How eigenvalues change when row is multiplied by a constant? Suppose I have an $n\times n$ real matrix $M$, not necessarily symmetric. Now I take a row $1 \le i \le n$ of this matrix, and I multiply it by a constant real $c$. 
Is there a way to understand the effects of the value of $c$ on the eigenvalues of the modified matrix in general?
 A: I fear that even for low dim the relation will be utterly complicated. 
If $B=DA$ with $D=diag(1,1,\cdots,c,\cdots,1)$ then $Bx=DAx=\lambda Dx$ 
but $Dx=(x_1,\cdots,cx_i,\cdots,x_n)$ has nothing to do with $x$ so eigenspaces of $A$ and $B$ will be completely different.
You can have fun with this nice matrix 
$\begin{bmatrix}1& 4& 5& 2\\5& 6& 1& 5\\ 4& 1& 4& 4\\ 3& 2& 3& 2\end{bmatrix}$ it has eigenvalues $\{-1,-3,4,-13\}$
Try to multiply a row by $-1$ or $2$ and you will see how different the eigenvalues become (some are not even real anymore).

Here is for instance a $2\times2$ example.
$A=\begin{bmatrix}2&2\\-1&5\end{bmatrix}$ with eigenvalues $\{3,4\}$ 
Let's multiply first row by $c$, then eigenvalues are now $\begin{cases}c+\frac 52+\frac 12\sqrt{4c^2-28c+25}\\ c+\frac 52-\frac 12\sqrt{4c^2-28c+25}\end{cases}$
We cannot call this some "easy relation", and this is just dimension $2$.
A: Very complicated. 
Just check this example:
let be  $A = \begin{bmatrix} 3 & 2 \\ 5 & 4 \end{bmatrix} $, with eigenvalues $\lambda_1,\lambda_2$, which are the solution of
$0=\det(A-\lambda I) = x^2-7x+2$ 
and now check 
$A = \begin{bmatrix} 3\alpha & 2\alpha \\ 5 & 4 \end{bmatrix} $ with characteristic polynomial 
$0=\det(\begin{bmatrix} 3\alpha & 2\alpha \\ 5 & 4 \end{bmatrix}-\lambda I) = x^2-(4+3\alpha)x+2\alpha$ 
there is no relation between those two equations (relations of shared roots, with a first look)
