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Please see my question at this page.

Now let $V=\mathbb{C}^n$. What is the rank of linear operator $T$? and what about its eigenvalues?

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SKMohammadi
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  • What have you tried? If you tell us this then we will be better able to help you. And it helps us feel that we are not just doing your homework for you. – user1729 Aug 07 '14 at 13:29
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    Scratch that - the OP has answered the question themselves, below, so I do not think that this question should be closed as "off topic". However, @MathMan, you should make your questions as self-contained as possible. So here, don't just link to the other question but include all the necessary information required to pose the problem, and then provide the link to give context. – user1729 Aug 07 '14 at 13:35

2 Answers2

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Hint. All elements of the range $T[V]$ of $T$ are multiples of the fixed vector $\gamma$. So $T[V] \subseteq \mathbb C \cdot \gamma$. Hence the rank is $0$ or $1$ depending on $\gamma$ and $\beta$.

For the eigenvalues, we allready know that zero is an eigenvalue with multiplicity at least $n-1$. The only vector which can be an eigenvector for another eigenvalue is $\gamma$ (if it is not equal to $0$). So look at $T\gamma$.

martini
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  • Can we use eigenvalue definition $T(v)=\lambda v$ for this problem? – SKMohammadi Aug 07 '14 at 10:59
  • You can $Tv = (v,\beta)\gamma$. When is this a multiple of $v$? (Note that it is always a multiple of $\gamma$! – martini Aug 07 '14 at 11:02
  • Dear @martini, i do not understand your hint! what is mean $T[V]\subseteq \mathbb{C}\cdot\gamma$? and why that zero is an eigenvalue with multiplicity at least $n−1$? – SKMohammadi Aug 07 '14 at 11:11
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With thanks to @martini, as he have mentioned above, $T[V]\subseteq \mathbb{C}\cdot \gamma$ and if we suppose $\gamma \neq 0 $, $T$ must be has rank $1$. Therefore zero is an eigenvalue of multiplicity $n-1$ for it. To find another eigenvalue, we must look at $T\gamma$:

$$ T[\gamma] = \lambda\gamma \quad \Rightarrow \quad (\gamma,\beta)\gamma = \lambda\gamma. $$ and so another eigv is $\lambda = (\gamma,\beta)$.

SKMohammadi
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