I'm having a little difficulty understanding Gram-Schmidt orthogonalization. I have a problem to apply Gram-Schmidt orthogonalization to the system of vectors $(1,1,1)^T, (1,2,1)^T$ then write the matrix of the orthogonal projection onto 2-dimensional subspace spanned by these vectors. After applying gram-schmidt I found $v_1 = (1,1,1)^T, v_2 = (-\frac{1}{3}, \frac{2}{3}, \frac{1}{3})^T$. I'm not sure what to do next to find the matrix of the orthogonal projection onto 2-dimensional subspace spanned by these vectors. Also, what exactly does gram-schmidt orthogonalization find? I'm just a little confused.
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asked Apr 26, 2015 at 17:45
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$\begingroup$ To start off , you should normalize your new vectors $\endgroup$– QualityCommented Apr 26, 2015 at 17:51
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$\begingroup$ okay so v_1 = $1/\sqrt{3}(1,1,1)^T and v_2 = $1/\sqrt{2/3}(-1/3,2/3,1/3)^T$ $\endgroup$– TheStrangeQuarkCommented Apr 26, 2015 at 18:01
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2 Answers
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Hint; your final vectors are not correct. The point of GS it to get an orthogonal set of vectors. Are yours orthogonal? You are starting off with two non orthogonal vectors , that is
$v_1=( 1 , 1 , 1)$ and $v_2= ( 1 , 2 ,1)$
The GS algorithm proceeds as follows;
let $w_1=(1,1,1)$
then we define $$w_2= v_2- \frac{\langle v_1 , w_1 \rangle}{\langle w_1 , w_1 \rangle} w_1$$
$$w_2=(1,2,1)-(4/3,4/3,4/3)=(-1/3,2/3,-1/3)$$ and it can be shown now that the set $$S=\{w_1,w_2\}$$ is orthogonal and also spans the same subspace as the original vectors v.
If we normalize S to say $$S_n=\{(1/3,1/3,1/3),(\frac{-1}{\sqrt6},\sqrt{\frac{2}{3}},\frac{-1}{\sqrt6})\}$$
In general to find the projection matrix P, you first consider the matrix A with your vectors from $S_n$ as columns, that is $$A=\begin{bmatrix} 1/3 & \frac{-1}{\sqrt6} \\ 1/3 & \sqrt{\frac{2}{3}} \\ 1/3 & \frac{-1}{\sqrt6} \\ \end{bmatrix}$$
that is, we will have the orthogonal projection matrix equal to,
$P=A(A^{T}A)^{-1}A^{T}$
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$\begingroup$ oops, I think I must have mistyped the second vector. And that makes more sense now the point of GS. how can you find the matrix for the projection though with these though? $\endgroup$ Commented Apr 26, 2015 at 18:04
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$\begingroup$ @TheStrangeQuark Great, and I will try to update it with more info soon $\endgroup$– QualityCommented Apr 26, 2015 at 18:05
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$\begingroup$ Yes, I want to write the matrix of the orthogonal projection onto 2-dimensional subspace spanned by these vectors $\endgroup$ Commented Apr 26, 2015 at 18:22
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$\begingroup$ Where does that final equation come from? $\endgroup$ Commented Apr 26, 2015 at 23:15
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$\begingroup$ I won't exactly post the derivation here, but you can find it on online. Anyways, it has to do with the orthogonal decomposition, $\endgroup$– QualityCommented Apr 26, 2015 at 23:43
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Gram-Schmidt finds, given a basis of a subspace, an orthonormal basis of this subspace. Its main tool is the following formula, which defines the orthogonal projection of a vector $\vec v$ onto another vector $\vec u$: $$p_{\vec u}(\vec v)=\frac{(\vec u,\vec v)}{(\vec u,\vec u)}\vec u.$$
What you should do:
Find an orthonormal basis of your subspace and complete it in an orthonormal basis of $\mathbf R^3$. In this orthonormal basis, the matrix of the orthogonal projection on the subspace (generated by the first two vectors) is simply: $$B=\begin{bmatrix} 1&0&0\\0&1&0\\0&0&0 \end{bmatrix}$$ Now the matrix of the projection in the original basis is given by: $$A=PBP^{-1}$$ where $P$ is the change of basis matrix from the original basis to the new one.
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$\begingroup$ But I only have 2 vectors, so P won't be a square matrix? $\endgroup$ Commented Apr 26, 2015 at 22:54
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$\begingroup$ You have to complete the orthogonal basis of the subspace into an orthogonal basis of $\mathbf R^3$ first. Thus you'll have a $3\times3$ matrix. $\endgroup$– BernardCommented Apr 26, 2015 at 23:22
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$\begingroup$ So I have to first find a change of basis matrix for this and the standard basis of R3? $\endgroup$ Commented Apr 26, 2015 at 23:24
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$\begingroup$ For some reason I'm blanking on how to do this. Why wouldn't that be a 2x3 matrix too, or it would have an all 0 column which makes it not invertable $\endgroup$ Commented Apr 26, 2015 at 23:35