Dimension of matrix spanning set

Studying for my final in linear algebra and came across this:

**Problem**: Let V be the vector space of all real n x n matrices over R. Let A be an element of V, and let W be the subspace of V spanned by . Prove that .

**Thoughts**: The statement seems intuitive, but I'm not sure exactly what theorem/properties it results from. I know that a subspace cannot have dimension larger than its finite parent space, so . However, my professor said that I should try to prove it by having it follow from the dimension of a matrix after raising it to higher powers; i.e. not from the obvious theorem.

Re: Dimension of matrix spanning set

The dimension of is , not .

Do you know Cayley-Hamilton theorem?

Re: Dimension of matrix spanning set

We touched on it in class, but have not gone through it rigorously. I'll read up on it now. Does this mean that there is a typo in the question, i.e. then?

Re: Dimension of matrix spanning set

No, the dimension of is at most , since we can write as

Re: Dimension of matrix spanning set

I read up on Cayley Hamilton, but am a little shaky as far as its implications. According to the theorem, the characteristic equation of any n x n matrix, A, satisfies , where are scalars, and is the identity matrix. But that creates, at most, equations.

Is the dimension related to the maximum number of solutions to this system? I don't see how to show that it would be at most n.

Re: Dimension of matrix spanning set

Does anyone have further insight to this question? I still haven't been able to solve it or relate it back to Cayley Hamilton.

Re: Dimension of matrix spanning set

if A satisfies ,

that is, if , so that p(A) = 0, then

which means the n-th power (and thus any higher power) of A

is a linear combination of .

but this means that this set spans W, so dim(W) ≤ n.

and the characteristic polynomial of A, det(xI - A), is just such a polynomial (of degree n).

the cayley-hamilton theorem says: if p(x) = det(xI - A), then p(A) = 0.

Re: Dimension of matrix spanning set

We can also assure that if is the minimal polynomial of and then, and is a basis of .

Re: Dimension of matrix spanning set

furthermore, μ(x) divides det(xI - A).

Re: Dimension of matrix spanning set

Furthermore, divides any annihilator polynomial of . :)

Re: Dimension of matrix spanning set

and furthermore...oh snap! merry christmas!