ is a nilpotent operator on $$V$$. Then there is a basis of $$V$$ with respect to which the matrix of $$N$$ has the form $\begin{bmatrix} 0 & & * \\ & \ddots & \\ 0 & & 0\end{bmatrix};$ here all entries on and below the diagonal are 0’s.

Proof. The proof is left for the reader.

Theorem 7.9 Suppose $$V$$ is a complex vector space and $$T\in \mathcal{L}(V)$$. Let $$\lambda_1,\ldots,\lambda_m$$ be the distinct eigenvalues of $$T$$. Then there is a basis of $$V$$ with respect to which $$T$$ has block diagonal matrix of the form $\begin{bmatrix} A_1 & & 0\\ & \ddots & \\ 0 & & A_m\end{bmatrix},$ where each $$A_j$$ is an upper-triangular matrix of the form $\begin{bmatrix} \lambda_1 & & * \\ & \ddots & \\ 0 & & \lambda_j\end{bmatrix}.$

Proof. The proof is left for the reader.

Theorem 7.10 Suppose $$N\in\mathcal{L}(V)$$ is nilpotent. Then $$I+N$$ has a square root.

Proof. The proof is left for the reader.

On real vector spaces there exist invertible operators that have no square roots. For example, the operator of multplication by $$-1$$ on $$\mathbb{R}$$ has no square root because no real number has its square equal to $$-1$$.

Theorem 7.11 Suppose $$V$$ is a complex vector space. If $$T\in \mathcal{L}(V)$$ is invertible, then $$T$$ has a square root.

Proof. The proof is left for the reader.

The minimal polynomial of $$T$$ is the monic polynomial $$p\in \mathcal{P}(\mathbb{F}$$ of smallest degree such that $$p(T)=0$$.

Theorem 7.12 Let $$T\in \mathcal{L}(V)$$ and let $$q\in \mathcal{P}(\mathbb{F})$$. Then $$q(T)=0$$ if and only if the minimal polynomial of $$T$$ divided $$q$$.

Proof. The proof is left for the reader.

Theorem 7.13 Let $$T\in \mathcal{L}(V)$$. Then the roots of the minimal polynomial of $$T$$ are precisely the eigenvalues of $$T$$.

Proof. The proof is left for the reader.

Every $$T\in \mathcal{L}(V)$$ where $$V$$ is a complex vector space, there is a basis of $$V$$ with respect to which $$T$$ has a nice upper-triangular matrix. We can do even better. There is a basis of $$V$$ with respect to which the matrix of $$T$$ contains zeros everywhere except possibly on the diagonal and the line directly above the diagonal.

Suppose $$N\in \mathcal{L}(V)$$ is nilpotent. For each nonzero vector $$v\in V$$, let $$m(v)$$ denote the largest nonnegative integer such that $$N^{m(v)}\neq 0$$.

Theorem 7.14 If $$N\in \mathcal{L}(V)$$ is nilpotent, then there exist vectors $$v_1,\ldots,v_k \in V$$ such that

• $$\left(v_1,N v_1,\ldots,N^{m(v_1)},\ldots,v_k, N v_k, \ldots, N^{m(v_k)}v_k\right)$$ is a basis of $$V$$;
• $$\left(N^{m(v_1)}v_1,\ldots,N^{m(v_k)}v_k\right)$$ is a basis of $$\text{ker} N$$.

Proof. The proof is left for the reader.

Example 7.1 Suppose $$T\in \mathcal{L}(V)$$. Prove that if $$U_1,\ldots,U_m$$ are subspaces of $$V$$ invariant under $$T$$, then $$U_1 + \cdots +U_m$$ is invariant under $$T$$. Suppose $$v\in U_1+\cdots + U_m$$. Then there exists $$u_1,\ldots,u_m$$ such that $$v=u_1+\cdots +u_m$$ with $$u_j\in U_j$$. Then $$Tv=T u_1+\cdots + T u_m$$. Since each $$U_j$$ is invariant under $$T$$, $$T u_j \in U_j$$, so $$T v \in U_1+ \cdots + U_m$$.

Example 7.2 Suppose $$T\in \mathcal{L}(V)$$. Prove that the intersection of any collection of subspaces of $$V$$ invariant under $$T$$ is invariant under $$T$$. Suppose we have subspaces $$\{U_j\}$$ with each $$U_j$$ invariant under $$T$$. Let $$v\in \cap_j U_j$$. Then $$Tv\in U_j$$ for each $$j$$, and so $$\cap_j U_j$$ is invariant under $$T$$.

