20211115 any n-order square matrix is similar to triangular matrix (upper triangle or lower triangle)

set up $\mathbf{A}$ by $n$ Order matrix , Its characteristic polynomial is
$$\phi (\lambda )=\det (\lambda \mathbf{I}-\mathbf{A})=\left(\lambda -{\lambda }_1\right)\left(\lambda -{\lambda }_2\right)\cdots \left(\lambda -{\lambda }_n\right)$$
For the order of the matrix $n$ Prove it by induction .
When $n=1$ when , The theorem clearly holds . Hypothetical pair $n-1$ The order matrix theorem holds , by The theorem is proved to be correct $n$ Order matrix also holds , set up ${\mathbf{x}}_1,{\mathbf{x}}_2,\cdots ,{\mathbf{x}}_n$ yes $n$ A linear independent The column vector ( Not all of them are eigenvectors ), among ${\mathbf{x}}_1$ It belongs to $\mathbf{A}$ The eigenvalues of the ${\lambda }_1$ Of Eigenvector , namely ${\mathbf{A}\mathbf{x}}_1={\lambda }_1{\mathbf{x}}_1$. remember
$${\mathbf{P}}_1=\left({\mathbf{x}}_1,{\mathbf{x}}_2,\cdots ,{\mathbf{x}}_n\right)$$
therefore
$${\mathbf{A}\mathbf{P}}_1=\left({\mathbf{A}\mathbf{x}}_1,{\mathbf{A}\mathbf{x}}_2,\cdots ,{\mathbf{A}\mathbf{x}}_n\right)=\left({\lambda }_1{\mathbf{x}}_1,{\mathbf{A}\mathbf{x}}_2,\cdots ,{\mathbf{A}\mathbf{x}}_n\right)$$
because ${\mathbf{A}\mathbf{x}}_i\in {\mathbf{C}}^n$, therefore ${\mathbf{A}\mathbf{x}}_i$ May by ${\mathbf{C}}^n$ Base ${\mathbf{x}}_1,{\mathbf{x}}_2,\cdots ,{\mathbf{x}}_n$ Uniquely linear Express , That is to say
$$\mathbf{A}{\mathbf{x}}_i=b_{1i}{\mathbf{x}}_1+b_{2i}{\mathbf{x}}_2+\cdots +b_{ni}{\mathbf{x}}_n(i=2,3,\cdots ,n)$$
therefore
$$\begin{array}{r}{\mathbf{A}\mathbf{P}}_1=\left({\lambda }_1{\mathbf{x}}_1,\mathbf{A}{\mathbf{x}}_2,\cdots ,\mathbf{A}{\mathbf{x}}_n\right)={\mathbf{P}}_1\left[\begin{array}{cccc}{\lambda }_1& b_{12}& \cdots & b_{1n}\\ 0& b_{22}& \cdots & b_{2n}\\ ⋮& ⋮& & ⋮\\ 0& b_{n2}& \cdots & b_{nn}\end{array}\right]\end{array}$$
namely
$${\mathbf{P}}_1^{-1}{\mathbf{A}\mathbf{P}}_1=\left[\begin{array}{cccc}{\lambda }_1& b_{12}& \cdots & b_{1n}\\ 0& & & \\ ⋮& & {\mathbf{A}}_1& \\ 0& & & \end{array}\right]$$
According to the theorem $1.14$, Can be set $n-1$ Order matrix ${\mathbf{A}}_1$ The characteristic polynomial of is ${\phi }_1(\lambda )=\det \left(\lambda \mathbf{I}-{\mathbf{A}}_1\right)=\left(\lambda -{\lambda }_2\right)\left(\lambda -{\lambda }_3\right)\cdots \left(\lambda -{\lambda }_n\right)$ Then assume by induction , Yes
$${\mathbf{Q}}^{-1}{\mathbf{A}}_1\mathbf{Q}=\left[\begin{array}{lll}{\lambda }_2& & \ast \\ & \ddots & \\ & & {\lambda }_n\end{array}\right]$$
remember
$${\mathbf{P}}_2=\left[\begin{array}{cc}1& 0^{\mathrm{T}}\\ 0& \mathbf{Q}\end{array}\right],\mathbf{P}={\mathbf{P}}_1{\mathbf{P}}_2$$
Then there are
$${\mathbf{P}}^{-1}\mathbf{A}\mathbf{P}={\left({\mathbf{P}}_1{\mathbf{P}}_2\right)}^{-1}\mathbf{A}\left({\mathbf{P}}_1{\mathbf{P}}_2\right)={\mathbf{P}}_2^{-1}\left({\mathbf{P}}_1^{-1}\mathbf{A}{\mathbf{P}}_1\right){\mathbf{P}}_2=$$
$${\mathbf{P}}_2^{-1}\left[\begin{array}{cccc}{\lambda }_1& b_{12}& \cdots & b_{1n}\\ \cdots & & & \\ ⋮& & {\mathbf{A}}_1& \\ 0& & & \end{array}\right]{\mathbf{P}}_2=\left[\begin{array}{cccc}{\lambda }_1& \ast & \cdots & \ast \\ & {\lambda }_2& \ddots & ⋮\\ & & \ddots & \ast \\ & & & {\lambda }_n\end{array}\right]$$

Empathy , It can be concluded that any n Order square matrix and triangular matrix （ Lower triangle ） be similar .

If for $\mathbf{A}$ , set up ${\mathbf{x}}_1,{\mathbf{x}}_2,\cdots ,{\mathbf{x}}_n$ yes $n$ A linear independent The column vector ( Not all of them are eigenvectors ), among ${\mathbf{x}}_n$ It belongs to $\mathbf{A}$ The eigenvalues of the ${\lambda }_n$ Of Eigenvector , namely ${\mathbf{A}\mathbf{x}}_n={\lambda }_n{\mathbf{x}}_n$

reference ： Matrix theory , Cheng Yunpeng