# Prove that $m$ is an eigenvalue of the matrix $\mathbf{A}$

We are given square matrix $\mathbf{A}$. Also, we know that sum along any column of $A$ equals $m$. Task is to prove, that $m$ is an eigenvalue of $\mathbf{A}$.

I'm stuck with that one. I will provide below my findings:

Since we know, that $\mathbf{A}$ has a least one eigenvalue, then $\det(A) \ne 0$, hence it's possible to bring $\mathbf{A}$ to the diagonal matrix $\mathbf{A}'$. Then eigenvalues would be roots of the polynomial $\prod_{i=1}^n (A_{ii}'-\lambda) = 0$
But we have nothing to go from here, and also I bothered by the fact, that in general case we loosing this 'sum over column = m' property, when diagonalizing $\mathbf{A}$.

I would appreciate some help here.

• "Since we know, that $A$ has a least one eigenvalue, then $\det(A) \neq 0$". That's false. Every $n\times n$ matrix has exactly $n$ eigenvalues counted with multiplicity, and every non-invertible matrix has an eigenvalue of zero. Jun 4 '14 at 12:27
• What is an obvious (perhaps) choice for an eigenvector for $A^T$? ($A$ and $A^T$ have the same eigenvalues.) Jun 4 '14 at 12:27
• @MichaelAlbanese, you claim after "That's false" depends decisively on what field is the matrix defined on. Jun 4 '14 at 12:40

An alternative hint: Every column sum of $A-mI$ is $0,$ so every linear combination of columns of $A-mI$ still has the sum of all its entries $0.$ Hence $A-mI$ does not have full column rank.
$$A^t\begin{pmatrix}1\\1\\\vdots\\1\end{pmatrix}=\begin{pmatrix}\sum_{k=1}^na_{k1}\\\sum_{k=1}^na_{k2}\\\vdots\\\sum_{k=1}^na_{kn}\end{pmatrix}=\begin{pmatrix}m\\m\\\vdots\\m\end{pmatrix}$$
And yet another alternative hint: given that $A$ and $A^T$ have the same eigenvalues, for what vector $b$ is $c=A^Tb$ a vector containing the sums of the columns of $A$ and what would $c$ be?