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So this is what I have so far. Just two vectors trying to make an orthogonal basis out of them using Gram-Schmidt but when I inner product the results it isn't zero.

Here is what I have so far. Given vectors (both already normalized) are: $$ \begin{align*} v_1=\frac{1}{\sqrt{2}}\left(\begin{array}{c} i\\1\\0 \end{array}\right) &&v_2=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\0\\1 \end{array}\right) \end{align*}$$

So let $x_1=v_1$ thus using Gram-Schmidt, $x_2$ is $$ \begin{align*} x_2&=v_2-\frac{v_2\cdot x_1}{x_1\cdot x_1}x_1\\ &=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\0\\1 \end{array}\right)- \Bigg[\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\0\\1 \end{array}\right)^*\cdot \frac{1}{\sqrt{2}}\left(\begin{array}{c} i\\1\\0 \end{array}\right) \Bigg] \frac{1}{\sqrt{2}}\left(\begin{array}{c} i\\1\\0 \end{array}\right)\\ &=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\0\\1 \end{array}\right)- \frac{i}{2\sqrt{2}}\left(\begin{array}{c} i\\1\\0 \end{array}\right)\\ &=\frac{1}{2\sqrt{2}}\left(\begin{array}{c} 3\\-i\\2 \end{array}\right) \end{align*}\\ $$

But when I do the inner product with $x_1$ and $x_2$ I don't get zero, instead I get

$$ \begin{align*} x_1\cdot x_2 &= \frac{1}{\sqrt{2}}\left(\begin{array}{c} i\\1\\0 \end{array}\right)^*\cdot \frac{1}{2\sqrt{2}}\left(\begin{array}{c} 3\\-i\\2 \end{array}\right)\\ &= \frac{1}{4}\left(\begin{array}{c} i\\1\\0 \end{array}\right)^*\cdot \left(\begin{array}{c} 3\\-i\\2 \end{array}\right)\\ &=-i \end{align*} $$

Anyone know what I am doing wrong here? Thanks in advance :)

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For the complex inner product, note that $v_2\cdot x_1$ is not the same as $x_1\cdot v_2$: they are complex conjugates of each other. Here, you ended up using the wrong one (apparently) when you found that dot product to be equal to $\frac i2$. If you instead use $-\frac i2$, you'll get a resulting $x_2$ that is indeed orthogonal to $x_1$.

(In particular, you used $v_2^*x_1$ to calculate $v_2\cdot x_1$, but I think the right definition is $x_1^*v_2$.)

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  • $\begingroup$ So you are saying the G-S should be defined as this instead? $$x_2=v_2-\frac{x_1\cdot v_2}{x_1\cdot x_1}x_1$$? (IE just reverse the order of the inner product on the projection term) $\endgroup$ – Wikkyd Oct 27 '15 at 6:57
  • $\begingroup$ You should either reverse the order of the inner product, or modify the definition of inner product, but not both! $\endgroup$ – Greg Martin Oct 27 '15 at 21:50

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