Example 7.3 Prove or give a counterexample: if $$U$$ is a subspace of $$V$$ that is invariant under every operator on $$V$$, then $$U=\{0\}$$ or $$U=V$$. We will prove the contrapositive: if $$U$$ is a subspace of $$V$$ and $$U\neq \{0\}$$ and $$U\neq V$$, then there exists an operator $$T$$ on $$V$$ such that $$U$$ is not invariant under $$T$$. Let $$(u_1,\ldots,u_m)$$ be a basis for $$U$$, which we extend to a basis $$(u_1,\ldots,u_m, v_1,\ldots,v_n)$$ of $$V$$. The assumption $$U\neq \{0\}$$ and $$U\neq V$$ means that $$m\geq 1$$ and $$n\geq 1$$. Define a linear map $$T$$ by $$Tu_1=v_1$$ and for $$j>1$$, $$T u_j=0$$. Since $$v_1\not \in U$$, the subspace $$U$$ is not invariant under the operator $$T$$.

Example 7.4 Suppose that $$S,T\in \mathcal{L}(V)$$ are such that $$S T= T S$$. Prove that $$\text{null }(T-\lambda I)$$ is invariant under $$S$$ for every $$\lambda \in F$$. Suppose $$v\in \text{ker}(T-\lambda I)$$. Then $$Tv = \lambda v$$ and using $$TS=ST$$, $$Sv$$ satisfies $T(S v)=S(T v)=S v)=(S v).$ Thus $$S v\in \text{ker}(T-\lambda I)$$ and so $$\text{ker} (T-\lambda I)$$ is invariant under $$S$$.

Example 7.5 Define $$T\in \mathcal{L}(F^2)$$ by $$T(w,z)=(z,w)$$. Find all eigenvalues and eigenvectors of $$T$$. Suppose $$(w,z)\neq (0,0)$$ and $$T(w,z)=(z,w)=\lambda(w,z)$$. Then $$z=\lambda w$$ and $$w=\lambda z$$. Of course this leads to $$w=\lambda z=\lambda^2w$$, $$z=\lambda w=\lambda^2 z$$. Since $$w\neq 0$$ or $$z\neq 0$$, we see that $$\lambda^2=1$$ so that $$\lambda =\pm 1$$. A basis of eigenvectors is $$(w_1,z_1)=(1,1)$$, $$(w_2,z_2)=(-1,1)$$ and they have eigenvalues 1 and $$-1$$ respectively.

Example 7.6 Define $$T\in \mathcal{L}(F^3)$$ by $$T(z_1,z_2,z_3)=(2z_2,0,5z_3)$$. Find all eigenvalues and eigenvectors of $$T$$. Suppose $$(z_1,z_2,z_3)\neq (0,0,0)$$ and $T(z_1,z_2,z_3)=(2z_2,0,5z_3)=\lambda (z_1,z_2,z_3).$ If $$\lambda=0$$ then $$z_2=z_3=0$$, and one checks that $$v_1=(0,0,0)$$ is an eigenvector with eigenvalue 0. If $$\lambda\neq 0$$ then $$z_2=0$$, $$2z_2=\lambda z_1=0$$, $$5z_3=\lambda z_3$$, so $$z_1=0$$ and $$\lambda =5$$. The eigenvector for $$\lambda=5$$ is $$v_2=(0,0,1)$$. These are the only eigenvalues and each eigenspace is one dimensional.

Example 7.7 Suppose $$n$$ is a positive integer and $$T\in \mathcal{L}(\mathbb{F}^n)$$ is defined by $T(x_1,\ldots,x_n)=(x_1+ \cdots + x_n,\ldots,x_1+\cdots +x_n).$ Find all eigenvalues and eigenvectors of $$T$$. First, any vector of the form $$v_1=(\alpha,\ldots,\alpha)$$, for $$\alpha\in \mathbb{F}$$, is an eigenvector with eigenvalue $$n$$. If $$v_2$$ is any vector $v_2=(x_1,,x_n),$ such that $$x_1+\cdots + x_n=0$$ then $$v_2$$ is an eigenvector with eigenvalue 0. Here are the independent eigenvectors: $$v_1=(1,1,\ldots,1)$$, and $$v_n=(1,0,\ldots,0)-E_n$$, for $$n\geq 2$$ where $$E_n$$ denoted the $$n$$-th standard basis vector.

