linear algebra: determinants and eigenvalues

[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

[matt grime]

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

[jhson114]

can you please explain it in more details?

[Galileo]

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

[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}

[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??

[Galileo]

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

[Muzza]

and why cant it be zero if n is even??


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

[jhson114]

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

[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)?

[jhson114]

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

[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...?

[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?

[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

[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?

[jhson114]

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