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Why are the matrices $\begin{pmatrix}0 & 1 \\0 & 0 \\\end{pmatrix}$ and $\begin{pmatrix}0 & 0 \\0 & 1 \\\end{pmatrix}$ not similar?

I can do a row operation. But on the other hand they don't have the same characteristic polynomial. How does it relates to: $A$ is similar to $B$ if and only if $A$ and $B$ have the same canonical form ?

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    $\begingroup$ Equivalent in what sense? They are certainly row equivalent (which forms an equivalence relation). $\endgroup$ – Sean Roberson Jun 21 at 10:16
  • $\begingroup$ What do you mean with equivalent? If you mean similarity, then the characteristic polynomial argument suffices. $\endgroup$ – Wojowu Jun 21 at 10:16
  • $\begingroup$ They are clearly not the same. As to whether they are equivalent depends on what equivalent mean here $\endgroup$ – Henry Jun 21 at 10:17
  • $\begingroup$ I meant similar, I changes it. thanks!!! $\endgroup$ – KIMKES1232 Jun 21 at 10:53
  • $\begingroup$ Obviously the matrices have different eigenvalues. $\endgroup$ – Michael Hoppe Jun 21 at 11:59
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The question seems to operate under the assumption that matrices which differ by elementary row operations should have the same characteristic polynomial.

This assumption is false (your matrices provide a counterexample). Elementary row operations preserve the row space and the kernel of a matrix, but not the eigenvectors.

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  • $\begingroup$ thats exactly what I was missing, thanks!!! $\endgroup$ – KIMKES1232 Jun 21 at 10:53
  • $\begingroup$ so it is not true to say: A is similar to B if and only if A and B have the same canonical form ? $\endgroup$ – KIMKES1232 Jun 21 at 11:03
  • $\begingroup$ @KIMKES1232 Reading your first comment under this answer makes me wonder why you did not upvote it? $\endgroup$ – Hanno Jun 21 at 15:10
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What you're looking for is matrix similarity, not matrix equivalence.

Matrix equivalence merely says:

Are two matrices equivalent, you can transform them into each other, multiplying them with nothing but regular matrices. This is true exactly iff they have same rank.

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  • $\begingroup$ so why it is not true to say: A is similar to B if and only if A and B have the same canonical form ? $\endgroup$ – KIMKES1232 Jun 21 at 11:09
  • $\begingroup$ You can deduce similarity by using a canonical form, just not one that is as simple as yours. Two matrices are similar iff they have the same Frobenius normal form. $\endgroup$ – Sudix Jun 21 at 11:15
  • $\begingroup$ what is a Frobenius normal form? $\endgroup$ – KIMKES1232 Jun 21 at 11:19
  • $\begingroup$ That's a rather involved question (for which I frankly lack some knowledge). An easier canonical form that is equal to similarity is this: Two matrices are similar, if their Jordan Normal Form is, up to a permutation of blocks, identical. (The Jordan Normal Form then again is a generalization of Diagonalization) $\endgroup$ – Sudix Jun 21 at 11:32
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The first matrix, $A$, satisfies $A^{2}=0$. For the second one, $B$, we see that its square does not equal $0$. This implies that we cannot have $A=S^{-1}BS$ for any $S$ because we would then have $A^{2}=S^{-1}B^{2}S$.

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Hint: Suppose they were equivalent then there had to exists a invertible matrix $S := \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ such that \begin{equation} \tag{1} S^{-1} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} S = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \end{equation} holds. We have $ S^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. $ Therefore, $(1)$ becomes \begin{align} & \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\ \iff & \frac{1}{ad - bc} \begin{pmatrix} c d & d^2 \\ -c^2 & -c d \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \end{align} This yields $cd = 0$ and $-cd = 1$, which clearly is a contradiction.

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