# DIMENSION OF A SUBSPACE - clarification

• Dec 14th 2012, 06:23 PM
MAX09
DIMENSION OF A SUBSPACE - clarification
Let C be a n x n matrix. Let W be the Vector space spanned by {I, C, C2, C3, .... C2n}. 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 anCn + an-1Cn-1+ an-2Cn-2+ +..... a0I = 0. which implies the n+1 elements, namely, I, C, C2, C3,...Cn 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?
• Dec 14th 2012, 10:50 PM
chiro
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.
• Dec 14th 2012, 11:07 PM
Deveno
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:

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

the characteristic polynomial of C is det(xI - C) = x3, so we have C satisfies the equation:

C3 = 0.

but {I,C,C2} is NOT a linearly independent set, because C2 = 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.