Comparing two Diagonalisable matrices with identical specific features

Question

Hello, I was wondering if anyone could help me with this Linear Algebra question that I'm stuck on.

So far this is what I've worked out:

I know that if a Matrix is Diagonalisable then it's diagonal entries are the eigenvalues of the matrix.

However having only been told that the matrices have the same set of eigenvalues, I believe this means that I cannot assume that they are ordered the same in each matrix and hence the determinants may differ. So b) cannot be correct.

However for the trace, I know that the trace for a Diagonalisable matrix is the addition of all the diagonal entries. Therefore the order of the eigen values along the diagonal doesn't matter and hence c) must be correct.

For a) I have read up in my notes that $X_A(\lambda)=det(A-(\lambda)I_n)$ and if I am saying that the determinants may not equal each other then I believe that $X_A(\lambda)$ cannot equal $X_b(\lambda)$.

And therefore I think my answer should be that c) and d) are correct.

Can anyone advise me if I'm correct or not/where I'm going wrong.

Thanks in advance.

Answer 1

First, a quick response to what you've "worked out".

I know that if a Matrix is Diagonalisable then it's diagonal entries are the eigenvalues of the matrix.

This is incorrect

However having only been told that the matrices have the same set of eigenvalues, I believe this means that I cannot assume that they are ordered the same in each matrix and hence the determinants may differ. So b) cannot be correct.

Also incorrect

However for the trace, I know that the trace for a Diagonalisable matrix is the addition of all the diagonal entries. Therefore the order of the eigen values along the diagonal doesn't matter and hence c) must be correct.

There is nothing special about diagonalizable (US spelling, hope you don't mind) matrices in this regard. It is true that c is correct.

For a) I have read up in my notes that $X_A(\lambda)=det(A-(\lambda)I_n)$ and if I am saying that the determinants may not equal each other then I believe that $X_A(\lambda)$ cannot equal $X_b(\lambda)$.

Also incorrect (but consistent with your earlier statement)

And therefore I think my answer should be that c) and d) are correct.

You haven't said anything about d).

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Now, to answer the questions. a,b,c are all correct, d is not correct.

If $A,B$ are diagonalizable with the same multiplicities, then there is a single diagonal matrix $D$ for which we have $A = PDP^{-1}$ and $B = QDQ^{-1}$ for some invertible matrices $P,Q$. In other words, $A$ and $B$ are similar to the same diagonal matrix (and are therefore similar to each other).

$A$ and $B$ have the same eigenvalues (the diagonal elements of $D$). The determinant is the product of these eigenvlaues. So, $A,B$ have the same determinant.

Likewise, the trace is the sum of these eigenvalues. So, $A,B$ have the same trace.

The algebraic multiplicity of the eigenvalue $\lambda$ of $A$ is the exponent of $(x - \lambda)$ in the characteristic polynomial of $A$. Since the eigenvalues of $A,B$ have the same algebraic multiplicities, $A,B$ must have the same characteristic polynomial.

The eigenspace $E_{\lambda}$ depends on the eigenvectors, not eigenvalues. $A$ and $B$ do not necessarily have the same eigenvectors. For example, if we consider the similar matrices $$ A = \pmatrix{2&1\\1&2}, \quad B = \pmatrix{1&2\\0&3}, $$ then we see that the eigenvalues of $A,B$ are $1,3$. $E_1(A)$ is spanned by $(1,-1)$, and this is not equal to $E_1(B)$, which is spanned by $(1,0)$.