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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?

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    Oh, I thought that was the easy part. ;-) If $Ax=\lambda x$, then $(A+bI)x = Ax+bx=(\lambda+b)x$.2012-02-19

2 Answers 2

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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 $s^Ts+b,b,\ldots,b$.

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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?

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    1. 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.2012-02-19