Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question

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 $B$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.