Gram-Schmidt Process: The given set is a basis for a...

Gram-Schmidt Process: The given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.\\ a) $\begin{bmatrix} 3\\0\\-1 \end{bmatrix}$ and $\begin{bmatrix} 8\\5\\-6 \end{bmatrix}$\\ b) $\begin{bmatrix} 0\\4\\2 \end{bmatrix}$ and $\begin{bmatrix} 5\\6\\-7 \end{bmatrix}$

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00:01 Okay, so in this problem we're given two vectors and i'm going to call them x1 x2 we're told that they form a basis for a vector subspace w so they're linearly independent and they span w and we're asked to find an orthogonal basis for w and we're going to do that using the graham schmidt procedure so how does this work? this is the way i like to think of it so here's x1 x2 let's say right and this is plane of the whiteboard is the vector space w.

00:37 Okay, so the whole point of doing this is to find a new set of basis v1, v2 of w, such that v1 and v2 are perpendicular to each other, right? that's what it means for them to be orthogonal.

00:54 So, first off, i can call x1 the same as v1, right? that can survive in the new basis, right? unchanged that's fine what the problem is is that this x2 might not be perpendicular to the x1 so i'm gonna make it perpendicular so what i want is for my v2 to look like this so this is what i'm aiming for my v2 to be so this is v1 and this is v2 okay so the question is how do i get to v2 it's clear that i can get to it by taking x2 and adding to it a suitable multiple of v1.

01:46 Okay, i should be able to do that anyway because v2 i wanted to sit in the span of x1 and x2 so how can i do that? well, imagine i take this x2 and i project it right, i project it onto v1 and i get this vector here.

02:06 Well if i subtract this vector from x1, sorry, if i subtract this vector right here the projected vector from x2 so x2 minus this projected vector i'm going to get v2 okay so let me call this projected vector proge right okay so v2 is going to be x2 minus some minus this vector and what is this vector right here well first of all it's going to be in the direction of v1 so i'll put the unit vector for v1 right and what is the what is the length of this going to be let the angle between this be theta so it's going to be definitely whatever the magnitude the length of x2 is times the cosine of theta right x2 cosine theta gives you this now i want to bring in v1 original like the original v1 as opposed to the unit vector so clearly this is the same thing as v1 divided by the magnitude okay now finally i'm going to multiply top and bottom by the magnitude of v1 and that's going to give me this and this my friends you can recognize as simply being the dot product of v1 and x2 which is very simple to calculate given the components of the vectors so v2 will just be given by x2 minus x2 .v1 divided by square of the magnitude of v1 times v1.

04:06 So that's the intuition that i have, at least, behind the grammshman procedure.

04:11 Right, and it gets more sophisticated than this, right? but at least wanted to show that to you, because it might be helpful...

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