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Let $$ B = \begin{bmatrix} a & 1 & & & \\ & \ddots & \ddots & \\ & & \ddots & 1 \\ & & & a \end{bmatrix} \qquad\text{and}\qquad A = \begin{bmatrix} a & & & \\ 1 & \ddots & & \\ & \ddots & \ddots & \\ & & 1 & a \end{bmatrix}. $$ I need to prove they are similar. I was thinking using the characteristic polynomial which is as definition $p(A) = t^n - t(\operatorname{tr}A) + \det A$ and $p(B) = t^n - t(\operatorname{tr}A) + \det B$ and then both have the same trace and same determinant so their characteristic polynomial is the same so they are similar to the same matrix and hence similar to each other?

Yet I came across another solution suggesting $A ,B$ are both same transformation matrices for different bases of $\mathbb R^n$. So I’m not sure if my solution is ok.

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    $\begingroup$ That isn't the characteristic polynomial. $\endgroup$ – Lord Shark the Unknown Aug 26 '18 at 14:16
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    $\begingroup$ As transformation matrix use the one that maps $e_k$ to $e_{n-k+1}$, $k=1,\ldots,n$. $\endgroup$ – amsmath Aug 26 '18 at 14:21
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Your approach is wrong. For instance, the matrices $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right)$ have the same characteristic polynomials ($\lambda^2$). However, they are not similar.

Of course, you can always use the fact that every matrix is similar to its transpose. But, in this case, the approach that you suggest at the end of your question (same transformation, different bases) is perhaps the best one.

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  • $\begingroup$ Isn't being the same transformation in different bases the very definition of similarity? $\endgroup$ – badatmath Aug 26 '18 at 14:43
  • $\begingroup$ No. Two $n\times n$ matrices $A$ and $B$ are similar if there is an invertible $n\times n$ matrix $P$ such that $B=P^{-1}AP$. $\endgroup$ – José Carlos Santos Aug 26 '18 at 14:45
  • $\begingroup$ Which says they are the same linear transformation in different bases; $P$ being the basis change marix. $\endgroup$ – badatmath Aug 26 '18 at 14:46
  • $\begingroup$ @itry I know. Nevertheless, the definition of similarity is not the one that you mentioned. $\endgroup$ – José Carlos Santos Aug 26 '18 at 14:54
  • $\begingroup$ So can you show how to prove it? According the the solution they chose $\ T(v) = Av $ and then $\ B_0 = (e_1,...,e_n) $ and $\ B_1 = (e_n,...,e_1)$ and then $\ [T]_{B_0} = A $ and $\ [T]_{B_1} = B $ but clearly $ [T]_{B_1} \not = B $ $\endgroup$ – bm1125 Aug 26 '18 at 15:08
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Similar matrices have the same characteristic polynomial, but matrices with the same characteristic polynomial are not necessarily similar.

They will have the same spectrum but the geometric multiplicities of the Eigenvalues can be different.

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Your approach of equality of characteristic polynomials to prove similarity is true for diagonalisable matrices. Here it is 'easy' to find the permutation matrix such that $P^{t}AP=B$.

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