Matrices

• Oct 25th 2006, 02:02 AM
Ideasman
Matrices
Doing some last minute homework, and stuck on this one. Although its due in 4 hours thought its worth a shot.

Our assignment is to "Use a property of determinants to show that A and A^T have the same characteristic polynomial."

I know that (A - LI)x = 0, where L = lambda.

Not sure how to prove that with something like

det(A-LI) = 0? No idea.

Also, one other: Show that if A and B are similar, then det A = det B
• Oct 25th 2006, 02:58 AM
CaptainBlack
Quote:

Originally Posted by Ideasman
Doing some last minute homework, and stuck on this one. Although its due in 4 hours thought its worth a shot.

Our assignment is to "Use a property of determinants to show that A and A^T have the same characteristic polynomial."

I know that (A - LI)x = 0, where L = lambda.

Not sure how to prove that with something like

det(A-LI) = 0? No idea.

Also, one other: Show that if A and B are similar, then det A = det B

Put B=A-LI, now det(B)=det(B'), so:

det(B')=det(A'+LI')=det(A'+LI)=det(B)=det(A+LI)

So the charateristic polynomial of A' (which is det(A'+LI)) is equal to
det(B'), which is equal to det(B) which is the charateristic polynomial of A
(that is: det(A-LI)).

RonL