Can a matrix satisfy all three of the following properties? Consider an $n \times n$ matrix of the form
$$
A = \begin{bmatrix}
a_1 &amp; a_2 &amp; \ldots &amp; a_{n-1} &amp; a_n \\
1 \\
&amp; 1 \\
&amp; &amp; \ddots \\
&amp; &amp; &amp; 1
\end{bmatrix}
$$
for certain $a_1, \ldots, a_n \in \mathbb{R}$ (and zeroes in all blank spaces). Can this matrix ever (for some $n \in \mathbb{N}$ and some $a_1, \ldots, a_n \in \mathbb{R}$) satisfy all of the following three properties:


$\lvert \lambda \rvert \leq 1$ for all eigenvalues $\lambda$ of $A$
For all eigenvalues $\lambda$ with $\lvert \lambda \rvert = 1$, the algebraic multiplicity is equal to the geometric multiplicity
There exists an eigenvalue $\lambda$ with $\lvert \lambda \rvert = 1$ and such that the multiplicity of $\lambda$ is greater than $1$?


Note that I do not require $A$ to be invertible. I don't immediately see why such a matrix can't exist, but if it could, that would lead to a counterexample to some unproven theorem in my course notes.

I calculated by hand that such a matrix can not exist for $n \leq 3$.
The answer is no. Your matrix $A$ is an avatar of a companion matrix. The geometric multiplicity of an eigenvalue of such a matrix is always $1$;
You can use the following property of a matrix:

Given some n by n matrix, A, the trace of the matrix is equivalent to the sum of its eigenvalues.

The sum of the eigenvalues of the matrix is:

trace (A) =  $a_{1}$ + 0(n-1)

= $a_{1}$ = $\lambda$

Using the above theorem, we can then clearly see the following must hold true:

1.) $\left | \lambda \right | \leq 1 $ iff  $\lambda = 0$ or $\lambda = 1$

2.) The  geometric multiplicity of the matrix is therefore equivalent to the multiplicity of the eigenvalue  $a_{1}$ = 1.

3.) It follows from our first conclusion that the if $\lambda = 1$, the multiplicity must be equal to the number of places in the diagonal that the number 1 occurs. In your example, this can be no more than 1 time. Therefore, the multiplicity of the non-zero value is at most 1.