How do I show that all remaining eigenvalues of this matrix is 0?

Question

Given some matrix $n\times n$ matrix $A = \lambda ee^T%$, where $e$ is one normalised $n \times 1$ eigenvector of A with eigenvalue $ \lambda$, I want to show that all remaining $n-1$ eigenvalues of A are zero.

My thoughts:

1. Determinant must be zero.

2. But this is only a necessary condition but not sufficient, since it only means any one of the eigenvalues is zero.

How should I proceed?

Answer 1

$A$ is a rank $1$ matrix, hence the nullity is $n-1$.

That is the dimension of nullspace is $n-1$. Hence we can find $n-1$ independent eigenvector corresponding to eigenvalue $0$.

Answer 2

Without loss of generality assume $\lambda\geq 0$; otherwise consider $-A$.

By $x^TAx = \lambda (e^Tx)^2\geq 0, \forall x\in \mathbb{R}^{n\times 1}$ we know that any eigenvalue of $A$ should be non-negative.

Now by $\lambda = \lambda\sum_{i=1}^ne_i^2=tr(A) = \sum_{i=1}^n \lambda_i = \lambda + \sum_{i=2}^n \lambda_i$, we know that $\lambda_i = 0$ for $2\leq i\leq n$.

Answer 3

Consider the vector space consisting of vectors orthogonal to $e$. This vector space has dimension $n-1$, and every vector in this space is an eigenvector with a $0$ eigenvalue. Therefore 0 is an eigenvalue with multiplicity $n-1$.