# Thread: Eigenvalues and Eigenvectors, PLEASE

1. ## Eigenvalues and Eigenvectors, PLEASE

a)Show that for any square matrix A, A^t (A transpose) and A have the same characteristic polynomial and hence the same eigenvalues.

b)Give an example of a 2x2 matrix A for which A^t and A have different eigenspaces.

Thank you.

2. Originally Posted by somestudent2
a)Show that for any square matrix A, A^t (A transpose) and A have the same characteristic polynomial and hence the same eigenvalues.

b)Give an example of a 2x2 matrix A for which A^t and A have different eigenspaces.

Thank you.
a) You should think about $\text{det}(A - \lambda I)$ and $\text{det}(A^T - \lambda I)$. Are they always the same?

b) Shouldn't be too hard to play around with a couple of 2x2 matrices, see what happens with them ......

3. would that be sufficient enough to say that since det(A) = det(A^t)

then det(A-xI)=det(A^t-xI) since we pretty much subtract the same number, I also considered this solution except it seemed too short to me to be true?

thanks again

4. Originally Posted by somestudent2
would that be sufficient enough to say that since det(A) = det(A^t) Mr F says: No! $det(A + B) \neq det(A) + det(B).$

then det(A-xI)=det(A^t-xI) since we pretty much subtract the same number, I also considered this solution except it seemed too short to me to be true?

thanks again
..

5. that is not what I meant mr F, I wrote that det(A)=det(A^t).....

6. Originally Posted by somestudent2
would that be sufficient enough to say that since det(A) = det(A^t)

then det(A-xI)=det(A^t-xI) since we pretty much subtract the same number, I also considered this solution except it seemed too short to me to be true?

thanks again
"then det(A-xI)=det(A^t-xI) since we pretty much subtract the same number .." is waaaay too vague ... you need to show, using a mathematical argument, how det(A) = det(A^t) leads to det(A-xI)=det(A^t-xI) ......

7. Originally Posted by somestudent2
would that be sufficient enough to say that since det(A) = det(A^t)

then det(A-xI)=det(A^t-xI) since we pretty much subtract the same number, I also considered this solution except it seemed too short to me to be true?

thanks again
Note 1: $(A - \lambda I)^T = A^T - \lambda I^T = A^T - \lambda I$.

Note 2: $\text{det} (A - \lambda I) = \text{det} (A - \lambda I)^T$ using $\text{det} B = \text{det} B^T$.

But $(A - \lambda I)^T = A^T - \lambda I$.

Therefore $\text{det} (A - \lambda I)^T = \text{det} (A^T - \lambda I)$.

Therefore ......