1. ## Gram-Schmidt Problem

Hey, I was wondering if any of you could look at this problem for me, to be specific, problem 2c (https://234ae6be-a-62cb3a1a-s-sites....attredirects=0)

is the problem I am having trouble on. I understand how Gram -Schmidt works, I am just wondering if I am doing something incorrectly because I've been working on this problem for the last couple hours and been turning up with ridiculous answers. Note my professor does not allow the use of calculators so I am surprised to these ridiculous answers for a final exam review.

2. ## Re: Gram-Schmidt Problem

in order to actually DO (2c) one needs to find a maximal linearly independent subset of the 4 vectors. it's clear that the 2nd vector is NOT a scalar multiple of the 1st, so the first 2 vectors form a linearly independent subset.

also, if a(2,3-1,4) + b(0,1,1,3) + c(1,0,0,3) = (0,0,0,0), then:

2a+c = 0
3a+b = 0
-a+b = 0
4a+b+3c = 0

subtracting the 3rd equation from the 2nd gives 4a = 0, so a = 0, and it is then obvious that b = 0 (from equation 2 or 3) and c = 0 (from equation 1). thus the first 3 vectors form a linearly independent set.

finally, observe that (2)(2,3,-1,4) + (1)(0,1,1,3) + (-3)(1,0,0,3) = (4,6,-2,8) + (0,1,1,3) + (-3,0,0,-9) = (1,7,-1,2) which is the 4th vector, so the first 3 vectors form a maximally independent subset of the span of your 4 vectors, and thus a basis.

so dim(V) = rank(A) = 3, so the dimension of the kernel is 1.

so...let's apply gram-schmidt to get an orthogonal basis for V. after, we've done that, we'll normalize each vector to unit length.

for our first vector, we can pick any one of the 3, so let's pick (1,0,0,3) (it has the most 0's, so the arithmetic may be simpler).

the find our second basis vector, let's find the projection of (0,1,1,3) in the direction of (1,0,0,3), this is:

$\displaystyle \frac{(1,0,0,3) \cdot (0,1,1,3)}{(1,0,0,3) \cdot (1,0,0,3)}(1,0,0,3) = \left(\frac{9}{10},0,0,\dfrac{27}{10}\right)$

now we need to subtract this from (0,1,1,3): (0,1,1,3) - (9/10,0,0,27/10) = (-9/10,1,1,3/10) <---this is our 2nd basis vector.

finally, we need to find the projections of our third vector in the directions of the other two. this gives:

$\displaystyle \frac{(1,0,0,3) \cdot (2,3,-1,4)}{(1,0,0,3) \cdot (1,0,0,3)}(1,0,0,3) = \left(\frac{7}{5},0,0,\frac{21}{5}\right)$ and

$\displaystyle \frac{(-9/10,1,1,3/10) \cdot (2,3,-1,4)}{(-9/10,1,1,3/10) \cdot (-9/10,1,1,3/10)}(-9/10,1,1,3/10) = \left( \frac{-63}{145},\frac{14}{29},\frac{14}{29},\frac{21}{145 } \right)$

we need to subtract both of these from (2,3,-1,4), so we may as well add them, first, and subtract the result. their vector sum is: (28/29,14/29,14/29,126/29)

so (2,3,-1,4) - (28/29,14/29,14/29,126/29) = (30/29,73/29,-43/29,-10/19) <---3rd basis vector

as a sanity check, let's verify that this is orthogonal to (1,0,0,3) and (-9/10,1,1,3/10):

(1,0,0,3).(30/29,73/29,-43/29,-10/19) = 30/29 + 0 + 0 - 30/29 = 0. check.

(-9/10,1,1,3/10).(30/29,73/29,-43/29,-10/19) = -270/290 + 730/290 - 430/290 - 30/290 = 0. check.

so the basis {(1,0,0,3),(-9/10,1,1,3/10),(30/29,73/29,-43/29,-10/29)} is orthogonal. now just normalize each vector (yes, the norms of the last two will be ugly, but you can leave them in "square root" form).

to make things easier, observe it makes no difference if you replace any basis vector with a scalar multiple of it before normalization, so you can use the basis: