Linear algebra: determinants and eigenvalues

1. May 26, 2005

jhson114

i'm reading and doing some work in introduction to linear algebra fifth edition, and i came across some problems that i had no clue.

1. An (n x n) matrix A is a skew symmetric (A(transposed) = -A). Argue that an (n x n) skew-symmetrix matrix is singular when n is an odd integer.

2. Prove that if A is nonsingular, then 1/(eigenvalue symbol) is an eigenvalue of A^-1

can someone explain some of these for me. thank you

2. May 26, 2005

matt grime

The second follows from considering the characteristic polynomial (as does the first now I imagine).

3. May 26, 2005

jhson114

can you please explain it in more details?

4. May 26, 2005

Galileo

For the first, you can show that det(A) is zero. Just use what you know about determinants and A.

5. May 26, 2005

matt grime

Yep, that seems better.

But the second is the char poly. e-values are the roots of... relate the char poly of A to that of A^{-1}

6. May 26, 2005

jhson114

i dont understand why det(A) has to be zero for the skew symmetric matrix. i know that if n is odd then det(-A)=-det(A) but i dont get why it has to equal zero. and why cant it be zero if n is even??

Last edited: May 26, 2005
7. May 26, 2005

Galileo

You're almost there. You haven't used the fact that A is skew symmetric yet.

8. May 26, 2005

Muzza

Eh? Nothing written in this thread implies that a skew-symmetric matrix of even order is always invertible (consider the zero matrix...).

9. May 26, 2005

jhson114

arg.. this is confusing. so since its skew-symmetric matrix det(-A)=-det(A)=det(A(transposed)). but why must it be zero :(

10. May 26, 2005

Muzza

Forget about skew-symmetric matrices for a while. In general, how are det(A') and det(A) related (A is any square matrix, A' the transpose of A)?

11. May 26, 2005

jhson114

det(A') and det(A) are equal. so that means the two determinants are zero?? man i suck

12. May 26, 2005

Muzza

Right:

det(A') = det(A).

But as already noted, for a skew-symmetric matrix of odd order, we also have that

det(A') = -det(A).

Hence...?

13. May 26, 2005

matt grime

Hopefully that's that cleared up for that one.

How about the other: if Av=tv (where t is the eigenvalue and v is the eigenvector) then v=A^{-1}tv=tA^{-1}v

can you complete that from there?

14. May 26, 2005

jhson114

i know that det(A') = det(A) = -det(A). i think i also mentioned this in the earlier post. But i still dont get why all the determinants are zero.

oh yeah. i figured out number 2. thanks

15. May 26, 2005

matt grime

Oh let D bet Det(A), you know D, a number, is equal to minus D. How many real (or complex) numbers satisfy D=-D? Or 2D=0?

16. May 26, 2005

jhson114

ah ha! that makes more sense. and thats so simple too. thanks alot!!