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My professor gave this question:

$V$ is an inner product space, $v$ and $w$ are fixed vectors in $V$. Let $T:V\to V$ be a linear map such that $T(u)=(u,v)w$. Find $T^*$.

Now I need to find the rank of $T$ and to find the matrix that represents $T$ by the standard basis of $\mathbb{C}^n$? So I found $T^*$ and according to my calculation it is $T^*(x)=\overline{(w,x)}v$.

Now assuming I am not wrong with $T^*$, how can I find rank$(T)$? I know that $$\text{rank}(T)=\text{rank}(TT^*)=\text{rank}(T*T)$$ but does this help me?

I do not know what $v$ and $w$ are and without that how can I calculate what $T$ is doing the vectors of the basis in order to see how the matrix would look?

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1 Answer 1

up vote 2 down vote accepted

HINT: $(u, v)$ is always a scalar. You should be able to conclude on $Rank(T)$ from that, assuming that neither $v$ nor $w$ is the zero vector.

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is my T* is true ? –  wantToLearn May 16 '13 at 20:37
    
Seems correct to me. –  Narut Sereewattanawoot May 16 '13 at 20:39
    
The way I see it is that assuming w and v are not the zero vector then we know that each time we look at T(u) while is also different from zero we get cw (while c must be different then zero because v is not zero and so as u). Therefore we will always get a vector that is different then zero meaning the rank(T) = dim (V). right ? –  wantToLearn May 16 '13 at 20:47
    
rank should always be $1$ assuming that neither $v$ nor $w$ is zero, because you can only map elements to a linear combination of one vector. –  Narut Sereewattanawoot May 16 '13 at 20:51
    
ohhh ok so I am understand that dimImT=sp{w}. that is really make sense. but what would happen if if v is the zero vector you can not conclude from it that (u,v)=0. am I wrong ? –  wantToLearn May 16 '13 at 20:58

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