91影视

Triangularizable matrices are those that are similar to triangular matrices. In a nutshell, this is the same as stabilising a flag: upper triangular matrices are the ones that keep the standard flag, which is determined by the standard ordered basis.

Consider an A is triangulizable matrix is identical to and upper triangular matrix . If $S^{-1}AS$is an upper triangular matrix then the first column of S is an eigen vector of A. Hence any matrix without real eigen vectors fails to be triangulizable vector over R .Consider the matrix as

$\left[\begin{array}{cc}1& 2\\ -3& 1\end{array}\right]$

With real entries. To find the eigen values are as follows

$$\begin{gathered} det\left(\left[\begin{array}{cc}1& 2\\ -3& 1\end{array}\right]-\left[\begin{array}{cc}位& 0\\ 0& 位\end{array}\right]\right)=0 \\ 鈬�\left|\begin{array}{cc}1-位& 2\\ -3& 1-位\end{array}\right|=0 \\ 鈬�{位}^2-2位+7=0 \\ 鈬�\left(位+(i脳\sqrt{6}-1)\right)脳\left(位-(i脳\sqrt{6}+1)\right)=0 \\ {位}_1=-i脳\sqrt{6}+1and{位}_2=i脳\sqrt{6}+1 \end{gathered}$$

The triangulizable matrix over R since it complex eigen values are

${位}_1=-i脳\sqrt{6}+1and{位}_2=i脳\sqrt{6}+1$

Consider the given problem for any$n脳n$ matrix with complex entries is triangulizable over C .True value for (n-1) .A real eigen value of$位$ for A and eigen vector of length 1 for$位$ .For every A there exists a complex invertible matrix P whose first column is an eigen vector of A .It can be represented as below

$P^{-1}AP=\left[\begin{array}{cc}位& \stackrel{鈫�}{V}\\ 0& B\end{array}\right]$

Hypothesis of induction if B is triangulizable matrix such that there exists$(n-1)脳(n-1)$ matrix Q that satisfies

$Q^{-1}BQ=T$

Consider

$R=\left[\begin{array}{cc}1& 0\\ 0& Q\end{array}\right]$

To evaluate

$$\begin{gathered} R^{-1}=\left[\begin{array}{cc}位& \stackrel{鈫�}{V}\\ 0& B\end{array}\right]R=\left[\begin{array}{cc}1& 0\\ 0& Q^{-1}\end{array}\right]\left[\begin{array}{cc}位& \stackrel{鈫�}{V}\\ 0& B\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& Q\end{array}\right] \\ =\left[\begin{array}{cc}位& \stackrel{鈫�}{v}Q\\ 0& Q^{-1}BQ\end{array}\right] \\ 鈬�R^{-1}\left[\begin{array}{cc}位& \stackrel{鈫�}{V}\\ 0& B\end{array}\right]R=\left[\begin{array}{cc}位& \stackrel{鈫�}{V}Q\\ 0& T\end{array}\right] \end{gathered}$$

Hence$R^{-1}\left[\begin{array}{cc}位& \stackrel{鈫�}{V}\\ 0& B\end{array}\right]R$ is upper triangular matrix

Consider as$R^{-1}\left[\begin{array}{cc}位& \stackrel{鈫�}{V}\\ 0& B\end{array}\right]R$

$$\begin{gathered} R^{-1}\left[\begin{array}{cc}位& \stackrel{鈫�}{V}\\ 0& B\end{array}\right]R=R^{-1}{P}^{-1}APR \\ =\left(PR\right){)}^{-1}APR \\ (S=PR) \\ 鈬�R^{-1}\left[\begin{array}{cc}位& \stackrel{鈫�}{V}\\ 0& B\end{array}\right]R=S^{-1}AS \end{gathered}$$

Hence, we can conclude that matrix with complex entries is triangulizable over C