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

  • 4
    $\begingroup$ 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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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