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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?

Thanks for advance

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    $\begingroup$ Give definition of "autovalues" please. Same as eigenvalues? $\endgroup$
    – coffeemath
    Commented Jan 14, 2018 at 17:17
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    $\begingroup$ @coffeemath yes, they are the same. $\endgroup$
    – YCB
    Commented Jan 14, 2018 at 17:19

3 Answers 3

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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.

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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 $A$.

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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 $$ p(\lambda) = (\lVert v \rVert^{2} - \lambda)(-1)^{n - 1}\lambda^{n - 1} = (-1)^{n}\lambda^{n - 1}(\lambda - \lVert v \rVert^{2}) $$

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