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 $$ B = P^{-1}AP \quad?$$
Best Answer
try thinking of a $P^{-1}$ which puts the rows in "reverse order", and then show that right-multiplication by $P$ puts "the columns in reverse order".