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Prove that $x-\dfrac{\langle a,x\rangle}{\langle a,a\rangle}a$ is orthogonal to $a$.

I know this has something to do with the QR algorithm, but I am unsure of where to start.I started with QR decomposition and I am unsure of where to head next

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    $\begingroup$ You don't need the QR algorithm. Just use the definition of orthogonality. You'll prove it almost immediately. $\endgroup$ – user137731 Nov 29 '16 at 2:58
  • $\begingroup$ Hint: the (real) inner product is linear in both of its arguments $\endgroup$ – eepperly16 Nov 29 '16 at 3:12
  • $\begingroup$ oh, so I could try proving that matrix a times the first expression is equal to I? $\endgroup$ – 12345 Nov 29 '16 at 3:13
  • $\begingroup$ That's not the definition of orthogonal vectors... $\endgroup$ – Steve D Nov 29 '16 at 3:14
  • $\begingroup$ Here $a$ is evidently a vector belonging to the same inner product space as $x$, perhaps Euclidean $n$-space. $\endgroup$ – hardmath Nov 29 '16 at 3:15
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Recall that two vectors $v,w$ are orthogonal if $\langle v,w\rangle=0$. So, to prove that $v:=x-\frac{\langle x,a\rangle}{\langle a,a\rangle} a$ is orthogonal to $a$, we compute

$$\langle v,a\rangle = \langle x-\frac{\langle x,a\rangle}{\langle a,a\rangle} a,a \rangle.$$

But, recall that the inner product is linear, so we have

$$\langle x-\frac{\langle x,a\rangle}{\langle a,a\rangle} a,a \rangle = \langle x, a \rangle - \frac{\langle x,a\rangle}{\langle a,a\rangle} \langle a,a\rangle = \langle x,a \rangle -\langle x,a\rangle = 0.$$

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