Skew symetric Matrix Proof

[ed84c]

M = -M^t where M is 3*3 prove that

det(M - kI) = -det(M+kI)

 

Ok for starters its a square matrix

 

=> M^t = M

 

=> M = -M for our matrix

 

So to get to our proof my solution would be;

 

det(M) = det(-M)

det(M)(I-kM^-1) = det(M)(I-kM^-1)

det(M - k I) = det(-M + k I)

 

 

Unfortunately as you can see this doesn't give the answer...

any ideas?

[timo]

I^t = I.

[ed84c]

I cant see how to use that?

 

M = -M^t where M is 3*3 prove that

det(M - kI) = -det(M+kI)

 

Ok for starters its a square matrix

 

=> M^t = M

 

=> M = -M for our matrix

 

So to get to our proof my solution would be;

 

det(M) = det(-M)

det(M)(I-kM^-1) = det(-M)(I-kM^-1)

det(M - k I) = det(-M + k I)

 

 

Unfortunately as you can see this doesn't give the answer...

any ideas?

 

btw it should look like that

[timo]

Your first step is wrong, already. M^t = -M, not M.

[ed84c]

So your suggesting I put (M - k I) to the power of T?

 

And then take into account that;

 

-M = M^t

and

I = I^t?

[timo]

Yes, but it's not taking something to the power of t but taking the transposed. If M^t = -M and I^t = I, what is (M-kI)^t ?

[ed84c]

M^t + (kI)^t

= -M + I^t k^T

= -M + Ik^t

 

Its wierd that the text book doesnt seem to have explained any of the ^t notation or rules thereof. Maybe it was assumed knowledge from FP1 but I can't remember it.

[timo]

I wouldn't assume anyone knowing what a matrix and a determinant is not knowing what the transpose of a matrix is, either. And I've never seen a different notation for the transpose, either. So yes, I strongly assume it was assumed knowledge. Anyways,

- your transformation is correct. The k^t is redundant. The transpose of a scalar is, depending on point of view, either not defined (in which case you wouldn't get the k^t in the first place but just k) or equals k. Either way, you end up with -M + kI.

- I later edited my post to ask what (M-kI)^t is, rather than (M+kI)^t. Of course, figuring that out goes exactly the same way as (M+kI)^t (alternatively, you could just replace k with -k). The reason why I later edited it to (M-kI)^t of course it that (M-kI) is the expression that appears on the left-hand side of your original question.

- Since you didn't know the notation of the transpose: Do you know what det(...) is and do you also know some calculation rules for it (you will need two of them)?

[ed84c]

I know what det(...) is and I know various rules relating to eigenvalue, eigenvectors and finding powers. Or are you talking about something else?

[timo]

You can probably get the rules I meant from those. Just see how far you get with your proof.