Eigenvalues of $s s^T + bI$ 
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?
 A: 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?
A: 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 $s^Ts+b,b,\ldots,b$.
