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Let $A$ and $B$ are real, square matrices with the same dimension. We know that $\text{rank } A = 1$ and we know the eigenvalues of $A$. Furthermore, we know that $B$ has only zeros in the diagonal, but we don't know anything else about $B$.

Can we say something about the eigenvalues of $A+B$ ?

Or in general, can we say anything about the eigenvalues of sum of two arbitrary non-symmetric matrices?

user153012
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2 Answers2

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First let us note that since $\text{rank}(A)=1$, $0$ is an eigenvalue of $A$ of multiplicity $n-1$. Then the eigenvalues of $A$ must be $0,0,0,..,0, \text{tr}(A)$.

Therefore, since $\text{rank}(A)=1$, saying that we know the eigenvalues of $A$ is equivalent to saying that we know $\text{tr}(A)$.

Moreover, as $\text{tr}(A+B)=\text{tr}(A)+\text{tr}(B)=\text{tr}(A)$, the sum of eigenvalues of $A+B$ must be $\text{tr}(A)$.

Let $a_1,\cdots,a_n\in\mathbb{R}:a_1+\cdots+a_n =\text{tr}(A)$. Then, $a_1,\cdots,a_n$ are the eigenvalues of $A+B$, where

$$A= \begin{bmatrix} a_1 & a_1 & \cdots & a_1 \\ a_2 & a_2 & \cdots & a_2 \\ \vdots & \vdots & \ddots & \vdots \\ a_n & a_n & \cdots & a_n \end{bmatrix} \\ B= \begin{bmatrix} 0 & -a_1 & \cdots & -a_1 \\ -a_2 &0 & \cdots & -a_2 \\ \vdots & \vdots & \ddots & \vdots \\ -a_n & -a_n & \cdots &0\end{bmatrix} $$

Moreover, because of the above comments, $A$ has the given eigenvalues.

We finish by covering the case $a_1=\cdots=a_n=0$. They are eigenvalues of $A+B$, where $$A= \begin{bmatrix} 0 & 0 & \cdots & 1 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix} \\ B= 0 $$

Pockets
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N. S.
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  • The fact that $A$ has rank 1 doesn't mean, that each column are the same. – user153012 May 25 '14 at 22:07
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    @user153012 First, this is an example, so it is irrelevant if the general matrix has this property or not.... . And second, if a matrix has rank $1$ you can prove that any two rows are proportional... And the same about the columns... So it actually means that the columns are proportional (not equal).... But as I said, this is irrelevant. – N. S. May 25 '14 at 22:15
  • Given $\text{rank}(A)=1$, how can $a_1=\cdots=a_n=0$ be a case? How do you get $a_{1,n}=1$ in that case? – Pockets May 25 '14 at 22:21
  • @SamuelLijin What is the rank of the matrix below? What are the eigenvalues? $$A= \begin{bmatrix} 0 & 0 & \cdots & 1 \ 0 & 0 & \cdots & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \cdots & 0 \end{bmatrix} \$$ – N. S. May 26 '14 at 01:25
  • @SamuelLijin And there are many other nilpotent matrices of rank one.... That is just one such example, you don't get $a_{1,n}=1$ in all examples. – N. S. May 26 '14 at 01:26
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We note that generally we have that:

$$\operatorname{Tr}(\mathbf{M})=\sum_{i=1}^{n}\lambda_{i},$$

Where $\lambda_{i}$ is the $i$th eigenvalue of $\mathbf{M}$ (a proof can be found here). Therefore in your case we have:

$$\operatorname{Tr}(\mathbf{A}+\mathbf{B})=\sum_{i=1}^{n}\lambda_{i}$$

Where $\lambda_{i}$ is the $i$th eigenvalue of $\mathbf{A}$. This is the case because the sum of eigenvalues of $\mathbf{B}$ must be $0$ as $\operatorname{Tr}(\mathbf{B})=0$ as we are told.

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Thomas Russell
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  • Yes, but sadly these are just the sum of the eigenvalues, but we don't know anything about what are the eigenvalues. – user153012 May 25 '14 at 21:27