If Matrix A, B are similar,Find the invertible matrix P such that $B=PAP^{-1}$

Question

Given Matrix $A \sim B$ , If Matrix A, B are similar,Find the invertible matrix P such that $B=PAP^{-1}$ $$ A = \begin{bmatrix} -2 & -2 & 1 \\ 2 & 3 & -2 \\ 0 & 0& -2 \\ \end{bmatrix},\qquad\qquad B = \begin{bmatrix} 2 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \\ \end{bmatrix} $$ Find the invertible matrix P such that $B=PAP^{-1}$ Seek the guidance of the great God,Best is a smarter solution

Answer 1

Here $A = \begin{bmatrix} -2 & -2 & 1 \\ 2 & 3 & -2 \\ 0 & 0& -2 \\ \end{bmatrix}$

Eigenvalues of $A$ are $\lambda=-2,-1,2$.

Now for eigenvectors $X$, we have to solve $(A-\lambda I)X=0$

For $\lambda = -2$, $\qquad(A+2 I)X=0\implies \begin{bmatrix} 0 & -2 & 1 \\ 2 & 5 & -2 \\ 0 & 0& 0 \\ \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\implies \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} -1 \\ 2 \\ 4 \end{bmatrix}$

For $\lambda=-1$, $\qquad(A+I)X=0\implies \begin{bmatrix} -1 & -2 & 1 \\ 2 & 4 & -2 \\ 0 & 0& -1 \\ \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\implies \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}$

For $\lambda=2$, $ \qquad(A-2 I)X=0\implies \begin{bmatrix} -4 & -2 & 1 \\ 2 & 1 & -2 \\ 0 & 0& -4 \\ \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\implies \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix}$

The $P$ matrix is just the matrix of eigenvectors of $A$.

Thus $P=\begin{bmatrix} -1 & -2 & -1 \\ 2 & 1 & 2 \\ 4 & 0& 0 \\ \end{bmatrix}$

Now $P^{-1}=-\frac{1}{12}\begin{bmatrix} 0 & 0 & -3 \\ 8 & 4 & 0 \\ -4 & -8& 3 \\ \end{bmatrix}=\begin{bmatrix} 0 & 0 & 1 \\ 2 & 1 & 0 \\ -4 & -8& 3 \\ \end{bmatrix}$

Therefore $PAP^{-1}=\begin{bmatrix} -1 & -2 & -1 \\ 2 & 1 & 2 \\ 4 & 0& 0 \\ \end{bmatrix}\begin{bmatrix} -2 & -2 & 1 \\ 2 & 3 & -2 \\ 0 & 0& -2 \\ \end{bmatrix}\begin{bmatrix} 0 & 0 & 1 \\ 2 & 1 & 0 \\ -4 & -8& 3 \\ \end{bmatrix}$

$=\begin{bmatrix} -28 & -44 & 13 \\ 14 & 31 & -14 \\ -32 & -40 & 4 \\ \end{bmatrix}$

After some operations, $$\begin{bmatrix} -28 & -44 & 13 \\ 14 & 31 & -14 \\ -32 & -40 & 4 \\ \end{bmatrix}\equiv\begin{bmatrix} 2 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{bmatrix}=B$$