# Can a matrix satisfy all three of the following properties?

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 $n \leq 3$.

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 $1$;

• Is there an easy explanation for this? May 26, 2016 at 14:07
– user9464
May 26, 2016 at 14:37

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.

• All clear but for one part: why should every application of $A- \lambda I$ lower the dimension of the image? May 26, 2016 at 15:10
• Nevermind, I got it - had to dig up my linear algebra course notes though. Thank you for your answer! May 26, 2016 at 16:03
• @Bib-lost: Good work. To solve linear algebra questions, sometimes knowing the course is useful. Actually this question appears to be concocted to test your knowledge of the theory (while throwing in a few confusing details), but if it came up more naturally, so much the better! May 26, 2016 at 16:30
• It came up 'naturally', i.e. not as an exercise in a linear algebra class. My linear algebra course was years ago, the question above comes from a problem in numerically solving differential equations (more specifically, analysing stability of linear multistep methods). May 26, 2016 at 16:37

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.

• Why should $a_1$ be an eigenvalue just because it is the sum of eigenvalues? Or am I misinterpreting your answer? May 26, 2016 at 14:51
• @Bib-lost $a_{1}$ is along the diagonal. Therefore it is an eigenvalue. What i am saying is, this will be the only nonzero eigenvalue based on the matrix you gave me, The other eigenvalue will be 0 wit a multiplicity of (n-1). May 26, 2016 at 15:44
• $\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ has a $1$ on the diagonal, surely this is not an eigenvalue? May 26, 2016 at 16:05
• @ChanHunt: Sorry, this answer is just entirely wrong. Please check the definition of an eigenvalue or that of a diagonal matrix. For information, the characteristic polynomial of this matrix is $X^n-a_1X^{n-1}-a_2X^{n-2}-\cdots-a_n$, which means the set of $n~$eigenvalues could be just anything, subject only (as for any real matrix) to the condition that for any complex eigenvalue, its complex conjugate is also an eigenvalue, May 26, 2016 at 16:24
• @MarcvanLeeuwen I am wrong, sorry about that. I am quite good at linear algebra. I just messed up. sorry about that. May 27, 2016 at 22:39