# Dimension of matrix spanning set

• Nov 30th 2011, 03:25 AM
jsndacruz
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 $I, A, A^2, A^3 ...$. Prove that $dim(W) \leq n$.

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 $dim(W) \leq dim(V) = n$. 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.
• Nov 30th 2011, 04:46 AM
girdav
Re: Dimension of matrix spanning set
The dimension of $V$ is $n^2$, not $n$.
Do you know Cayley-Hamilton theorem?
• Nov 30th 2011, 07:37 AM
jsndacruz
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. $dim(W) \leq n^2$ then?
• Nov 30th 2011, 08:11 AM
girdav
Re: Dimension of matrix spanning set
No, the dimension of $W$ is at most $n$, since we can write $A^n$ as $a_0\cdot I+a_1\cdot A+\ldots+a_{n-1}A^{n-1}$
• Nov 30th 2011, 02:12 PM
jsndacruz
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 $A^n + c_1 A^{n-1} c_2 +...+ c_{n-2} A + c_{n-1} I_n = 0$, where $c_i$ are scalars, and $I_n$ is the identity matrix. But that creates, at most, $n^2$ 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.
• Dec 22nd 2011, 05:21 PM
jsndacruz
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.
• Dec 22nd 2011, 07:01 PM
Deveno
Re: Dimension of matrix spanning set
if A satisfies $A^n + c_1 A^{n-1} + c_2 A^{n-2}+...+ c_{n-2} A + c_{n-1} I_n = 0$,

that is, if $p(x) = x^n + c_1x^{n-1} + c_2x^{n-2} + \dots + c_{n-1}$, so that p(A) = 0, then

$A^n = -(c_1 A^{n-1} c_2 +...+ c_{n-2} A + c_{n-1} I_n)$ which means the n-th power (and thus any higher power) of A

is a linear combination of $\{I_n,A,A^2,\dots,A^{n-1}\}$.

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.
• Dec 23rd 2011, 01:20 AM
FernandoRevilla
Re: Dimension of matrix spanning set
We can also assure that if $\mu$ is the minimal polynomial of $A$ and $m=\textrm{deg}\;(\mu)$ then, $\dim V=m$ and $B=\{I,A,A^2,\ldots, A^{m-1}\}$ is a basis of $V$ .
• Dec 23rd 2011, 04:52 AM
Deveno
Re: Dimension of matrix spanning set
furthermore, μ(x) divides det(xI - A).
• Dec 23rd 2011, 06:05 AM
FernandoRevilla
Re: Dimension of matrix spanning set
Furthermore, $\mu (x)$ divides any annihilator polynomial of $A$ . :)
• Dec 24th 2011, 02:22 PM
Deveno
Re: Dimension of matrix spanning set
and furthermore...oh snap! merry christmas!