Notes on linear algebra

(x) (2 3) (a)
(y-2x) = (0 1) (b)
Or 2a + 3b = x; 0a + 1b = y

Therefore b = y.

2a + 3b = x  a = (x-3b)/2 = (x-3y)/2

So, given a vector (x,y) we can find a,b such that (x,y) = aW1 + bW2. So W1, W2 is a basis!

SECOND EXAMPLE


(2) (4)
W1 = (4) W2 = (8)
Then we must solve


(x) (2) (4)
(y) = a (4) + b (8)
Then


(x) (2a) (4b)
(y) = (4a) + (8b)

Hence
(x) (2 4) (a)

(y) = (4 8) (b)
NOW WE DO GAUSSIAN ELIMINATION:

So, what should we multiply the first row by? We need 2m + 4 = 0, so m = -4/2 = -2.

Hence we get

(x) (2 4) (a)
(y-2x) = (0 0) (b)

So, the two equations we must solve are:

2a + 4b = x and 0a + 0b = y - 2x

Now, regardless of what a and b are, 0a + 0b is always zero. If y - 2x is not zero, it will be impossible to solve the second equation. So, can we find x and y such that y - 2x  0? Sure. Take x = 0, y non-zero. Or take x nonzero, y = 0. Or take y = 22, x = 12. Almost any choice works.

So we see that W1, W2 are not a basis. We ended up with a row of zeros. Let’s look at our two vectors again:

(2) (4)
W1 = (4) W2 = (8)

Notice that

(2*2) (2)
W2 = (2*4) = 2 (4) = 2W1

Not only are W1 and W2 not a basis, but they lie on the same line!

THIRD EXAMPLE


(1) (2) (0)
W1 = (2) W2 = (2) W3 = (1)
(1) (2) (1)
Are W1, W2, W3 a basis?
(x) (1) (2) (0)
(y) = a (2) + b (2) + c (1)
(z) (1) (2) (1)
(x) (1a) (2b) (0c)