• Jan 22nd 2008, 01:26 PM
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.(Whew)
• Jan 22nd 2008, 05:34 PM
mr fantastic
Quote:

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.(Whew)

a) You should think about $\displaystyle \text{det}(A - \lambda I)$ and $\displaystyle \text{det}(A^T - \lambda I)$. 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 ......
• Jan 22nd 2008, 06:07 PM
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
• Jan 22nd 2008, 08:22 PM
mr fantastic
Quote:

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

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

..
• Jan 22nd 2008, 09:15 PM
somestudent2
that is not what I meant mr F, I wrote that det(A)=det(A^t).....
• Jan 22nd 2008, 10:38 PM
mr fantastic
Quote:

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) ......
• Jan 23rd 2008, 01:27 AM
mr fantastic
Quote:

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

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

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

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

Therefore ......