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Given

$A=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$ &

$B=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$

Show that $A$ & $B$ are similar matrices.

I just know these two matrices are nilpotent matrices of index 3. How do I show them similar? Is there is any easy way? I know by the definition of similarity that I had to find matrix $P$ such that $B = P^{-1}AP$. But I am unable to find it. Please help me.

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  • 2
    $\begingroup$ A matrix is always similar to its transpose. $\endgroup$ – Lord Shark the Unknown Sep 11 '17 at 14:32
  • $\begingroup$ Maybe helps to rewrite the equation to $PB=AP$ $\endgroup$ – mathreadler Sep 11 '17 at 14:33
  • $\begingroup$ @mechanodroid : "Two matrices are equivalent if and ..." $\endgroup$ – jibounet Sep 11 '17 at 14:45
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You can explicitly show that $A$ and $B$ are similar. Let $a$ be the linear map from $\mathbb{R}^3$ to itself which is represented, into the canonical basis $(e_1, e_2, e_3)$, by the matrix $A$. Then we have :

$$ a(e_1) = 0, \quad a(e_2) = e_1, \quad a(e_3) = e_2. $$

Showing that $A$ is similar to $B$ is equivalent to finding another basis, say $(\varepsilon_1, \varepsilon_2, \varepsilon_3)$, of $\mathbb{R}^3$ such that $a$ is represented in this basis by the matrix $B$. So, we would like to have : $a(\varepsilon_1) = \varepsilon_2$, $a(\varepsilon_2) = \varepsilon_3$ and $a(\varepsilon_3) = 0$.

It follows that a possible choice is :

$$ \varepsilon_1 = e_3, \quad \varepsilon_2 = e_2, \quad \varepsilon_3 = e_1. $$

Here, the matrix $P$ is simply :

$$ P = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}. $$

By definition, the $j$-th row $(1 \leq j \leq 3)$ of $P$ contains the coordinates of the vector $\varepsilon_j$ in the basis $(e_1, e_2, e_3)$. Note that in this case, the change of basis matrix $P$ (invertible) such that $AP = PB$ is a permutation matrix.

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  • $\begingroup$ Thanks this make me more clear. $\endgroup$ – Akash Patalwanshi Sep 11 '17 at 14:45
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$$\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix} = \begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}^{-1}\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix} \begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}$$

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Any matrix $P$ of the following form satisfies this equation.

$$ \begin{bmatrix} x & y & z \\ y & z & 0 \\ z & 0 & 0 \\ \end{bmatrix} $$

where $x,y,z \in\Bbb R$

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