1. Let A\in M_3(\mathbb C) with \operatorname{tr}A=1,\operatorname{tr}A^2=-3,\operatorname{tr}A^3=4. Find \operatorname{det}A.

  • Consider there exists Q such that Q^{-1}AQ is upper-triangular. Suppose that the diagonal is \lambda_1,\lambda_2,\lambda_3. We have \operatorname{tr}(A^k)=\mathrm{tr}(A^kQ^{-1}Q)=\mathrm{tr}((Q^{-1}AQ)^k)=\lambda_1^k+\lambda_2^k+\lambda_3^k, so \sum_{i=1}^3\lambda_i^k=1,-3,4,k=1,2,3. Notice that \sum_{i=1}^3\lambda_i^3-3\prod_{i=1}^3\lambda_i=\frac{1}{2}\sum_{i=1}^3\lambda_i[3\sum_{i=1}^3\lambda_i^2-(\sum_{i=1}^3\lambda_i)^2], therefore \mathrm{det}A=\prod_{i=1}^3\lambda_i=\frac{4-\frac{1[3\times (-3)-1^3]}{2}}{3}=3.

2. Let A=\begin{pmatrix} 2&3&2\\ 1&8&2\\ -2&-14&-3\end{pmatrix}.

(1) Find the Jordan canonical form J of A and a matrix Q s.t. A=QJQ^{-1}.

  • |\lambda-A|=(x-3)^2(x-1), therefore J=\begin{pmatrix}1&0&0\\0&3&1\\0&0&3\end{pmatrix}. Notice that \mathcal (A-I)\begin{pmatrix}-2\\0\\1\end{pmatrix}=0,(A-3I)\begin{pmatrix}1\\-1\\2\end{pmatrix}=0,(A-3I)\begin{pmatrix}-1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\-1\\2\end{pmatrix}, therefore Q=\begin{pmatrix}-2&1&-1\\0&-1&0\\1&2&0\end{pmatrix}.

(2) Compute A^{10}.

  • A^{10}=QJ^{10}Q^{-1}=Q\begin{pmatrix}1&0&0\\0&3^{10}&10\times 3^9\\0&0&3^{10}\end{pmatrix}Q^{-1}=\begin{pmatrix}-137781&-747958&-275564\\196830&1043199&393660\\-393660&-2086396&-787319\end{pmatrix}.

3. Let V be a finate-dimensional vector space over a field F, T\in\mathcal L(V). Let W be a T-invariant subspace of V. Suppose that V has its generalized eigenspace decomposition of T, V=G(\lambda_1)\oplus G(\lambda_2)\oplus\cdots\oplus G(\lambda_k) [1], where \lambda_1,\dots,\lambda_k are distinct eigenvalues of T and G(\lambda_i) are the generalized eigenspace decomposition of T corresponding to the eigenvalue \lambda_i. Prove that the generalized eigenspace decomposition of T|_W for W can be induced from [1], that is W=(G(\lambda_1)\cap W)\oplus(G(\lambda_2)\cap W)\oplus\cdots\oplus(G(\lambda_k)\cap W). (Note that for an arbitrary subspace decomposition V=V_1\oplus V_2, then in general V\cap W\neq(V_1\cap W)\oplus(V_2\cap W).)

  • On one side, x\in G_{T|_W}(\lambda_i)\Rightarrow(T|_W-\lambda_i)^{l_i}x=0 for some positive integer l_i \xRightarrow{x\in W} (T-\lambda_i)^{l_i}x=(T|_W-\lambda_i)^{l_i}x=0\Rightarrow x\in G(\lambda_i), therefore G_{T|_W}(\lambda_i)\subseteq G(\lambda_i)\cap W.
  • On the other side, x\in G(\lambda_i)\cap W\Rightarrow(\exists l_i\in \Z^+,(T-\lambda_i)^{l_i}x=0)\wedge(x\in W)\Rightarrow ((T|_W-\lambda_i)^{l_i}x=(T-\lambda_i)^{l_i}x=0)\wedge(x\in W)\Rightarrow x\in G_{T|_W}(\lambda_i), which means G(\lambda_i)\cap W\subseteq G_{T|_W}(\lambda_i).
  • Therefore G_{T|_W}(\lambda_i)=G(\lambda_i)\cap W. Substitute it into [1], we have W=G_{T|_W}(\lambda_1)\oplus\cdots\oplus G_{T|_W}(\lambda_k)=(G(\lambda_1)\cap W)\oplus(G(\lambda_2)\cap W)\oplus\cdots\oplus(G(\lambda_k)\cap W)

4. Let V be a finite-dimensional vector space over a field F, and L,T\in\mathcal L(V) be both diagonalizable. Show that L and T are simultaneously diagonalizable (i.e., there exists a basis \alpha=(v_1,v_2,\cdots,v_n) for V such that both A=[L]_\alpha and B=[T]_\alpha are diagonal) if and only if LT=TL. (This is important in quantum mechanics. This means that if two operators do not commute, then they do not have a common eigenbasis. Hence we have the uncertainty principle.)