Example 7.8 Suppose $$T\in \mathcal{L}(V)$$ is invertible and $$0\neq \lambda \in F$$. Prove that $$\lambda$$ is an eigenvalue of $$T$$ if and only if $$\frac{1}{\lambda}$$ is an eigenvalue of $$T^{-1}$$. Suppose $$v\neq 0$$ and $$T v =\lambda v$$. Then $$v=T^{-1}T v=\lambda T^{-1}v$$, or $$T^{-1}v=\frac{1}{\lambda}v$$, and the other direction is similar.

Example 7.9 Suppose $$S,T\in \mathcal{L}(V)$$. Prove that $$S T$$ and $$T S$$ have the same eigenvalues. Suppose $$v\neq 0$$ and $$STv=\lambda v$$. Multiply by $$T$$ to get $$T S(Tv)=\lambda T v$$. Thus if $$t\neq 0$$ then $$\lambda$$ is also an eigenvalue of $$TS$$, with nonzero=o eigenvector $$Tv$$. On the other hand, if $$T v=0$$, then $$\lambda =0$$ is an eigenvalue of $$ST$$. But if $$T$$ is not invertible, then $$\text{im} TS \subset\text{im} T$$is not equal to $$V$$, so $$TS$$ has a nontrivial null space, hence 0 is an eigenvalue of $$TS$$.

Example 7.10 Suppose $$T\in \mathcal{L}(V)$$ is such that every vector in $$V$$ is an eigenvector of $$T$$. Prove that $$T$$ is a scalar multiple of the identity operator. Pick a basis $$(v_1,\ldots,v_N)$$ for $$V$$. By assumption, $$T v_n=\lambda_n v_n$$. Pick any two distance indices, $$m,n$$. We also have $$T(v_m+v_n)=\lambda(v_m+v_n)=\lambda(v_m+v_n)=\lambda_m v_m + \lambda_n v_n$$. Write this as $$0=(\lambda-\lambda_m)v_m+(\lambda-\lambda_n)v_n$$. Since $$v_m$$ and $$v_n$$ are independent, $$\lambda=\lambda_m=\lambda_n$$, and all the $$\lambda_n$$ are equal.

Example 7.11 Suppose $$S, T\in \mathcal{L}(V)$$ and $$S$$ is invertible. Prove that if $$p \in \mathcal{P}(\mathbb{F})$$ is a polynomial, then $$p(S T S^{-1})=S p(T) S^{-1}$$. First let’s show that for positive integers $$n$$, $$(STS^{-1})^n=S T^n S^{-1}$$. We may do this by induction, with nothing to show if $$n=1$$. Assume it’s true for $$n=k$$, and consider $(STS^{-1})^{k+1}=(STS^{-1})^k(STS^{-1})=ST^k S^{-1}STS^{-1}=S T^{k+1}S^{-1}.$ Now suppose $$P(z)=a_n z^N+\cdots + a_1 z+a_0.$$ Then \begin{align*} p(STS^{-1})&=\sum_{n=0}^N a_n (STS^{-1})^n=\sum_{n=0}^N a_n ST^n S^{-1}\\ & =S\left( \sum_{n=0}^N a_n T^n \right) S^{-1}=Sp(T)S^{-1}. \tag*{} \end{align*}

