Dimension of matrix spanning set

**jsndacruz** · November 30th 2011, 03:25 AM

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.

**girdav** · November 30th 2011, 04:46 AM

The dimension of $V$ is $n^2$ , not $n$ .
Do you know Cayley-Hamilton theorem?

**jsndacruz** · November 30th 2011, 07:37 AM

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?

**girdav** · November 30th 2011, 08:11 AM

No, the dimension of $W$ is at most $n$ , since we can write $A^n$ as $a_0\cdot I+a_1\cdot A+\ldots+a_{n-1}A^{n-1}$

**jsndacruz** · November 30th 2011, 02:12 PM

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.

**jsndacruz** · December 22nd 2011, 05:21 PM

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

**Deveno** · December 22nd 2011, 07:01 PM

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.

**FernandoRevilla** · December 23rd 2011, 01:20 AM

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 $V$ .

**Deveno** · December 23rd 2011, 04:52 AM

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

**FernandoRevilla** · December 23rd 2011, 06:05 AM

Furthermore, $\mu (x)$ divides any annihilator polynomial of $A$ .

**Deveno** · December 24th 2011, 02:22 PM

and furthermore...oh snap! merry christmas!

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