Proof: Showing A is Similar to Upper-Triangular Matrix

[ksinclair13]

Homework Statement

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,...,x_n} is a basis of R$$^{n}$$ such that Ax_k $$\in$$ span{x₁,...,x_k} for k = 1,2,...,n. You may find the Cayley-Hamilton Theorem helpful.

Homework Equations

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

The Attempt at a Solution

I have already done the following problem:
Let A be an n x n matrix and let {x₁,...,x_n} 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 Ax_k $$\in$$ span{x₁,...,x_k} for k = 1,2,...,n.

I think the matrix [x₁,...,x_n] 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!

 

[mighty2000]

Hello!

To show that A is similar to an upper-triangular matrix, we need to find a basis with the property stated in the problem. We can approach this by using the Cayley-Hamilton Theorem, which states that p(A) = 0. This means that A satisfies its own characteristic polynomial.

Since p(t) = (\lambda-t)^{n}, we know that \lambda is an eigenvalue of A with multiplicity n. This means that there exists a basis {x1,...,xn} such that Axk = \lambda xk for k = 1,2,...,n. This is because the characteristic polynomial of A is the product of its eigenvalues, and in this case, all eigenvalues are equal to \lambda.

Now, let us consider the matrix [x1,...,xn]. Since Axk = \lambda xk for k = 1,2,...,n, we can see that the first column of this matrix will be [x1,0,0,...,0], the second column will be [0,x2,0,...,0], and so on. Therefore, this matrix is upper-triangular.

Furthermore, since Axk = \lambda xk for k = 1,2,...,n, we can see that Axk \in span{x1,...,xk}. This satisfies the property stated in the problem, and thus A is similar to an upper-triangular matrix.

I hope this helps! Let me know if you need any further clarification.