Eigenvalues of $I + v v^T$

Question

I'm trying to calculate the eigenvalues of an $N\times N$ matrix $M$ whose matrix elements are $m_{ij} = \delta_{ij} + c$, where c is a real constant. Anybody knows if there is a closed formula for this?

Notice that the matrix can also be written as $M=I + v v^T$, where the vector $v$ has elements $v_i = \sqrt{c}$.

I have search through the stack.exchange questions with the words "diagonalization" and "low-rank" in the title, but they seem to concern more general cases than the simple one I'm interested in.

Thank you!

Answer 1

Note that if $\langle v, w \rangle = 0$ then $Mw = w$. This accounts for $N-1$ eigenvalues $1$. Finally $M v = (1+\lVert v \rVert^2) v$ so this gives the final eigenvalue.

Answer 2

Let $\mathbf 1$ be the vector with each element being $1$. $$(I+c\mathbf 1\mathbf 1^T)\mathbf 1=(1+cN)\mathbf 1$$ For each one $x$ of the $N-1$ indepedent vectors orthogonal to $\mathbf 1$ $$(I+c\mathbf 1\mathbf 1^T)x = x$$ So the eigenvalues of $I+c\mathbf 1\mathbf 1^T$ are $1+cN$ with multiplicity $1$, and $1$ with multiplicity $N-1$.

Answer 3

The $N\times N$ matrix $M=I+cvv^T$ (where $v$ is a vector and $c$ a scalar) has eigenvalue $1$ with multipilicity $N-1$ and the eigenvalue $\lambda=1+c\|v\|^2$ with multipicity $1$. The corresponding eigenvector is $v$, which upon multiplication by $M$ yields $$Mv = v + (cvv^T)v= v + cv(v^Tv)=v+cv\|v\|^2 = (1+c\|v\|^2)v.$$ In the special case where $v$ is the vector of all $1$ recovers the formula for your $m_{ij}=\delta_{ij}+c$ case.