# Eigenvalues of $A=vv^T$

Let $v\in\mathbb{R}^n$ a column vector and consider the matrix $A=v\cdot v^T$.

I need find all of auto-values of $A$. For the case $n=2$ and $n=3$, I was obtain that the only auto-values are $\lambda=x_1^2+x_2^2$ and $\lambda=x_1^2+x_2^2+x_3^2$ respectively (Where $v:=(x_1,x_2,\dots,x_n)$).

I think that this holds for any case, i.e., $\lambda=\sum_{i=1}^nx_i^2$ foe $n=1,2,...$ but I can't prove it.

My conjecture is right? And if this is the case, can someone give me a hint to solution the problem?

Note that $v$ is an eigenvector of $vv^T$ with eigenvalue $v^Tv=\sum_{i=1}^nx_i^2$. Also, note that the matrix $vv^T$ has rank at most $1$, so all other eigenvalues of $vv^T$ are $0$ by the rank-nullity theorem.
Let us compute $$A^2$$. We have $$A^2=(v v^T)(vv^T)=v(v^Tv)v^T.$$ Note that $$v^Tv\in \mathbb{R}$$ is scaler. Actually, $$v^Tv=|v|^2$$. Thus $$A^2=|v|^2vv^T=|v|^2A.$$ Thus $$A$$ satisfies the equation $$x^2=|v|^2x$$, which implies that its minimal polynomial is a factor of $$x^2-|v|^2x$$. Thus the two possible eigenvalues are $$0$$ and $$|v|^2$$. Moreover, if $$|v|^2\ne 0$$, $$A$$ is diagonalizabl. If $$|v|=0$$, $$A$$ is nilpotent. Furthermore, one can check that $$|v|^2=tr(A)$$. Thus the two possible eigenvalues of $$A$$ are $$0$$ and $$tr(A)$$. If $$tr(A)=0$$, $$0$$ is the only eigenvalue of $$A$$.
Note that the matrix $$A = vv^{T}$$ is symmetric. So, by Spectral Theorem for symmetric matrices $$A$$ has real eigenvalues and it is diagonalizable by an orthogonal matrix $$P$$ and a diagonal matrix $$D = \text{diag}(\lambda_{1}, \dots, \lambda_{n})$$, where the $$\lambda_{i}'s$$ are the eigenvalues of $$A$$. Also note that $$\lVert v \rVert^{2}$$ is an eigenvalue of $$A$$, since $$v \neq 0$$ and $$Av = (vv^{T})v = v(v^{T} v) = v(\lVert v \rVert^{2}) = \lVert v \rVert^{2} v.$$ Diagonalizing $$A$$ we have $$P^{-1}AP = P^{-1}(v v^{T})P = (P^{T}v)(P^{T}v)^{T} = ww^{T} = D,$$ where $$w = P^{T}v$$ is a column vector whose components are $$w_{1}, \dots, w_{n}$$. Since $$ww^{T}$$ is equal to a diagonal matrix we have that $$ww^{T} = \text{diag}(w_{1}^{2}, \cdots, w_{n}^{2}).$$ Also note that $$vv^{T}$$ and $$ww^{T}$$ have the same characteristic polynomial $$p(\lambda)$$, so we can write $$p(\lambda) = (w_{1}^{2}- \lambda) \cdots (w_{n}^{2} - \lambda) = (\lVert v \rVert^{2} - \lambda)q(\lambda),$$ for some polynomial $$q(\lambda)$$ of degree $$n - 1$$. This implies that $$w_{i}^{2} = \lVert v \rVert^{2}$$ for some $$i$$. In addition, since $$P$$ is orthogonal, the operator $$v \mapsto P^{T}v$$ is an isometry, then we have that $$\lVert w \rVert^{2} = w_{1}^{2} + \cdots + w_{n}^{2} = \lVert v \rVert^{2}$$. This implies that $$w_{j}^{2} = 0$$ for all $$j \neq i$$. So $$\lambda = \lVert v \rVert^{2}$$ is an eigenvalue of multiplicity 1 and $$\lambda = 0$$ is an eigenvalue of multiplicity $$n - 1$$. Then we can write $$p(\lambda) = (\lVert v \rVert^{2} - \lambda)(-1)^{n - 1}\lambda^{n - 1} = (-1)^{n}\lambda^{n - 1}(\lambda - \lVert v \rVert^{2})$$