  • Suppose that there exists a basis \alpha=(v_1,v_2,\cdots,v_n) for V such that both A=[L]_\alpha and B=[T]_\alpha are diagonal, we have AB=BA, which exactly means LT=TL.
  • Now suppose that LT=TL. Notice that LTv_{Li}=TLv_{Li}=\lambda_{Li}Tv_{Li},v_{Li}\in E(\lambda_{Li},L), therefore E(\lambda_{Li},L) is an invariant subspace of T. Because T is diagonalizable in V, we know T|_{E(\lambda_{Li},L)} is also diagonalizable in E(\lambda_{Li},L). So let \alpha_i be the basis for E(\lambda_{Li},L) such that [T|_{E(\lambda_{Li},L)}]_{\alpha_i} is diagonal, \alpha=(\alpha_1,\cdots,\alpha_k), we have [T]_\alpha is diagonal. And as \alpha_i is the basis for E(\lambda_{Li},L), [L]_\alpha is also diagonal.

5. Let p(t),q(t)\in\mathbb C[t] be relatively prime, A\in M_n(\mathbb C). Show that \operatorname{rank}(p(A))+\operatorname{rank}(q(A))\ge n.

  • Let J be the Jordan canonical form of A, and J=Q^{-1}AQ. We have \mathrm{rank}(p(A))=\mathrm{rank}(Qp(J)Q^{-1})=\mathrm{rank}(p(J)). So \operatorname{rank}(p(A))+\operatorname{rank}(q(A))=\operatorname{rank}(p(J))+\operatorname{rank}(q(J)). Notice that p(J)_{ii}=p(J_{ii}),q(J)_{ii}=q(J_{ii}), and because p(t),q(t) are relatively prime, p(x),q(x) can't be zero together, so the all together non-zero diagonal elements of p(J) and q(J) must be more than n. As the rank of an upper-triangular matrix equals its amount of non-zero diagonal elements, \operatorname{rank}(p(A))+\operatorname{rank}(q(A))\ge n.

6. Let V be a finite-dimensional vector space over a field F, and f\in\mathcal L(V). Show that V=\operatorname{Ker}(f)\oplus\operatorname{Im}(f) iff \operatorname{Im}(f^2)=\operatorname{Im}(f).

  • V=\mathrm{Ker}(f)\oplus\mathrm{Im}(f)\Rightarrow\forall x\in\mathrm{Im}(f)\exists\alpha\in\mathrm{Ker}(f),\beta=f(\gamma)\in\mathrm{Im}(f)\ni x=f(\alpha+\beta)=f^2(\gamma)\Rightarrow\mathrm{Im}(f)\subseteq\mathrm{Im}(f^2)\xRightarrow{\mathrm{Im}(f)\subseteq V\Rightarrow\mathrm{Im}(f^2)\subseteq\mathrm{Im}(f)}\mathrm{Im}(f^2)=\mathrm{Im}(f).
  • \mathrm{Im}(f^2)=\mathrm{Im}(f)\Rightarrow\forall v\in V\exists u\in V\ni fv=f^2u\xRightarrow{v-fu\in\mathrm{Ker}(f),fu\in\mathrm{Im}(f)}V=\mathrm{Ker}(f)+\mathrm{Im}(f)\xRightarrow{\dim V=\dim\mathrm{Ker}(f)+\dim\mathrm{Im}(f)}V=\operatorname{Ker}(f)\oplus\operatorname{Im}(f).

7. Let A\in M_n(F) is a nilpotent matrix. Prove that the matrix A+I_n is invertible and find its inverse.

  • Suppose A^k=0. Notice that (A+I_n)\sum_{i=0}^{k-1}(-1)^iA^i=(-1)^{k-1}A^k+I_n=I_n. Therefore A^{-1}=\sum_{i=0}^{k-1}(-1)^iA^i.

8. Let n be a positive integer, and let A=J_n(\lambda) be the Jordan block of size n\times n corresponding to the eigenvalue \lambda. Find the Jordan canonical form of A^2.

  • The eigenvalue of A is \lambda means the eigenvalue of A^2 is \lambda^2. If \lambda=0, r_i=\mathrm{rank}((A^2-0)^i)=n-2i,i\leq[\frac{n}{2}];0,i>[\frac{n}{2}]. So the Jordan canonical form of A^2 is \mathrm{diag}\{J_{[\frac{n}{2}]}(0),J_{[\frac{n-1}{2}]+1}(0)\}. Else if \lambda\neq0, \mathrm{rank}(A^2-\lambda^2)=n-1. So the Jordan canonical form of A^2 is J_n(\lambda^2).