Spectrum of matrix with single scaled row

Question

Let $M$ be a real symmetric positive-definite matrix and $D_a$ the diagonal matrix $$D_a = \left[\begin{array}{ccccc}a & & & &\\& 1 & & &\\& & 1 & &\\& & & \ddots &\\& & & & 1\end{array}\right].$$

If the eigenvectors and eigenvalues of $M$ are known, can anything be said about the spectrum of $D_a M$, without recomputing its spectral decomposition?

Answer 1

1. Generally speaking, $D_aM=D_aM^{1/2}M^{1/2}$ has same eigenvalues as $M^{1/2}D_aM^{1/2}$. Thus its signature is $(n,0,0)$ if $a>0$, $(n-1,1,0)$ if $a<0$ and $(n-1,0,1)$ if $a=0$.

2. Now we assume that $a-1$ is small. Let $\chi(x,a)$ be the charac. polynomial of $D_aM$ , $(\lambda_i)_i$ (resp$(\mu_i)_i)$ be the (positive) eigenvalues of $M$ (resp. $D_aM$). Note that $\chi(x,a)=\det(xI-M)+(a-1)\det(xI-M-xE_{1,1})$ where $E_{1,1}$ is the matrix, the entries of which are $0$ except one of index $(1,1)$ which is $1$. Above all, we assume that the eigenvalues of $M$ are simple (then the $(\mu_i)_i$ are real analytic functions of $a$). Then $\dfrac{dx}{da}=-\dfrac{\dfrac{\partial\chi}{\partial a}}{\dfrac{\partial\chi}{\partial x}}$ and let $\rho_i=\dfrac{dx}{da}_{a=1,x=\lambda_i}$ (note that $\dfrac{\partial\chi}{\partial a}$ is constant). Finally $\mu_i\approx \lambda_i+(a-1)\rho_i$.

Answer 2

Multiplying with $D_a$ from left makes one singe elementary row operation on $M$. More specific $D_aM$ is different from $M$ just in the first row, and if I denote the first row with index 1, then if $M$ has the first row $M_1$ then $D_aM$ has $aM_1$. This property comes from the construct of matrix multiplication. You read more about elementary row operations here and here.

In general case elementary row operations will not preserve eigenvalues. There are special situations, for example if you get a similar matrix $B$ with row operations from $A$, so you get $A \sim B$. In this case the eigenvalues are the same, but eigenvectors not. But in general case row operations may completely change eigenvalues. You can find examples for this here. Because for small sizes we get the eigenvalues from determinants in general, you can find results in this topic here. Also this article could be interesting, but it is a little bit far from here.

In your special case you can get results only if $M$ has a propery that you can handle. For example if $M$ is a triangular matrix then one of its $\lambda$ eigenvalue changes to $a \lambda$. There are other matrix classes which you can treat by algebraic tricks, but in general I think you can say nothing about this mathematics.

Answer 3

Definitely, we can say that the eigenvalues of $D_aM$ approach those of $M$ as $a$ gets close to $1$. But I think one hardly can expect something beyond that. For instance, the matrix $$M=\begin{pmatrix}0&0&0&a\\1&0&0&b\\0&1&0&c\\0&0&1&d\\\end{pmatrix}$$ has charateristic polynomial $g(t)=t^4+at^3+bt^2+ct+d$ while scaling the first column of $M$ by $h$ results in the matrix with characteristic polynomial $g_h(t)=t^4+aht^3+bt^2+ct+d$. There seems no deeper relation between the roots of $g$ and $g_h$ than that they are close enough if $h\rightarrow1$.

The situation with eigenvectors can be even more complicated, say, the matrix $\left(\begin{smallmatrix}a&-1\\a&-1\end{smallmatrix}\right)$ has a unique eigenvector if $a=1$ and a pair of them otherwise.