# Proof involving char. poly., a basis, and similarity to an upper-triangular matrix

1. Apr 8, 2009

### ksinclair13

1. The problem statement, all variables and given/known data
Let A be an n x n matrix and suppose that the characteristic polynomial of A is
p(t) = ($$\lambda$$-t)$$^{n}$$ for some fixed $$\lambda$$ $$\in$$ R. Show that A is similar to an upper-triangular matrix. Do this by finding a basis with the following property: {x1,...,xn} is a basis of R$$^{n}$$ such that Axk $$\in$$ span{x1,...,xk} for k = 1,2,...,n. You may find the Cayley-Hamilton Theorem helpful.

2. Relevant equations
p(t) = det(A-t*I) = ($$\lambda$$-t)$$^{n}$$
p(A) = 0 (Cayley-Hamilton Theorem)

3. The attempt at a solution
I have already done the following problem:
Let A be an n x n matrix and let {x1,...,xn} be a basis of R$$^{n}$$. Show that the matrix of the linear transformation associated to A with respect to this basis is upper-triangular if and only if Axk $$\in$$ span{x1,...,xk} for k = 1,2,...,n.

I think the matrix [x1,...,xn] must itself be upper-triangular. Such a basis would not be difficult to find, however I don't understand how to connect this with the Cayley-Hamilton Theorem or the fact that $$\lambda$$ is an eigenvalue of A with multiplicity n. I also don't have any good ideas of how to go about the proof in general.

Any help would be greatly appreciated!