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I am trying to find out whether $A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \\ \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 1 \\ -2 & 1 \\ \end{bmatrix}$ are similar. Both matrices have the same trace, same rank, same determinant and the same eigenvalues. So I think they are similar. The eigenvalues are $3, 2$. Now I'm not sure how to find if they are indeed similar.

I am trying to find if there is a basis for $\mathbb{R^2}$ that contains only eigenvectors but when trying to solve $(3 I - A)v = 0$ there is no solution... What should I do from here?

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Similarity is an equivalence relation (prove it). So if $A$ and $B$ are similar to the same matrix, then they are similar.

Since the eigenvalues of $A$ and $B$ are the same (and distinct), they both are similar to $$ \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} $$ so they're similar. No other computation is needed.

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  • $\begingroup$ Why are they both similar to this matrix? I think I need to show that the eigenvectors are a valid basis for $M_[x]^{\mathbb{R^2}}$ first $\endgroup$ – Joe928 May 14 '16 at 13:23
  • $\begingroup$ @Joe928 Both matrices have $3$ and $2$ as eigenvalues, so they're diagonalizable; the order of eigenvalues along the diagonal is immaterial, because it just corresponds to a permutation of the basis of eigenvectors. $\endgroup$ – egreg May 14 '16 at 13:25
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Of course there is a solution for $(3I-A)v=0$: all vectors $(x,0) $.

In general, if an $n\times n $ matrix has $n $ distinct eigenvalues, then it is diagonalizable, so your matrices are indeed similar.

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Sketch:

(1) Find the eigenvectors for each of the matrices.

(2) Consider the map $M$ that takes the eigenvectors for $A$ to the eigenvectors for $B$ with corresponding eigenvalues.

(3) Consider $M^{-1}BM$.

Why does this work: Let $\lambda_1\not=\lambda_2$ be the eigenvalues of $A$ and $B$. Let $v_1,v_2$ be the (corresponding) eigenvectors of $A$ and $w_1,w_2$ be the (corresponding) eigenvectors of $B$. Then $Av_i=\lambda_iv_i$. On the other hand, $M^{-1}BMv_i=M^{-1}Bw_i=M^{-1}\lambda_iw_i=\lambda_iM^{-1}w_i=\lambda_iv_i$. Basically $A$ and $B$ act the same way on the corresponding eigenvectors, so by mapping one to the other, they act the same way.

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If $A=PDP^{-1}$, and $B=QDQ^{-1}$, where $D$ is the diagonal matrix with diagonal entries $2$ and $3$, then $A=PQ^{-1}BQP^{-1}$, and you have done.

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