How to prove matrix similarity?

Question

I couldn't find the answer to this but, is it possible to prove 2 matrices are similar by their properties ("if and only if" - works both way)?

i.e – how do u prove those matrices are similar

$$A = \begin{pmatrix} 0 & 1 & -2\cr 0 & 1 & 5 \cr 0 & 0 & 4 \end{pmatrix} , \qquad B = \begin{pmatrix} 1 & 0 & 0\cr -1 & 4 & 0 \cr 3 & 0 & 0 \end{pmatrix}$$

if the only way to solve this is by using $$A=P^{-1}BP$$

I'd love to see how to solve this (finding a possible P) EDIT: What I tried it putting both of them in the same matrix and starting doing eliminations but I'm not sure whats the way to go

thanks in advance

Answer 1

Both matrices have the same distinct eigenvalues and so have a basis of eigenvectors.

We have $A u_0 = 0, A u_1 = u_1, A u_4 = 4 u_4$ where $u_0,u_1,u_4$ are a basis for $\mathbb{R}^3$.

Similarly we have $v_0,v_1,v_4$ for $B$.

Let $U = \begin{bmatrix} u_0 &u_1 &u_4 \end{bmatrix}$ and similarly for $V$.

Then we have $U^{-1}AU = V^{-1} B V$ or $A = (V U^{-1})^{-1} B (V U^{-1})$.

Answer 2

Along the lines of the first comment:

The matrices $A$ and $B$ are diagonalizable. In other words, there exist matrices $P_1$ and $P_2$ such that $$ P_1^{-1}A P_1 = \pmatrix{0&0&0\\0&1&0\\0&0&4}, \qquad P_2^{-1}B P_2 = \pmatrix{0&0&0\\0&1&0\\0&0&4}. $$ It follows that $$ P_1^{-1}AP_1 = P_2^{-1}BP_2 \implies\\ (P_1P_1^{-1}) A( P_1 P_1^{-1}) = P_1(P_2^{-1}BP_2)P_1^{-1} \implies\\ P_1(P_1^{-1} A P_1 )P_1^{-1} = (P_1P_2^{-1})B(P_2P_1^{-1}) \implies\\ A = (P_2P_1^{-1})^{-1}B(P_2P_1^{-1}). $$ So, the desired result holds if we take $P = P_2P_1^{-1}$.

Answer 3

Let $D$ be the diagonal matrix $\operatorname{diag}(0,1,4)$. You may try to prove that $X^{-1}AX=D=Y^{-1}BY$ for some matrices $X$ and $Y$. When this is done, we have $A=P^{-1}BP$ where $P=YX^{-1}$.

Let $X=[x_1,x_2,x_3]$. The equation $X^{-1}AX=D$ implies that $AX=XD$, i.e. $Ax_1=0,\,Ax_2=x_2$ and $Ax_3=4x_3$. The first two equations can be solved by inspection: just set $x_1=(1,0,0)^T$ and $x_2=(1,1,0)^T$. The last one isn't hard and I'll leave it to you. (But remember, we need $X$ to be invertible. So, you cannot take $x_3$ as the trivial solution $0$.)

Similarly, let $Y=[y_1,y_2,y_3]$. Again, we need $By_1=0,\,By_2=y_2$ and $By_3=4y_3$. The first and the third equations are easy: just set $y_1=(0,0,1)^T$ and $y_2=(0,1,0)^T$. You may continue from here.