# Proving that all eigenvalues are $0$ bar one.

A symmetric $n\times n$ matrix is given by $\mathbf{A}=\lambda \mathbf{ee}^{\text{T}}$ where $\mathbf{e}$ is a unit vector. Show that $\mathbf{A}$ has an eigenvector of $\mathbf{e}$ with eigenvalue $\lambda$ and that all $n-1$ other eigenvalues are $0$.

Working: $\mathbf{Ae}=\lambda \mathbf{e}(\mathbf{e}^{\text{T}}\mathbf{e})=\lambda\mathbf{e}$ so the first part is satisfied.

• I am stuck on showing that all other eigenvalues are $0$. I can't see why $\lambda$ cannot have multiplicity more than $1$, for instance.

Any help would be lovely.

• it's sufficient to prove that all $Ax=a\mathbf{e}$ May 16, 2014 at 12:47

Show that all vectors orthogonal to $e$ are in the kernel of $A$. If you pick a basis for that subspace, you get $n-1$ eigenvectors with eigenvalue $0$.

If $\lambda=0$ it is easy.

If $\lambda \neq 0$ then

$$rank(A) = rank(\lambda \mathbf{ee}^{\text{T}}) = rank(\mathbf{ee}^{\text{T}}) \leq rank(e) =1$$

For a matrix, if the rank of $A$ is $k$ then $0$ is an eigenvalue of multiplicity at least $n-k$ (result a.k.a. geometric multiplicity $\leq$ algebraic multiplicity).

Consider the matrix $A=ee^t$ for $e=(e_1,\dots,e_n)$ some no-zero column vector. Since $e\neq0$, there is an $i\in\{1,\dots,n\}$ such that $e_i\neq0$.

For each $j\in\{1,\dots,n\}$ which is not $i$, substract from the $j$th column of $A$ the result of multiplying the $i$th column by $e_j/e_i$. The resulting matrix has exactly one non-zero column, the $i$th, and obviously has rank $1$.

Since column operations do not change the rank, the rank of $A$ itself is $1$. If $\lambda\neq0$, then $\lambda A$ and $A$ have the same rank, so we are done.

• Or: Each column of $ee^T$ is a multiple of $e$, and since $e^T\ne0$, at least one is a non-zero multiple. This means that $e$ is an eigenvector of $A$, and $A$ has rank $1$ or $0$, depending on $\lambda$. The rank-nullity theorem gives us the rest. May 16, 2014 at 6:12

The matrix $A=ee^t$ is idempotent: $A^2=AA=(ee^t)(ee^t)=e(e^te)e^t=ee^t=A$ because $e^te=1$. It follows that the eigenvalues of $A$ are all $0$ or $1$.

Now immediate computation shows that $\operatorname{tr}A=\lVert e\rVert^2=1$, so that the eigenvalues, whose sum is the trace, have to be all zero except exactly one which is $1$.

Suppose $e$ is a unit column vector and let $A=ee^t$.

Let $\mathcal B=\{e_1,\dots,e_n\}$ be an orthonormal basis with $e_1=e$, and let $B$ be the matrix whose columns are $e_1$, $\dots$, $e_n$. Since $\mathcal B$ is an orthonormal basis, one can easily check that $B^tB=I$ so that $BB^t=I$ also and, therefore, $B$ is invertible with inverse $B^t$.

Now one can easily check that $B^{-t}AB=B^tAB=B^tee^tB$ is a matrix with all zeroes except one $1$ on the top left corner: it follows that $B^{-1}AB$ has eigenvalues $1$ with multiplicity $1$ and $0$ with multipicity $n-1$. . Since $A$ and $B^{-1}AB$ have the same eigenvalues, we get what we wanted.

The rank of $A=ee^t$ is the same thing a the dimension of the image of the map $f:v\in\mathbb R^n\mapsto Av\in\mathbb R^n$.

Now if $v\in\mathbb R^n$, we have $f(v)=Av=(ee^t)v=e(e^tv)$ (because matrix products are associative!) and $e^tv$ is a scalar, so $f(v)$ is a scalar multiple of $e$. It follows that the whole image of $f$ is contained in the subspace spanned by $e$ and therefore the dimension of that image of $f$, which is the rank of $A$, is at most $1$. Since it is not zero (because $A\neq0$), we see that it must be $1$.