# showing that 2 matrices are not similar

There are two $3\times 3$ matrices: $$A = \begin{bmatrix} 2 &-1 &-1\\ 0& 1 &1\\ 0 &0 &2 \end{bmatrix}$$ $$B = \begin{bmatrix} 2 &-1 &1\\ 0& 1 &1\\ 0& 0& 2 \end{bmatrix}$$ I need to show that these are not similar. They have the same determinant, rank and trace. I've tried to subtract with a matrix of the form $xI$ so that $x$ is a real number but that didn't work. Thanks in advance!

• Use MathJax! See meta.math.stackexchange.com/questions/5020/… – aGer May 11 '15 at 14:10
• I have corrected the notation for you – marco11 May 11 '15 at 14:13
• What do you mean by "similar"? – aGer May 11 '15 at 14:16
• en.wikipedia.org/wiki/Matrix_similarity – NotSure May 11 '15 at 14:17
• @5xum I corrected it before Casteels did, but he introduced his own edit, simultanously rejecting mine. – marco11 May 11 '15 at 14:21

Hint. Look at the rank of $A - 2I$ and $B- 2I$.

• I have no idea how i've missed this, thanks ! – NotSure May 11 '15 at 14:28
• In this example, can we calculate the eigenvectors of one matrix to construct a change of basis matrix and then compare it with the second matrix? – marco11 May 11 '15 at 14:29
• @marco11 We can. But there is now need to, when we start to compare the eigenspaces, we see that the $2$-eigenspace of $A$ is one-dimensional, while the $2$-eigenspace of $B$ is two-dimensional. – martini May 11 '15 at 14:32
• @martini Because in case of $A$ for eigenvalue $\lambda=2$ the eigenvector is $(1,0,0)$ and in case of $B$ for eigenvalue $\lambda=2$ the eigenvectors are $(1,0,0)$ and $(0,1,1)$. Am I right? – marco11 May 11 '15 at 14:41
• @marco11 You are right. – martini May 11 '15 at 15:04

this is what we get by row reducing the two matrices $A-2I = \begin{bmatrix} 0 &-1 &-1\\ 0& -1 &1\\ 0 &0 &0 \end{bmatrix} \to \begin{bmatrix} 0 &1 &0\\ 0& 0 &1\\ 0 &0 &0 \end{bmatrix}$ and

$B-2I = \begin{bmatrix} 0 &-1 &1\\ 0& -1 &1\\ 0 &0 &0 \end{bmatrix} \to \begin{bmatrix} 0 &1 &-1\\ 0& 0 &0\\ 0 &0 &0 \end{bmatrix}$

so that the null space of $A-2I$ has dimension one and the null space of $B-2I$ has dimension two. that is the matrices $A$ and $B$ are not similar.