Eigenvalues of $s s^T + bI$

Question

Possible Duplicate:
eigenvalues and eigenvectors of $vv^T$

I'm reading an article concerning the matrix

$$s s^T + bI,$$

where $s$ is a vector of length $N$, $b$ is a real scalar and $I$ is the unit matrix. $s^T$ is the transpose of $s$.

The article states that the first eigenvalue is $E_s+b$, and the rest are $b$. $E_s$ denotes the signal energy of $s$, i.e.

$$E_s = s^2(1)+s^2(2)+\cdots+s^2(n)$$

How can these eigenvalues be found?

@Hans: The main problem I had was to understand how the second term affected the eigenvalues. — Mark, Feb 19 '12 at 16:35
Oh, I thought that was the easy part. ;-) If $Ax=\lambda x$, then $(A+bI)x = Ax+bx=(\lambda+b)x$. — Hans Lundmark, Feb 19 '12 at 21:23

score 1 · Answer 1 · answered Feb 19 '12 at 12:09

1

Can you see that each column of $ss^T$ is a multiple of the 1st column? Can you use this to find $n-1$ eigenvectors, each with eigenvalue zero? Do you know about the trace of a matrix, and its relation to the eigenvalues? Can you use that to find the last eigenvalue of $ss^T$? Do you understand the effect on the eigenvalues of adding $b$ to each diagonal entry of a matrix?

answered Feb 19 '12 at 12:09

Gerry Myerson

179,216

What do you mean by being a multiple of the first column? Each column of $ss^T$ is a multiple of $s$. For e.g. $s=(a,b)$, the columns are $(a^2, ab)$ and $(ab, b^2)$, where the latter is not necessarily an integer multiple of the former. – Mark Feb 19 '12 at 12:16
The sum of the eigenvalues should be $Tr(ss^T+bI)=E_s+nb$. I realise that $ss^T$ only has one non-zero eigenvalue -- please disregard the above comment. How does adding the diagonal matrix affect the eigenvalues except for changing their sum? – Mark Feb 19 '12 at 12:20
I wrote "multiple", not "integer multiple". This is linear algebra, not number theory. 2. For the result of adding $bI$, see Hans Lundmark's second comment.

Gerry Myerson

Feb 19 '12 at 22:57

score 1 · Accepted Answer · answered Feb 19 '12 at 14:26

Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $ss^T$. Note that $$ \ker((ss^T+bI)-\lambda I)=\ker(ss^T-(\lambda-b)I). $$ So $\lambda$ is an eigenvalue of $ss^T+bI$ if and only if $\lambda-b$ is an eigenvalue of $ss^T$.

Assume first that $s^Ts=1$. Then $ss^T$ satisfies $$ (ss^T)s=s(s^Ts)=s, $$ so $s$ is an eigenvector of $ss^T$ with eigenvalue 1. Now construct an orthonormal basis with $s$ as its first element. For any $t$ in the orthonormal basis, we have that $t$ is orthogonal to $s$, i.e. $s^Tt=0$. But then, $$ (ss^T)t=s(s^Tt)=0, $$ so we have obtained an orthonormal basis of eigenvectors, with eigenvalues $1,0,\ldots,0$.

Now, for general $s$, the above tells us that $\frac1{s^Ts}\,ss^T$ has eigenvalues $1,0,\ldots,0$, so $ss^T$ has eigenvalues $s^Ts,0,\ldots,0$.

In conclusion, $ss^T+bI$ has eigenvalues $s^Ts+b,b,\ldots,b$.

Eigenvalues of $s s^T + bI$

2 Answers2

Linked