Range of a non-diagonalizable matrix

Question

Suppose $T ∈ L(C^3)$is such that 11 and 17 are eigenvalues of T. Furthermore, suppose that T is not diagonalizable. Prove that there exists a vector $(x, y, z) ∈ C^3$ such that: $T(x, y, z) = (55 + 23x, π2 + 23y, √111 + 23z)$.

I've been working through this problem for a while now and cannot figure out where to start. It seems like most of the information is irrelevant except for the fact that we are working with an operator over a three dimensional space that is non diagonalizable and has two guaranteed eigenvalues. I've been trying to prove that the range of this operator is all of $C^3$, but don't know what tools I have to work with. Any hints would be appreciated.

Answer 1

This is a $3 \times 3$ matrix, and two eigenvalues $11$ and $17$ are given. If the matrix is not diagonalizable, at least one eigenvalue has algebraic multiplicity $> 1$. But there are only $3$ eigenvalues counted by algebraic multiplicity, so there can be no others.

If you write $v = (x,y,z)$ and $w = (55, \pi2, \sqrt{111})$ (is that $\pi2$ supposed to be $\pi^2$? it doesn't matter), your question is saying $(T-23 I) v = w$. But $23$ is not an eigenvalue, so $T-23I$ is surjective.