Example 7.12 Suppose $$F=\mathbb{C}$$, $$T\in \mathcal{L}(V)$$, $$p\in \mathcal{P}(\mathbb{C})$$, and $$a\in \mathbb{C}$$. Prove that $$a$$ is an eigenvalue of $$p(T)$$ if and only if $$a=p(\lambda)$$ for some eigenvalue  of $$T$$. Suppose first that $$v\neq 0$$ is an eigenvalue of $$T$$ with eigenvalue $$\lambda$$; that is $$T v = \lambda v$$. Then for positive integers $$n$$, $$T^n v=T^{n-1} \lambda v = \cdots \lambda^n v$$, and so $$p(T)v=p(\lambda) v$$. That is $$\alpha=p(\lambda)$$ is an eigenvalue of $$p(T)$$ if $$\lambda$$ is an eigenvalue of $$T$$. Conversely, suppose now that $$\alpha$$ is a eigenvalue of $$p(T)$$, so there is a $$v\neq 0$$ with $$p(T)v =\alpha v$$, or $$(p(T)-\alpha I)v=0$$. Since $$\mathbb{F}=\mathbb{C}$$, we may factor the polynomial $p(T)-I$ into linear factors $0=(p(T)-\alpha I)v=\prod (T-\lambda_n I)v.$ At least one of the factors is not invertible, so at least one of the $$\lambda_n$$, say $$\lambda_1$$, is an eigenvalue of $$T$$. Let $$w\neq 0$$ be an eigenvector for $$T$$ with eigenvalue $$\lambda_1$$. Then $0=(T-\lambda_N I)\cdots (T-\lambda_1 I)w=(p(T)-\alpha I) w,$ so $$w$$ is an eigenvalue for $$p(T)$$ with eigenvalue $$\alpha$$. But by the first part of the argument, $$p(T)w=p(\lambda_1)w=\alpha w$$ and $$\alpha=p(\lambda_1)$$.

Example 7.13 Show that the previous exercise does not hold with $$F=\mathbb{R}$$. Take $$T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ given by $$T(x,y)=(-y,x)$$. We’ve seen perviously that $$T$$ has no real eigenvalues. On the other hand, $$T^2(x,y)=(-x,-y)=-1(x,y)$$.

Example 7.14 Suppose $$V$$ is a complex vector space and $$T\in \mathcal{L}(V)$$. Prove that $$T$$ has an invariant subspace of dimension $$j$$ for each $$j=1,\ldots, \text{dim} V$$. Let $$(v_1,\ldots,v_N)$$ be a basis with respect to which $$T$$ has an upper triangular matrix. Then by a previous proposition, $$T: \text{span}(v_1,\ldots,v_j) \rightarrow \text{span}(v_1,\ldots,v_j)$$.

Example 7.15 Give an example of an operator whose matrix with respect to some basis contains only 0’s on the diagonal, but the operator is invertible. Consider $$T=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$.

Example 7.16 Give an example of an operator whose matrix with respect to some basis contains only nonzero numbers on the diagonal, but the operator is not invertible. Taking $$T=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$. If $$v=[1,-1]$$, then $$Tv=0$$.

Example 7.17 Give an example of an operator on $$\mathbb{C}^4$$ whose characteristic and minimal polynomials both equal $$z (z-1)^2(z-3)$$.

Example 7.18 Give an example of an operator on $$\mathbb{C}^4$$ whose characteristic polynomial equals $$z (z-1)^2(z-3)$$ and whose minimal polynomial equals $$z (z-1)(z-3)$$.

Example 7.19 Suppose $$a_0, \ldots, a_{n-1}\in \mathbb{C}$$. Find the minimal and characteristic polynomials of the operator on $$\mathbb{C}^n$$ whose matrix is $\begin{bmatrix} 0 & & & & & -a_0 \\ 1 & 0 & & & & -a_1 \\ & 1 & \ddots & & & -a_2 \\ & & \ddots & & & \vdots \\ & 1 & & & 0 & -a_{n-2} \\ & & & & 1 & -a_{n-1} \end{bmatrix}$ with respect to the standard bases.

## 7.2 Jordan Canonical Form

A basis of $$V$$ is called a Jordan basis for $$T$$ if with respect to this basis $$T$$ has block diagonal matrix $\begin{bmatrix} A_1 & & 0 \\ & \ddots & \\ 0 & & A_m \\ \end{bmatrix}$ where each $$A_j$$ is an upper triangular matrix of the form $A_j = \begin{bmatrix} \lambda_j & 1 & & 0 \\ & \ddots & \ddots & \\ & & \ddots & 1 \\ 0 & & & \lambda_j \\ \end{bmatrix}$ where the diagonal is filled with some eigenvalue $$\lambda_j$$ of $$T$$.

Because there exist operators on real vector spaces that have no eigenvalues, there exist operators on real vector spaces for which there is no corresponding Jordan basis.

Theorem 7.15 Suppose $$V$$ is a complex vector space. If $$T\in \mathcal{L}(V)$$, then there is a basis of $$V$$ that is a Jordan basis for $$T$$.

Proof. The proof is left for the reader.

