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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!

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  • $\begingroup$ Use MathJax! See meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – aGer
    Commented May 11, 2015 at 14:10
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    $\begingroup$ I have corrected the notation for you $\endgroup$
    – marco11
    Commented May 11, 2015 at 14:13
  • $\begingroup$ What do you mean by "similar"? $\endgroup$
    – aGer
    Commented May 11, 2015 at 14:16
  • $\begingroup$ en.wikipedia.org/wiki/Matrix_similarity $\endgroup$
    – NotSure
    Commented May 11, 2015 at 14:17
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    $\begingroup$ @5xum I corrected it before Casteels did, but he introduced his own edit, simultanously rejecting mine. $\endgroup$
    – marco11
    Commented May 11, 2015 at 14:21

2 Answers 2

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

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  • $\begingroup$ I have no idea how i've missed this, thanks ! $\endgroup$
    – NotSure
    Commented May 11, 2015 at 14:28
  • $\begingroup$ 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? $\endgroup$
    – marco11
    Commented May 11, 2015 at 14:29
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    $\begingroup$ @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. $\endgroup$
    – martini
    Commented May 11, 2015 at 14:32
  • $\begingroup$ @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? $\endgroup$
    – marco11
    Commented May 11, 2015 at 14:41
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    $\begingroup$ @marco11 You are right. $\endgroup$
    – martini
    Commented May 11, 2015 at 15:04
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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.

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