几道英文数学题，会的帮忙解一下啊。。。 这次分开寻求帮助-5/6

Theorem 6.1 (The Fundamental Theoremof Algebra by Gauss (1779) Consider a polynomial of degree n in z shownbelow.

Where ∈ℂ. Then,equation (∗) has n solutions with the property that for each i=1,…,n. Thisincludes the case in which for some i≠j.

Exercise Find the eigenvalues and the associatedeigenvectors of the matrices, A and B.

In fact, it is convenient to write the characteristic function as apolynomial in −λ:

The zeros of this characteristic polynomial are precisely the eigenvaluesof A.

图片格式题目：

参考：

希望楼主早日解答。

顶~~~

$|\lambda I_{2}-A|=\begin{vmatrix}\lambda-1&-2\\-3&\lambda\end{vmatrix}=\lambda^2-\lambda-6=(\lambda-3)(\lambda+2)$，所以A有两个特征值 3，-2。
那么特征值3对应的特征向量，要解齐次线性方程 $(3I_2-A)x=0 \iff \begin{pmatrix}2&-2\\-3&3\end{pmatrix}x=0\iff \begin{pmatrix}1&-1\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=0\iff x_1=x_2=t $
令t=1，去特征向量 $\begin{pmatrix}1\\1\end{pmatrix}$，有$A\begin{pmatrix}1\\1\end{pmatrix}=3\begin{pmatrix}1\\1\end{pmatrix}$
。
同理求特征值3对应的特征向量，要解齐次线性方程 $(-2I_2-A)x=0 \iff \begin{pmatrix}-3&-2\\-3&-2\end{pmatrix}x=0\iff \begin{pmatrix}3&2\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=0\iff 3x_1=-2x_2=6t $
令t=1，取特征向量 $\begin{pmatrix}2\\-3\end{pmatrix}$，有$A\begin{pmatrix}2\\-3\end{pmatrix}=-2\begin{pmatrix}2\\-3\end{pmatrix}$

B请你模仿自己求吧

北固隐 发表于 2018-10-15 22:32
$|\lambda I_{2}-A|=\begin{vmatrix}\lambda-1&-2\\-3&\lambda\end{vmatrix}=\lambda^2-\lambda-6=(\lambda ...

非常感谢，其他不会的题目我再找找资料。