A basis of $$V$$ is called a Jordan basis for $$T$$ if with respect to this basis $$T$$ has block diagonal matrix $\begin{bmatrix} A_1 & & 0 \\ & \ddots & \\ 0 & & A_m \\ \end{bmatrix}$ where each $$A_j$$ is an upper triangular matrix of the form $A_j = \begin{bmatrix} \lambda_j & 1 & & 0 \\ & \ddots & \ddots & \\ & & \ddots & 1 \\ 0 & & & \lambda_j \\ \end{bmatrix}$ where the diagonal is filled with some eigenvalue $$\lambda_j$$ of $$T$$.

An operator $$T$$ can be put into Jordan canonical form if its characteristic and minimal polynomials factor into linear polynomials. this is always true if the vector space is complex.

Theorem 7.16 Let $$T\in\mathcal{L}(V)$$ whose characteristic and minimal polynomials are, respectively, $c(t)=(t-\lambda_1)^{n_1} \cdots (t-\lambda_r)^{n_r}) \qquad \text{and} \qquad m(t)=(t-\lambda_1)^{m_1} \cdots (t-\lambda_r)^{m_r})$ where the $$\lambda_i$$ are distinct scalars. Then $$T$$ has block diagonal matrix representation $$J$$ whose diagonal entries are of the form $J_{ij}= \begin{bmatrix} \lambda_i & 1 & 0 & \cdots & 0 & 0 \\ 0 & \lambda_i & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_i & 1 \\ 0 & 0 & 0 & \cdots & 0 & \lambda_i \end{bmatrix}$ For each $$\lambda_i$$ the corresponding blocks have the following properties:

• There is at least one $$J_{ij}$$ of order $$m_i$$; all other $$J_{ij}$$ are of order $$\leq m_i$$
• The sum of the orders of the $$J_{ij}$$ is $$n_i$$.
• The number of $$J_{ij}$$ equals the geometric multiplicity of $$\lambda_i$$.
• The number of $$J_{ij}$$ of each possible order is uniquely determined by $$T$$.

Proof. The proof is left for the reader.

The matrix $$J$$ in the above proposition is called the Jordan canonical form of the operator $$T$$. A diagonal block $$J_{ij}$$ is called a Jordan block belongng to the eigenvalue $$\lambda_i$$. Observe that $\begin{bmatrix} \lambda_i & 1 & 0 & \cdots & 0 & 0 \\ 0 & \lambda_i & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_i & 1 \\ 0 & 0 & 0 & \cdots & 0 & \lambda_i \end{bmatrix} = \begin{bmatrix} \lambda_i & 0 & 0 & \cdots & 0 & 0 \\ 0 & \lambda_i & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_i & 0 \\ 0 & 0 & 0 & \cdots & 0 & \lambda_i \end{bmatrix} + \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 \end{bmatrix}$ That is, $$J_{ij}=\lambda_i I+N$$ where $$N$$ is the nilpotent block.

Example 7.20 Suppose the characteristic and minimum polynomials of an operator $$T$$ are, respectively, $c(t)=(t-2)^4(t-3)^3 \qquad \text{and} \qquad m(t)=(t-2)^2(t-3)^2.$ Then the Jordan canonical form of $$T$$ is one of the following matrices: $\begin{bmatrix} 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix} \qquad \text{or} \qquad \begin{bmatrix} 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix}$ The first matrix occurs if $$T$$ has two independent eigenvectors belonging to the eigenvalue 2; and the second matrix occurs if $$T$$ has three independent eigenvectors belonging to 2.

Example 7.21 Suppose $$N\in \mathcal{L}(V)$$ is nilpotent. Prove that the minimal polynomial of $$N$$ is $$z^{m+1}$$, where $$m$$ is the length of the longest consecutive string of $$1'\text{s}$$ that appears on the line directly above the diagonal in the matrix of $$N$$ with respect to any Jordan basis for $$N$$.

Example 7.22 Suppose $$V$$ is a complex vector space and $$T\in \mathcal{L}(V)$$. Prove that there does not exist a direct sum decomposition of $$V$$ into two proper subspaces invariant under $$T$$ if and only if the minimal polynomial of $$T$$ is of the form $$(z-\lambda)^{\text{dim} V}$$ for some $$\lambda \in \mathbb{C}$$.

Example 7.23 Suppose $$T\in \mathcal{L}(V)$$ and $$(v_1,\ldots,v_n)$$ is a basis of $$V$$ that is a Jordan basis for $$T$$. Describe the matrix of $$T$$ with respect to the basis $$(v_n,\ldots,v_1)$$ obtained by reversing the order of the $$v$$’s.

