For a $n \times n$ matrix A, the columns sum to a real value $c$. How do I show that $c$ is an eigenvalue of the Matrix.
My attempt, I can see that the matrix $A-cI$ has columns summing to $0$, why would that mean that $A-cI=0$ in general for the condition to be satisfied?
On
A matrix is similar to its transpose. For the matrix $A$,we see that $A^T$ has all rows adding to $c.$Thus if $v$ is the $n-$place column vector with all entries equal to 1, then $$A^Tv=cv$$, so $c$ is an eigenvalue of $A^T$. Since similar matrices have the same eigenvalues and $A$ is simiar to $A^T$ , $c$ is also an eigenvalue of $A.$
Let $A$ be matrix such that$$\sum_{j=1}^n a_{ij}=c\quad\forall i\tag{1}$$ We now need to prove $det (A-cI)=0$. To prove this, let's sum all the lines in $A-cI$. Then for all $i$ and for all $j$, we have $$\sum_{j\neq i}a_{ij}+a_{ii}-c$$ $$\sum_{j=1}^n a_{ij} -c$$ By equation $(1)$ $$c-c=0$$ Since we now have a line of $0$, the determinant is $0$ $$det (A-cI)=0$$ and $c$ is an eigenvalue of $A$.