Can a matrix satisfy all three of the following properties?

Question

Consider an $n \times n$ matrix of the form $$ A = \begin{bmatrix} a_1 & a_2 & \ldots & a_{n-1} & a_n \\ 1 \\ & 1 \\ & & \ddots \\ & & & 1 \end{bmatrix} $$ 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:

1. $\lvert \lambda \rvert \leq 1$ for all eigenvalues $\lambda$ of $A$
2. For all eigenvalues $\lambda$ with $\lvert \lambda \rvert = 1$, the algebraic multiplicity is equal to the geometric multiplicity
3. 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 $n \leq 3$.

Answer 1

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 $1$;

Answer 2

To expand upon the answer by loup blanc, the minimal polynomial of$~A$ is of degree$~n$, namely it is $X^n-a_1X^{n-1}-a_2X^{n-2}-\cdots-a_n$. If some eigenvalue$~\lambda$ had geometric multiplicity${}>1$ then there would to the contrary exist a monic polynomial of degree less than$~n$ that annihilates$~A$ (for instance the one obtained by dividing its characteristic polynomial by one factor $X-\lambda$). The conditions involving $|\lambda|$ are red herrings and can be left out; just that condition that some$~\lambda$ has geometric multiplicity${}>1$ is imposisble.

To prove that no monic polynomial$~P$ with $\deg(P)<n$ annihilates$~A$, it suffices to compute that $(0~0~\ldots~0~1)\cdot P[A]\neq0$. To see why geometric multiplicity${}>1$ for$~\lambda$ means that the exponent$~d$ of $X-\lambda$ in the characteristic polynomial can be lowered while still obtaining an annihilating polynomial, consider the action of $A-\lambda I$ restricted to the generalised eigenspace $E_\lambda$ for$~\lambda$, which has dimension$~d$ (by definition some power of $A-\lambda I$ will act as$~0$ on$~E_\lambda$). The (ordinary) eigenspace for$~\lambda$ is the kernel of this action, and if it has dimension at least$~2$, then the image of this restriction of $A-\lambda I$ has dimension at most $d-2$, and so at most $d-2$ more applications of $A-\lambda I$ (each lowering the dimension by at least$~1$) suffice to reduce that image to$~0$. All in all that is at most $d-1$ applications of $A-\lambda I$, and so a power $(X-\lambda)^{d-1}$ suffices for an annihilating polynomial.

Answer 3

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.