Finding eigenvector of linear map defined by matrix whose columns each add to $k$

Question

Let $M$ be an $n \times n$ matrix over $\mathbb F$, where $\mathbb F\in \{\mathbb C, \mathbb R\}$. Define the linear map $T: \mathbb F^n\rightarrow \mathbb F^n$ by $Tv= Av$.

Suppose the sum of the entries in each column of $M$ is equal to $k$.

I want to determine that $k$ is an eigenvalue and find a corresponding eigenvector.

I am able to show that it's an eigenvalue indirectly, namely by showing that the map $T-kI$ is not invertible.

But I think that I should show that $k$ is an eigenvalue by obtaining an eigenvector. This way I find the eigenvector I want, too.

However, I cannot come up with one. How could I find such an eigenvector? Note that I am following Axler's book, which avoids using determinants when dealing with eigenvalues. So I would appreciate an explanation that does the same.

Edit: I was linked to a possible duplicate. The question is similar to mine, but the only direct answers involved this fact: $[1, \ldots, 1]A= [k, \ldots, k]=k[1, \ldots, 1]$. The problem here is that the operator $T$ I defined above has the form $Av$. So I would want $Av =kv$ for some $v$. This means that $v$ must be a column vector, not a row vector as described in the possible duplicate.

I am specifically looking for a nonzero vector $v$ such that $Tv=kv$, were $v$ is considered as a column vector.

Answer 1

What follows isn't obtaining an eigenvector, but I thought maybe it would be helpful.

The eigenvalues of $A$ and its transpose are the same. To see this in Axler-approved non-determinant terms, consider that $A$ and $A^T$ have the same minimal polynomial (which follows from the property $p(A^T) = p(A)^T$ for any polynomial $p$ and square matrix $A$). It's fundamental to Axler's presentation that the minimal polynomial completely determines all of the generalized eigenvalues of $A$, so that $A$ and its transpose have the same multiset of eigenvalues.

Unfortunately this correspondence between eigenvalues of a matrix and its transpose does not carry over to eigenvectors. From the correspondence of eigenvalues we are assured the existence of eigenvectors that realize them, but that's it.

So, as you noted, the all $1$ vector $(1,1,1,1, \ldots)$ does not help you find an eigenvector of $A$. But it is clearly an eigenvector of $A^T$ with eigenvalue $k$, and that immediately shows that $A$ has an eigenvalue of $k$.

One can go a little further by noting that a matrix is always conjugate to its transpose (see e.g. here), so that $A^T = P^{-1} A P$ for some change of basis $P$. If you can figure out such a $P$, then you know that given an eigenvector $x$ of $A^T$, $Px$ is an eigenvector of $A$ with the same eigenvalue.

In your case, if $P$ is any basis which conjugates $A$ to $A^T$, then the sum of the basis vectors in $P$ is an eigenvector of $A$ with eigenvalue $k$.