How to find a non-diagonalizable matrix $T \in L(\mathbb{C}^3)$ such that $6, 7$ are its eigenvalues? I am currently reading Linear Algebra Done Right by Axler on my own without a teacher, and am stuck on the exercise in the title. My current progress is as follows:
Suppose that $x, y, z$ is a basis of $\mathbb{C}^3$, then let $T(x) = 6x, T(y) = 7y$, and we guaranteed that $6, 7$ are eigenvalues, so we just have to figure out what $T(z)$ is so that $6, 7$ are the only eigenvalues and $\dim E(6, T) = \dim E(7, T) = 1$. Let $T(z) = k_1x + k_2y + k_3z$.
Let $\alpha = a_1x + a_2y + a_3z$. Suppose that $\lambda(a_1x + a_2y + a_3z) = T(\alpha) = 6a_1x + 7a_2y + a_3(k_1x + k_2y + k_3z)$. Simplifying gives us the following system of equations: $a_1(\lambda - 6) = a_3k_1, a_1(\lambda - 7) = a_3k_2$ and $k_3 = \lambda$. But I do not know how to proceed from here. Any hint will be helpful!
 A: You ask for a general method... the Jordan Canonical Form, mentioned, would give you an easy way to get at transformations like this (if I had to guess, that's how TheSilverDoe came up with the example). But here is another way to get at it in the absence of the Jordan Canonical Form.
Presumably, yoo have seen an example of an operator on a $2$-dimensional vector space that has eigenvectors but is not diagonalizable. This would require a single doubled eigenvalue. The simplest is the case where the eigenvalue is $0$, but the nullspace has dimension $1$. The simplest example is $U\colon\mathbb{C}^2\to\mathbb{C}^2$ given by $U(a,b) = (b,0)$. It is easy to verify that the nullspace has dimension $1$, but the only eigenvalue is $0$, so the operator is not diagonalizable.
Now we use the following useful fact:
Proposition. Let $T$ be an operator on a vector space $V$, let $\alpha,\beta$ be scalars. If $\mathbf{x}$ is an eigenvector of $T$ corresponding to $\lambda$, then $\mathbf{x}$ is an eigenvector of $\alpha T+\beta I$ corresponding to $\alpha\lambda + \beta$.
Proof. Evaluating, we have
$$(\alpha T+\beta I)(\mathbf{x}) = \alpha T(\mathbf{x})+\beta\mathbf{x} = \alpha(\lambda \mathbf{x})+\beta\mathbf{x} = (\alpha\lambda+\beta)\mathbf{x}.\quad\Box$$
So now that we have a linear operator $U$ on a $2$-dimensional space that is not diagonalizable, with eigenvalue $0$, we can easily obtain one that is not diagonalizable and with eigenvector $\lambda$: just take $U+\lambda I$. In this case, to get one with unique eigenvalue $6$, we can take $T=U+6I$. This will give $T(a,b) = (b,0)+6(a,b) = (6a+b,6b)$.
(You could do other things to get it to work, of course, using the proposition.)
Now we extend it to a linear operator on $\mathbb{C}^3$ that has the desired properties: not diagonalizable, has only two eigenvalues ($6$ and $7$). We take a basis $x,y,z$, and think of $x$ as corresponding to $(1,0)$ and $y$ as corresponding to $(0,1)$ in our definition above. So we let $T(x)=6x$, $T(y)=x+6y$, $T(z)=7z$. Or,
$$T(a,b,c) = (6a+b,6b,7c).$$
