## DSQ 特徵分析1

(1) If $A$ is a 10 by 10 matrix, what are the eigenvalues of $A$?

1. $A^2=I$
2. $\text{trace}A=4$

(2) Let $A$ be a $2\times 2$ matrix. What is the determinant of $A$?

1. The characteristic polynomial of $A$ is $p(t)=t^2-2t-3$.
2. It is known that $\text{trace}A=2$ and $\text{trace}(A^2)=10$.

(3) Let $A$ be a $3\times 3$ real matrix. What is the rank of $A$?

1. $A$ has two different eigenvalues $0, 1$.
2. $\text{trace}A=1$

(4) What is the determinant of $A^2+A$?

1. $A$ is known to have eigenvalues $1, 1, 2$.
2. $\det A=2$

(5) Let $A$ be a 2 by 2 matrix. Does $A$ have at least one nonzero entry?

1. $A$ is known to have eigenvalues $0, 0$.
2. $A^2=0$

(6) Let $A$ be an $n\times n$ matrix. Is $A$ diagonalizable?

1. It is known that the only eigenvalue of $A$ is $1$.
2. $A+2I$ is diagonalizable.

(7) If $-2, 1, k$ are the eigenvalues of a $3\times 3$ matrix $A$, is $A$ invertible?

1. $\text{trace}A=0$
2. $A$ is not diagonalizable.

(8) Let $A$ and $B$ be $n\times n$ matrices. Is $A=B$?

1. $A$ and $B$ have the same four fundamental subspaces.
2. $A$ and $B$ have the same eigenvalues $\lambda_1,\ldots,\lambda_n$ with the same independent eigenvectors $\mathbf{x}_1,\ldots,\mathbf{x}_n$.

(9) Let $A$ be a $2\times 2$ matrix. What is $A$?

1. The eigenvalues of $A$ are $1,3$, and the corresponding eigenvectors are $(1,1), (1,-1)$.
2. It is known that $A$ has positive eigenvalues and $A^2=\left[\!\!\begin{array}{rr} 5&-4\\ -4&5 \end{array}\!\!\right]$.

(10) Let $A$ be a square matrix and let $a$ and $b$ be two scalars. Is $N(A-aI)\neq N(A-bI)$?

1. It is known that $a$ and $b$ are different eigenvalues of $A$.
2. $a\neq b$

(1) C
$\lambda_i$$A$ 的一個特徵值，則存在非零向量 $\mathbf{x}$ 使得 $A\mathbf{x}=\lambda_i\mathbf{x}$。左乘 $A$，可得 $A^2\mathbf{x}=\lambda_i A\mathbf{x}=\lambda_i^2\mathbf{x}$，故 $\lambda_i^2$$A^2$ 的一個特徵值，對應特徵向量 $\mathbf{x}$。陳述 (1) $A^2=I$ 說明 $\lambda_i=\pm 1$$1\le i\le 10$。矩陣跡數是特徵值的和，陳述 (2) 說 $\text{trace}A=\sum_{i=1}^{10}\lambda_i=4$。如果二個陳述同時成立，設 $A$$k$ 個特徵值 $1$，則 $\text{trace}A=k-(10-k)=4$，解出 $k=7$，故 $A$ 有相重數為 $7$ 的特徵值 $1$，相重數為 $3$ 的特徵值 $-1$

(2) D
$A$ 有特徵值 $\lambda_1$$\lambda_2$，則特徵多項式為 $p(t)=(t-\lambda_1)(t-\lambda_2)=t^2-(\lambda_1+\lambda_2)t+\lambda_1\lambda_2$。因為 $\det A=\lambda_1\lambda_2$，由陳述 (1) 可知 $\det A=-3$。平方 $\text{trace}A=\lambda_1+\lambda_2$，並使用 $\text{trace}(A^2)=\lambda_1^2+\lambda_2^2$，可得 $(\text{trace}A)^2=\lambda_1^2+\lambda_2^2+2\lambda_1\lambda_2=\text{trace}(A^2)+2\det A$，解出 $\det A=\frac{1}{2}((\text{trace}A)^2-\text{trace}(A^2))$，由陳述 (2) 可得 $\det A=\frac{1}{2}(2^2-10)=-3$

(3) E
$A$ 有相異特徵值 $0, 1$$A$ 實際可能有特徵值 $0,0,1$$0,1,1$。若 $\text{trace}A=1$，表明特徵值的和等於 $1$。陳述 (1) 或 (2) 皆不足以斷定 $A$ 的行空間的維數，即 $\text{rank}A$。如果二個陳述同時成立，可知 $A$ 有特徵值 $0,0,1$，說明 $\text{rank}A\ge 1$，但仍無法確定其值。舉例來說，若 $A=\begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&0&1 \end{bmatrix}$，則 $\text{rank}A=1$；若 $A=\begin{bmatrix} 0&1&0\\ 0&0&0\\ 0&0&1 \end{bmatrix}$，則 $\text{rank}A=2$

(4) A

(5) E

(6) B

(7) D

(8) B

\displaystyle\begin{aligned} A\mathbf{x}&=\sum_{i=1}^nA(c_i\mathbf{x}_i)=\sum_{i=1}^nc_iA\mathbf{x}_i=\sum_{i=1}^nc_i\lambda_i\mathbf{x}_i\\ B\mathbf{x}&=\sum_{i=1}^nB(c_i\mathbf{x}_i)=\sum_{i=1}^nc_iB\mathbf{x}_i=\sum_{i=1}^nc_i\lambda_i\mathbf{x}_i. \end{aligned}

(9) D

$\displaystyle A=\left[\!\!\begin{array}{cr} 1&1\\ 1&-1 \end{array}\!\!\right]\begin{bmatrix} 1&0\\ 0&3 \end{bmatrix}\left[\!\!\begin{array}{cr} 1&1\\ 1&-1 \end{array}\!\!\right]^{-1}=\left[\!\!\begin{array}{rr} 2&-1\\ -1&2 \end{array}\!\!\right]$

(10) A
$a$$b$$A$ 的相異特徵值，則特徵向量 $\mathbf{x}\in N(A-aI)$$\mathbf{y}\in N(A-bI)$ 必定線性獨立，因此 $N(A-aI)\neq N(A-bI)$。陳述 (2) $a\neq b$ 不能斷定 $N(A-aI)$ 是否等於 $N(A-bI)$。例如，$A=\begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}$，當 $a=2$$b=3$$N(A-2I)=N(A-3I)=\{\mathbf{0}\}$

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