Eigenvalues of the following matrix

Question

I have to find the eigenvalues of the matrix

$$ I+xx^T/||x||^2 $$

With x being a column vector of size n and I an identity matrix of size nxn. The eigenvalues are supposed to be easily found and be $\lambda_n= 2$ and $\lambda_1=1$, with the rest of eigenvalues inside that range

Any help would be highly appreciated

Answer 1

Note that $xx^T/||x||^2$ is a matrix with rank $1$, so there is exactly a non-zero eigen-value and zero eigen-values with multiciplity $n-1$. Since $(xx^T/||x||^2)x=x$ then the non-zero eigen value of $xx^T/||x||^2$ is $1$. Moreover, since for any matrix $A$, $\lambda$ is eigen value of $A$ iff $1+\lambda$ is eigen value of $I+A$ then the eigen values of $I+xx^T/||x||^2$ are $2$ with multiplicity $1$ and $1$ with multiplicity $n-1$.

Answer 2

We can divide the space into $x$ and $n-1$ vectors that are orthogonal to $x$.

For $x$ we have that $(I+\frac{xx^T}{||x||^2})x = x + \frac x {||x||^2} x^Tx = 2x$.

For any $x^\perp$ we have $(I+\frac{xx^T}{||x||^2})x^\perp = x^\perp + \frac x {||x||^2} x^Tx^\perp = x^\perp$.

So we have one eigenvalue 2 with eigenvector $x$ and we have (n-1) eigenvalues 1 with corresponding eigenvectors perpendicular to $x$.