# operator over inner product

Let $V$ be a finite-dimensional inner product space over $\mathbb{R}$ and let $u,v \in V$ be given. Define a linear operator $u\otimes v: V \rightarrow V$ by $(u\otimes v)x=<v,x>u$, where $<\cdot ,\cdot>$ denotes the inner product on $V$.

(a) Find the rank of $u\otimes v$.

(b)Find all eigenvalues of $u\otimes v$ and a basis for each eigenspace.

(c)Find the characteristic polynomial, determinant, and trace of $u\otimes v$.

I know the rank is 1, because $<v,x>$ is scalar. But I am confused about how to solve (b) and (c).

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Hint: For (b) try to describe $ker(u\otimes v)$ in terms of $v$: $$x\in ker (u\otimes v) \Leftrightarrow <v,x>u=0\Leftrightarrow x \in ...$$ And for (c) take the the Basis $B$ of eigenvectors that you computed in (b) and consider the matrix of $u\otimes v$ with respect to $B$.
I think we should consider describing $ker(u\otimes v-\lambda x)$. @walcher – 81235 Aug 16 '13 at 23:17
Theoretically yes, but we know $ker(u\otimes v)$, which is the eigenspace of the eigenvalue $\lambda =0$ has dimension $n-1$ (by rank-nullity), so this already covers almost the entire space. For the remaining eigenvalue try plugging in $u$ and see what happens. – walcher Aug 16 '13 at 23:30
$x \in ker(u\otimes v)\Leftrightarrow <v,x>u=0 \Leftrightarrow x\in {v}^{\perp}$, so how can you say dimension is $n-1$. @walcher – 81235 Aug 17 '13 at 0:15
$dim(v^{\perp})=?$ Also, we know $n=dim V=rank(f)+dim(ker(f))$ and $rank(u\otimes v)=1$. – walcher Aug 17 '13 at 0:21
A basis for $Eig(0)$ is a basis for $v^{\perp}$, I don't think you can be more specific about that. If $u \notin v^{\perp}$ (otherwise $Eig(<v,u>)=Eig(0)$), then $Eig(<v,u>)$ has basis $\{ u\}$. The characteristic polynomial is almost correct, except it is $\lambda ^{n-1}(\lambda -<v,u>)$ with trace $<v,u>$. – walcher Aug 17 '13 at 0:34