# Why are $\left(\begin{smallmatrix}0&1\\0&0\\\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&0\\0&1\\\end{smallmatrix}\right)$ not similar?

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 ?

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

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.

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

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.

• 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 ? – KIMKES1232 Jun 21 at 11:09
• 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. – Sudix Jun 21 at 11:15
• what is a Frobenius normal form? – KIMKES1232 Jun 21 at 11:19
• 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) – Sudix Jun 21 at 11:32

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 $$A^{2}=S^{-1}B^{2}S$$.

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.