Let A and B be nxn matrices over the field F. Show that if A is invertible there are at most n scalars c in F for which the matrix cA + B is not invertible.

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So I'm thinking that I need to find a c such that there exists a 0 in the diagonal of the triangularized sume cA+B, because then the determinant is zero, and it will be singular. Since you need to pick an entry in the diagonal of A to work with to make zero in the end, you have n choices of c.

...I just need help formalizing that into a proof.

Anyone?

3/20/2006 9:35:08 PM

I don't think that's how you want to prove that. Think about Eigenvalues, this is just a more "general" form and *hint* make use of the invertability of A.

3/20/2006 10:32:19 PM

something like..

want det(cA+B)=0 --> c det (A) = - det (B) --> det(A) = -det(B) * 1/c

you can get a negative determinant by switching the order of the rows an odd number of times, and 1/c is simple enough, but this way i'm not seeing the number of possibilities of c being n.

3/20/2006 11:15:30 PM

^^ sounds like a good idea. Maybe look at the determinant of cA + B. You know that will be an n-th order polynomial in c. You also know that if cA + B is not invertible then the determinant is zero. I guess you also know something about solving polynomial equations in a field F. I know we are done if F is the complex numbers...

3/20/2006 11:16:08 PM

I should clarify, think about the fundamental theorem of algebra. Every n-th order complex polynomial
has n-roots in the complex numbers. If F=real numbers this same theorem guarantees that there can be no more than n-roots still because it's easy to prove that the polynomial can be completely factored in the complex case and in the case there are complex roots they allway come in conjugate-pairs if the polynomial has real coefficients... so even in the real case there can be no more than n-roots to a n-th order polynomial equation. In the case of finite fields I'm not so clear on this question...

[Edited on March 20, 2006 at 11:24 PM. Reason : .]

3/20/2006 11:23:24 PM

I'm thinking this question has to be solved using properties of determinants, given we haven't done eigenvalues in this class yet, and the fundamental algebra theorem isn't expressly given in the text.

I..just don't see it happening given the handful of properties we've talked about, unless I discuss similarity between a row operated triangular matrix and the original form, like I was talking about in my original post.

3/20/2006 11:42:11 PM

Well, I suppose you could multiply on the right by A^-1

cA+ B not invertible implies (cA+ B)A^-1 = cI + BA^-1 also not invertible.

det ( cI + BA^-1 ) = 0 pretty clearly will give a n-th order polynomial in c.

If a n-th order polynomial had (n+1)-distinct roots then you could prove that it was a (n+1)-th order
polynomial, which would be a contradiction. That is not hard to prove, the fun. thm. of alg. is profound in that it says that there exist n-roots. The fact that there cannot be more than n is easier, even I can prove that

But, I don't know what tools you are allowed to use in your course. The algebra I'm describing should really be prerequisite...

[Edited on March 20, 2006 at 11:51 PM. Reason : .]

3/20/2006 11:50:50 PM

Quote :

"det ( cI + BA^-1 ) = 0 pretty clearly will give a n-th order polynomial in c."

^yeah that's what I had in mind. The key of the "at most n" statement is realized when looking at the field your on. Like you mentioned in your post ^^^.

Quote :

"In the case of finite fields I'm not so clear on this question..."

But recall this Thm:
Let f(x) in F[x] have degree n, then f(x) can have at most n roots in any extension, K of F.

Thia makes no assumption on the field. So you don't have to worry about any contrived example you can come up with.

[Edited on March 21, 2006 at 12:23 AM. Reason : .]

To Virga:
If you don't know about abstract fields just think about the complex, real, and rationals.
Only in the complex case are you guarenteed n roots by FundThm.
Others will have <=n.

[Edited on March 21, 2006 at 12:35 AM. Reason : .]

3/21/2006 12:21:17 AM

Might need to be a little careful though about the statement cA+B not invertable => (cA+B)A^-1 not invertable. It's certainly true but it might need to be shown why. You can bypass this by just using the det. Actually your kinda proving it by using the following hint.

hint: det(A)*det(B)=det(AB)

[Edited on March 21, 2006 at 12:59 AM. Reason : .]

3/21/2006 12:58:46 AM

^I guess I should have expected that in the case of finite fields we can only loose roots in comparison to the complex case. Thanks for the reminder, I don't run into finite fields to much in my work. I agree he probably needs to argue that multiplying by A^-1 keeps it non-invertible, but it's not really an essential step. I just recommended it to help see more clearly that the determinant will give an n-th order polynomial in c.

I don't think there is anyway around using the fundamental theorem of algebra ( or some close cousin at least ) to do this problem. It's still linear algebra because you'll have to use properties of the determinant to prove that it produces a n-th order polynomial in c. Although, if you know the formula for the determinant it's more or less clear, but proving it well that depends on who your professor is.

3/21/2006 2:12:47 AM

Quote :

"proving it well that depends on who your professor is."

haaha, true.

I agree with you about multiplying by A^-1. here is what I was thinking,

cA+B Singular <=> 0=det(cA+B)=det(cA+B)*det(A^-1)=det(cI+BA^-1) = p(c) nth order over F.
<=> p(c)=0 i.e. c is a root of p which can have at most n solutions by the thm I qouted.

[Edited on March 21, 2006 at 11:16 AM. Reason : .]

3/21/2006 11:12:13 AM