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Let $v \cdot w$=$x_1y_1+2x_2y_2+3x_3y_3+4x_4y_4$, for every $v,w\in \mathbb{R}^4$. This is a dot product. Give the orthonormal basis of the linear subspace vct $\{(1,0,1,0),(1,0,0,1)\}$ of $\mathbb{R}^4$ (relative to this dot product).

So I found a basis, which is $\left(\frac{1}{\sqrt2},0,\frac{1}{\sqrt2},0\right)$, $\left(\frac{1}{\sqrt6},0,\frac{-1}{\sqrt6},\frac{2}{\sqrt6}\right)$.

I don't know how I can make this relative to that dotproduct.

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    $\begingroup$ What does "vct" mean? $\endgroup$
    – 5xum
    Jan 18, 2017 at 13:23
  • $\begingroup$ You can always get an orthonormal by applying the Gram-Schmidt orthogonalization method to any basis and then normalizing the vectors of the obtained orthogonal basis! $\endgroup$ Jan 18, 2017 at 13:26

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Hint:

$\{(1,0,1,0),(1,0,0,1)\}$ is already a basis on its own. Now, you need to make it orthonormal. For that, you first need to make it orthogonal.

Use the Gramm-Schmidt procedure for that. But use the new dot product to calculate the appropriate dot products.


Reminder:

The Gramm Scmidt process takes a linearly independent set $\{w_1,w_2\dots, w_n\}$ and returns a set of pairwise orthogonal vectors $\{u_1,u_2,\dots, u_n\}$ that spans the same space.

It calculates $u_1$ as $u_1 = w_1$, now try to remember how $u_2$ is calculated.

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Let us do a Gram-Schmidt: $$ a_1 = (1,0,1,0)^T \quad a_2 = (1,0,0,1)^T $$ then $$ \lVert a_1 \rVert^2 = a_1 \cdot a_1 = 1^2 + 3\cdot 1^2 = 4 $$ and $$ b_1 = a_1 / \lVert a_1 \rVert = (1,0,1,0)^T/2 $$ Further we take the part in $b_1$ direction away: $$ b_2 = a_2 - (a_2 \cdot b_1) b_1 =(1,0,0,1)^T - (1^2/2)(1,0,1,0)^T/2 = (3/4,0,-1/4,1)^T $$ and we also normalize: $$ \lVert b_2 \rVert^2 = 9/16 + 3\cdot (1/16) + 4 \cdot 1^2 = 12/16 + 4 = 3/4 + 4 = 19/4 $$ and get $$ c_2 = b_2 / \lVert b_2 \rVert = (3/4,0,-1/4,1)^T (2/\sqrt{19}) = (3/2,0,-1/2,2)^T / \sqrt{19} $$ and your sought ON basis is $\{ b_1, c_2 \}$.

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