DIMENSION OF A SUBSPACE - clarification

Let C be a *n x n* matrix. Let W be the Vector space spanned by {I, C, C^{2}, C^{3}, .... C^{2n}}. What is the dimension of the vector space W?

My argument is as follows.

Argument- By Cayley hamilton theorem, we have a degree of equation 'n' satisfied by the matrix C.

This is of the form a_{n}C^{n} + a_{n-1}C^{n-1}+ a_{n-2}C^{n-2}+ +..... a_{0}I = 0. which implies the n+1 elements, namely, I, C, C^{2}, C^{3},...C^{n} are linearly dependent or (one-less) no. of elements is linearly independent.

Therefore, n+1-1 = n is the basis of the vector space.

Are there any other possible explanations? answers?

Re: DIMENSION OF A SUBSPACE - clarification

Hey MAX09.

Consider that C^n is an operator with det(C^n) != 0.

Now consider an eigen-decomposition of the matrix C^n.

Each matrix will have different eigenvalues assuming they are not all 1 or 0.

But this implies linear independence given the properties of the eigen-values.

If you can show how many distinct eigen-values there are you get the number of independent operators and thus the dimension of the entire space spanned by the set of operators.

I haven't given a complete proof but hopefully it is an alternative viewpoint to ponder for more ideas.

Re: DIMENSION OF A SUBSPACE - clarification

the cayley-hamilton theroem is a step in the right direction...but it doesn't go far enough. for example consider the 3x3 matrix:

$\displaystyle C = \begin{bmatrix}0&0&1\\0&0&0\\0&0&0 \end{bmatrix}$

the characteristic polynomial of C is det(xI - C) = x^{3}, so we have C satisfies the equation:

C^{3} = 0.

but {I,C,C^{2}} is NOT a linearly independent set, because C^{2} = 0.

the trouble is, C might satisfy some factor of its characteristic polynomial (the degree n one guaranteed by cayley-hamiltion). what we need to know (and what is not supplied by the problem) is the degree of the MINIMAL polynomial for C.