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Assume $V$ is a real $n$-dimensional vector space, and $v,w \in V $. Define $ T \in L(V)$ by $ T(u) = u - (u,v)w$. Find a formula for Trace(T)

All I know about this is that trace is sum of the diagonal entries of the matrix. So how do I find the diagonal entries? I don't really know what steps to follow.

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  • $\begingroup$ You could apply $T$ to $e_i$ and then see what the coefficient of $e_j$ is in the result; that's $m_{ij}$. Do this for each $i,j$ and you've got the matrix. (Fortunately, you need only the $i,i$ entries!) $\endgroup$
    – John Hughes
    Dec 4, 2014 at 21:44
  • $\begingroup$ Do you mean to plug in $e_i$ to the definition of $T$? Thanks! $\endgroup$
    – user179313
    Dec 4, 2014 at 21:48
  • $\begingroup$ Yes; mfl's answer does this nicely. $\endgroup$
    – John Hughes
    Dec 5, 2014 at 3:38

2 Answers 2

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Despite mfl's nice answer, I want to propose an alternative one: trace is invariant under an orthonormal change of basis, so I can pick a different basis and compute using that basis instead.

My choice: $$b_1 = v / \| v \|$$ while $b_2, \ldots, b_n$ form an orthonormal basis for the hyperplane perpendicular to $v$.

Clearly $T(b_i) = b_i$ for $i = 2, \ldots n$. Hence there aren $n-1$ 1's on the diagonal. And $T(b_1) = b_1 - (b_1, v) w$; inner-producting with $b_1$ gives \begin{align} m_{11} &= (b_1, b_1) - \frac{1}{\|v \|} (v, v) (w, b_1) \\ &= 1 - \frac{1}{\|v \|} \|v\|^2 (w, v/\|v\|) \\ &= 1 - (w, v) \end{align} So the trace is $(n-1) + (1 - (w, v) ) = n - (w, v)$.

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  • $\begingroup$ Thanks for showing me another way to solve this problem! $\endgroup$
    – user179313
    Dec 5, 2014 at 4:13
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    $\begingroup$ It's a nice trick in general; can also be applied, in more or less the same form, to compute $\det (I + ab^t)$, where $a$ and $b$ are column vectors. Closely related to the Physics folks' rule of thumb that you should "always work in the right frame of reference." $\endgroup$
    – John Hughes
    Dec 5, 2014 at 11:36
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Let $(e_i)$ be an orthonormal basis. Then, it is

$$\mathrm{trace}(T)=\sum_{i=1}^n \langle Te_i,e_i\rangle =\sum_{i=1}^n\langle e_i-\langle e_i,v\rangle w,e_i\rangle =\sum_{i=1}^n(\langle e_i,e_i\rangle-\langle e_i,v\rangle \langle w,e_i\rangle)=n-\langle v,w\rangle.$$

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  • $\begingroup$ Thanks a lot. I'd like to ask you why we need inner product? $\endgroup$
    – user179313
    Dec 4, 2014 at 22:02
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    $\begingroup$ There is an inner product in the definition of $T:$ $T(u) = u - (u,v)w.$ $\endgroup$
    – mfl
    Dec 4, 2014 at 22:11
  • $\begingroup$ Thanks!, one last question why we have $\langle e_i,e_i \rangle = n $? $\endgroup$
    – user179313
    Dec 4, 2014 at 22:16
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    $\begingroup$ It is $\langle e_i,e_i \rangle = 1,i=1,\cdots,n.$ So, $\sum_{i=1}^n\langle e_i,e_i \rangle = n.$ $\endgroup$
    – mfl
    Dec 4, 2014 at 22:20
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    $\begingroup$ You're welcome. $\endgroup$
    – mfl
    Dec 4, 2014 at 22:27

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