Matrices and eigenvalues

Question

Can there exist a 3x3 matrix which has only one eigenvalue? How is that possible for a 3x3 indicating all three variables are free variables?

Moreover, what is a rank of 3x3 matrix having 2 eigenvalues (2 same, 1 different) ?

Answer 1

The identity matrix has its three eigenvalues equal to 1. That doesn't have to do with the independence of the variables because you can find three independent eigenvectors associated with the multiple eigenvalue.

A 3x3 matrix with exactly 2 equal eigenvalues can have rank equal to 1, 2 or 3.

Example rank = 1:

$$ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} $$

Example rank = 2:

$$ \begin{matrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{matrix} $$

Example rank = 3:

$$ \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{matrix} $$

Answer 2

This is just a remark on rank-eigenvalue relation. First, denote the set of all $n\times n$ matrices with complex entries by $M_n(\mathbb{C})$. Let $A \in M_n(\mathbb{C})$ be given with $\mathrm{rank}(A) = r$. Then, $A$ has at most $r$ non-negative eigenvalues.

To see this, note two facts. First, one can identify $r$ linearly independent columns of $A$ such that every column of $A$ is a linear combination of these $r$ columns. Denote the matrix whose columns corresponds to these $r$ columns of $A$ as $X$ (hence, $X \in M_{n,r}(\mathbb{C})$). Now, $A=XY^{T}$ for some $Y \in M_{r,n}$. Then, an important theorem in linear algebra states that $AB$ and $BA$ has the same set of eigenvalues (with a possibility of additional zeroes in favor of $AB$ or $BA$). Hence, $A$ and $Y^{T}X \in M_r(\mathbb{C})$ has the same eigenvalues, with the addition of $n-r$ extra zeroes to the set of eigenvalues of $A$.

Hence, $\mathrm{rank}(A)=r$ implies that $A$ has at most $r$ non-negative eigenvalues.