I have the following two matrices:
A = \begin{pmatrix}1&0&...&0&0\\ 0&2&...&0&0\\ 0&0&...&n-1&0 \\ 0&0&...&0&n\end{pmatrix} B = \begin{pmatrix}n&0&...&0&0\\ 0&n-1&...&0&0\\ 0&0&...&2&0 \\ 0&0&...&0&1\end{pmatrix}
I am asked if A is similar to B and if so then find a P such that $ B = P^{-1}AP .$ If the two matrices are not similar than I should justify it.
So far I think the two matrices are similar as they have:
- the same eigenvalues (just reversed) $ A: \lambda1 = 1, \lambda2 = 2, \lambda3 = 3, ..., \lambda n = n $ $ A: \lambda1 = n, \lambda2 = n-1, \lambda3 = n-2, ..., \lambda n = 1 $
- the same rank $ rank(A) = n = rank(B) $
- the same determinant as both matrices are diagonal so the product of the diagonal is the determinant. Since A and B both have the same elements (in opposing order) than their product is the same $ det(A) = det(B) = 1 * 2 * 3 * ... * (n-1) * n $
- the same characteristic polynomial (with the same argument as the determinant). $ A: (\lambda - 1)(\lambda -2) ... (\lambda - n - 1)(\lambda - n) $ $ B: (\lambda - n)(\lambda - n - 1)...(\lambda - 2)(\lambda - 1) $
Considering the above I have concluded that these two matrices are similar and thus I am trying to determine P.
For A using the characteristic polynomial: $ (\lambda - 1)(\lambda -2) ... (\lambda - n - 1)(\lambda - n)$
I can go on to determine that P should be \begin{pmatrix}1&0&...&0&0\\ 0&2&...&0&0\\ 0&0&...&n-1&0 \\ 0&0&...&0&n\end{pmatrix}
Yet through B I get a P which is \begin{pmatrix}n&0&...&0&0\\ 0&n-1&...&0&0\\ 0&0&...&2&0 \\ 0&0&...&0&1\end{pmatrix}
These two $\mathbf{P}$s are not the same and I know are not right.
Where am I going wrong in trying to find a P such that $ B = P^{-1}AP \quad?$