# What can be said about eigenvalues from the sum of elements in a row?

Here I'm struggling with my math. Now I have this question which seems unmanageable. Let $A \in M_n(\mathbb{R})$, with $n\ge2$. Assume that the sum of any row is $1$. What can be said about eigenvalues and eigenvectors of $A$? The alternatives are irrelevant but say something like: $A$ can never have real eigenvalues or $1$ and $0$ are always eigenvalues of $A$. I can't find the relation between the sum of elements per row and eigenvalues/eigenvectors! Can someone please, give me a direction?

• Observe that $(1, 1, \ldots, 1)^T$ (the column vector containing $1$ in every coordinate) is an eigenvector, with eigenvalue $1$. (Why?) – Bungo Oct 11 '17 at 18:05
• I assume you mean $M_n$, not $M_2$. – Ian Oct 11 '17 at 18:06
• Do you mean the sum of any row, or the sum of every row? – Fly by Night Oct 11 '17 at 18:08
• FlybyNight, the sum of i-th row is 1, i = 1, 2, 3,...,n. – daniel.franzini Oct 11 '17 at 18:19

Finally, I think I completely understand the question. Here is a proof that if $k \in \mathbf{R}, k \neq 0$ is the sum of the elements of each row in a square matrix A, then $k$ is an eigenvalue of A.

Let $v = (1,1,1,...,1)$. Calculating $Av$ will result in $(k,k,k,...,k) = k(1,1,1,...,1) = kv$. By the definition, $k$ is an eigenvalue with associated eigenvector $v$.

Now, for the particular question, $k=1$ and the only information we can get is that $1$ is an eigenvalue associated with eigenvector $(1,1,1,...1)$.

Thank you all for the help.

With just what you said, you can say that there is an eigenvalue of $1$, by writing down the corresponding right eigenvector (just a vector of $1$s). For the other eigenvalues, the best you can do is Gerschgorin's theorem, which only gives an estimate. You can say a bit more if you assume the elements are nonnegative, but you didn't say that here.

• Elements don't have to be non-negative. The question is exactly the way I posted. What amazes me is how magically these things pop out. For me it is completely non-trivial and non-obvious why 1 is an eigenvalue with $(1,1,1,1,...1)^T$ as associated eigenvector. – daniel.franzini Oct 11 '17 at 18:25
• @daniel.franzini It actually is easy if you think of matrix multiplication properly. $(Ax)_i=\sum_j a_{ij} x_j$. If x is constant then each entry of Ax is proportional to the corresponding row sum. This also has a probabilistic interpretation again in the case of nonnegative entries. – Ian Oct 11 '17 at 18:33
• If $x = (1,1,1,...,1)$, then $\sum_{j} a_{ij}x_j$ became $\sum_{j} a_{ij}$ which is always 1, according to the question. But shouldn't it work for e.g. $x = (2,2,2,2,2,...,2)$ also? – daniel.franzini Oct 11 '17 at 19:56
• I got it partially. For a vector $x$ be an eigenvector, it should verify $A-\lambda \mathbf{I} = \mathbf{0}$, with $\lambda$ being the associated eigenvalue, right? So, this works for $\lambda = 1$ and $x = (1,1,\cdots,1)$. Does it work for other real numbers? – daniel.franzini Oct 11 '17 at 20:28
• Actually, it should be $(A-\lambda \mathbf{I})x = \mathbf{0}$. – daniel.franzini Oct 11 '17 at 20:37