Example 7.24 Consider a 2-by-2 matrix of real numbers $\begin{bmatrix} a & c \\ b & d \end{bmatrix}.$

Example 7.25 Suppose $$A$$ is a block diagonal matrix $A=\begin{bmatrix} A_1 & & 0 \\ & \ddots & \\ 0 & & A_m \end{bmatrix},$ where each $$A_j$$ is a square matrix. Prove that the set of eigenvalues of $$A$$ equals the union of the eigenvalues of $$A_1,\ldots,A_m$$.

Example 7.26 Suppose $$A$$ is a block upper-triangular matrix $A= \begin{bmatrix} A_1 & & * \\ & \ddots & \\ 0 & & A_m \end{bmatrix},$ where each $$A_j$$ is a square matrix. Prove that the set of eigenvalues of $$A$$ equals the union of the eigenvalues of $$A_1$$,,$$A_m$$.

Example 7.27 Suppose $$V$$ is a real vector space and $$T\in \mathcal{L}(V)$$. Suppose $$\alpha, \beta \in \mathbb{R}$$ are such that $$T^2+\alpha T+\beta I=0$$. Prove that $$T$$ has an eigenvalue if and only if $$\alpha^2 \geq 4 \beta$$.

Example 7.28 Suppose $$V$$ is a real inner-product space and $$T\in \mathcal{L}(V)$$. Prove that there is an orthonormal basis of $$V$$ with respect to which $$T$$ has a block upper-triangular matrix $\begin{bmatrix} A_1 & & * \\ & \ddots & \\ 0 & & A_m \end{bmatrix}.$ where each $$A_j$$ is a 1-by-1 matrix or a 2-by-2 matrix with no eigenvalues.

Example 7.29 Prove that if $$T\in \mathcal{L}(V)$$ and $$j$$ is a positive integer such that $$j \leq \text{dim} V$$, then $$T$$ has an invariant subspace whose dimension equals $$j-1$$ or $$j$$.

Example 7.30 Prove that there does not exist an operator $$T\in \mathcal{L}(\mathbb{R}^7)$$ such that $$T^2+T+I$$ is nilpotent.

Example 7.31 Give an example of an operator $$T\in \mathcal{L}(\mathbb{C}^7)$$ such that $$T^2+T+I$$ is nilpotent.

Example 7.32 Suppose $$V$$ is a real vector space and $$T\in \mathcal{L}(V)$$. Suppose $$\alpha, \beta \in \mathbb{R}$$ are such that $$\alpha^2< 4\beta$$. Prove that null $$(T^2+\alpha T + \beta I)^k$$ has even dimension for every positive integer $$k$$.

Example 7.33 Suppose $$V$$ is a real vector space and $$T\in \mathcal{L}(V)$$. Suppose $$\alpha, \beta \in \mathbb{R}$$ are such that $$\alpha^2< 4\beta$$ and $$T^2+\alpha T+\beta I$$ is nilpotent. Prove that $$\text{dim} V$$ is even and $$(T^2+\alpha T+\beta I)^{\text{dim} V/2}=0.$$

Example 7.34 Prove that if $$T\in \mathcal{L}(\mathbb{R}^3)$$ and 5, 7 are eigenvalues of $$T$$, then $$T$$ has no eigenpairs.

Example 7.35 Suppose $$V$$ is a real vector space with $V =n$ and $$T\in \mathcal{L}(V)$$ is such that null $T^{n-2}$ null $$T^{n-1}$$. Prove that if $$T$$ has at most two distinct eigenvalues and that $$T$$ has no eigenpairs.

Example 7.36 Prove that 1 is an eigenvalue of every square matrix with the property that the sum of the entries in each row equals 1.

Example 7.37 Suppose $$V$$ is a real vector space with $V =2$. Prove that if $\begin{bmatrix} a & c \\ b & d \end{bmatrix}$ is the matrix of $$T$$ with respect to some basis of $$V$$, then the characteristic polynomial of $$T$$ equals $$(z-a)(z-d)-b c$$.

Example 7.38 Suppose $$V$$ is a real inner-product space and $$S\in \mathcal{L}(V)$$ is an isometry. Prove that if $$(\alpha, \beta)$$ is an eigenpair of $$S$$, then $$\beta=1$$.