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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.

José Carlos Santos
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bm1125
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3 Answers3

3

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.

2

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.

José Carlos Santos
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  • Isn't being the same transformation in different bases the very definition of similarity? –  Aug 26 '18 at 14:43
  • No. Two $n\times n$ matrices $A$ and $B$ are similar if there is an invertible $n\times n$ matrix $P$ such that $B=P^{-1}AP$. – José Carlos Santos Aug 26 '18 at 14:45
  • Which says they are the same linear transformation in different bases; $P$ being the basis change marix. –  Aug 26 '18 at 14:46
  • @itry I know. Nevertheless, the definition of similarity is not the one that you mentioned. – José Carlos Santos Aug 26 '18 at 14:54
  • 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 $ – bm1125 Aug 26 '18 at 15:08
  • @bm1125 Why do you say that? If the matrix of a linear transformation with respect to a basis $(e_1,\ldots,e_n)$ is $A$, the the matrix of the same linear transfrmation with respect to the basis $(e_n,\ldots,e_1)$ is $B$. – José Carlos Santos Aug 26 '18 at 15:27
  • Because we defined $\ T(v) = Av $ so $\ [T]_{B_1} = \begin{matrix} 0 & 0 & ... & a \ 0 & ... & a & 1 \ ... \ a & 0 & ... & 0 \end{matrix} $ – bm1125 Aug 26 '18 at 17:13
  • @bm1125 Not at all. For instance, $T.e_1=ae_1$. Therefore, since $e_1$ is the last vector of $B_1$, the last column of $[T]_{B_1}$ must be$$\begin{array}{c}0\0\\vdots\0\ a.\end{array}$$ – José Carlos Santos Aug 26 '18 at 17:19
  • I don't understand. We set $\ T(v) $ as $\ Av $ so for every vector $\ x $ we choose the transformation should just multiplying $\ A \cdot x $. Therefore the transformation for $\ e_1 $ will result $\ T e_1 = a\cdot e_1 + 1\cdot e_2 = [a,1,0,...0]^T $ ?? if it will be the last vector of $\ B_1 $ it should be exactly as I said. – bm1125 Aug 26 '18 at 17:32
  • @bm1125 Where did you get that $Te_1=ae_1+e_2$? Look at the first column of $A$! It just an $a$ followed by $0$s. Therefore, $Te_1$ is just $ae_1$. – José Carlos Santos Aug 26 '18 at 17:38
  • Sorry, Fixed the question. Look at the matrices again. – bm1125 Aug 26 '18 at 17:44
  • @bm1125 They are the same matrices; you just exchanged their names. Therefore, my answer remains the same as before. – José Carlos Santos Aug 26 '18 at 17:48
1

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 $P^{t}AP=B$.

Toni Mhax
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