Why Are My Gram-Schmidt Results Different on Wolfram|Alpha and MATLAB?

[nedf]

Wolframalpha provides this solution:
Wolfram|Alpha: Computational Knowledge EngineHowever when i compute the second term v2 = x2 - (x2.v1)/(v1.v1) * v1
The result is different from that of above. What's wrong?
Wolfram|Alpha: Computational Knowledge Engine)

 

[I like Serena]

Wolframalpha provides this solution:
Wolfram|Alpha: Computational Knowledge EngineHowever when i compute the second term v2 = x2 - (x2.v1)/(v1.v1) * v1
The result is different from that of above. What's wrong?
Wolfram|Alpha: Computational Knowledge Engine)

Hi nedf! Welcome to MHB! ;)

Did you take into account that the dot product for complex numbers requires a conjugation?
That is, $\mathbf a \cdot \mathbf b = \sum a_i\overline{b_i}$.

 

[nedf]

Hi nedf! Welcome to MHB! ;)

Did you take into account that the dot product for complex numbers requires a conjugation?
That is, $\mathbf a \cdot \mathbf b = \sum a_i\overline{b_i}$.

Thanks.
At the end of the page on https://www.mathworks.com/help/matlab/ref/dot.html
Why is the dot product defined this way instead? Wouldnt the answer be different?
$\mathbf a \cdot \mathbf b = \sum\overline a_i{b_i}$

Also, i computed the orth on matlab:

Why is it different from Wolfram|Alpha: Computational Knowledge Engine?

Is the orthonormal (normalized) basis unique for a given matrix?

 

[I like Serena]

Thanks.
At the end of the page on https://www.mathworks.com/help/matlab/ref/dot.html
Why is the dot product defined this way instead? Wouldnt the answer be different?
$\mathbf a \cdot \mathbf b = \sum\overline a_i{b_i}$

It's a matter of convention. Both dot products are valid inner products. And one is the conjugate of the other.
However, your formula for the Gram-Schmidt process assumes the $\sum a_i\overline {b_i}$ version.
Otherwise the dot product in the fraction should have been the other way around.
So the results will be the same - if we use the proper formula.
And your Gram-Schmidt formula is incompatible with mathworks's variant.

Note that with the standard $\sum a_i\overline{b_i}$ we have:
$$\mathbf v_1\cdot \mathbf v_2 = \mathbf v_1 \cdot\left( \mathbf x_2-\frac{\mathbf x_2 \cdot \mathbf v_1}{\mathbf v_1 \cdot \mathbf v_1}\mathbf v_1\right)
= \mathbf v_1 \cdot \mathbf x_2- \mathbf v_1 \cdot \left(\frac{\mathbf x_2 \cdot \mathbf v_1}{\mathbf v_1 \cdot \mathbf v_1}\mathbf v_1\right)
= \mathbf v_1 \cdot \mathbf x_2- \left(\frac{\mathbf x_2 \cdot \mathbf v_1}{\mathbf v_1 \cdot \mathbf v_1}\right)^*\mathbf v_1 \cdot \mathbf v_1 \\
= \mathbf v_1 \cdot \mathbf x_2- \left(\frac{\mathbf v_1 \cdot \mathbf x_2}{\mathbf v_1 \cdot \mathbf v_1}\right)\mathbf v_1 \cdot \mathbf v_1
= \mathbf v_1 \cdot \mathbf x_2 - \mathbf v_1 \cdot \mathbf x_2 = 0
$$

Also, i computed the orth on matlab:Why is it different from Wolfram|Alpha: Computational Knowledge Engine?Is the orthonormal (normalized) basis unique for a given matrix?

And no, an orthonormal basis is typically not unique.
Consider for instance $\mathbb R^3$.
The standard orthonormal basis is $\{(1,0,0),(0,1,0),(0,0,1)\}$.
But $\{(1,0,0),(0,1/\sqrt 2,1/\sqrt 2),(0,1/\sqrt 2,-1/\sqrt 2)\}$ is also an orthonormal basis.