number of matrices in $GL_{3}(\mathbb{Z}_{3})$ such that $[1 \ 1 \ 1]^{T}$ is an eigenvector.

Question

Find the number of matrices in $GL_{3}(\mathbb{Z}_{3})$ such that $[1 \ 1 \ 1]^{T}$ is an eigenvector.
This question was asked to me in an interview which I wasn't able to answer.
I approached as follows,
If $A$ is a matrix in $GL_{3}(\mathbb{Z}_{3})$ such that $[1 \ 1 \ 1]^{T}$ is an eigenvector then $0$ can't be an eigenvalue as $A$ is invertible. So $A$ can have $1$ and $2$ as eigenvalues. Let $v=[1 \ 1 \ 1]^{T}$.
If $1$ is an eigenvalue of $A$ then we have $Av=v$ and if $2$ is an eigenvalue of $A$ then $Av=2v$.
So we need to find the number of matrices $A$ in $GL_{3}(\mathbb{Z}_{3})$ such that $Av=v$ and $Av=2v$.
Then I tried to find out such A's but I got stuck at a point.
Can you please let me know how to solve this question ? Thanks.

Answer 1

Hint: instead of thinking of the matrix, think of the corresponding (invertible) linear transformation $\mathbb{F}_3^3 \to \mathbb{F}_3^3$. Then, use the fact that $\{ (1, 1, 1), (1, 0, 0), (0, 1, 0) \}$ is a basis of $\mathbb{F}_3^3$. So, the linear transformations are uniquely determined by the images of these basis vectors (and each combination comes from a linear transformation); and the transformation is invertible if and only if the image of the basis is another basis (which is further equivalent to the image of the basis being linearly independent).

So:

• How many choices are there for the image of $(1, 1, 1)$?
• Given a choice from the previous step, how many choices are there for the image of $(1, 0, 0)$? (Further hint: the only restriction is that it must not be a scalar multiple of the image of $(1, 1, 1)$.)
• Given a choice from the previous two steps, how many choices are there for the image of $(0, 1, 0)$? (This must not be in the span of the previous two vectors.)

Answer 2

Daniel Schepler's +1 answer maps probably the simplest route to the destination.

As an alternative I proffer the use of group actions on the projective plane $\Bbb{P}^2(\Bbb{Z}_3)$. There are $3^2+3+1=13$ elements on $\Bbb{P}^2(\Bbb{Z}_3)$ — the lines through the origin in $\Bbb{Z}_3^3$.

It is known (and easy to show) that $G=GL_3(\Bbb{Z}_3)$ acts transitively on that projective plane. The question asks about the stabilizer of the line spanned by $L=(1,1,1)^T$. By the orbit-stabilizer theorem $$ \operatorname{Stab}_G(L)=\frac{|G|}{13}. $$ I trust you know $|G|$ already :-)