## Friday, December 16, 2011

### Linear Algebra Solutions

These are from the last two problem sets, first ...from:
http://brane-space.blogspot.com/2011/12/more-linear-algebra-vectors-spaces.html

1) Show the following vectors are linearly independent:

a) (π, 0) and (0, 1)

Solution: We write out: a (π, 0) + b (0, 1) = (0,0)

i.e. to show linear independence we require the sum of the linear combination to be zero, and hence a = 0, b = 0.

Then:

πa + b(0) = 0

a(0) + b = 0

from which we see: b = 0 and also a = 0

Hence, linear independence applies.

b) (1, 1, 0), (1, 1, 1) and (0, 1, -1)

Solution:

We are to generalize to 3 dimensions here so":

a(1,1,0) + b(1,1,1) + c(0, 1, -1) = (0, 0 , 0)

Whence we find:

1) a + b = 0

2) a + b + c = 0

3) 0 + b - c = 0

Subtract (1) from (2): c = 0

From (3): b = c = 0

From (1): a = 0

Hence, linear independence applies

2) Express X as a linear combination of the given vectors A, B and find the coordinates of X with respect to A, B:

a) X = (1, 0), A = (1, 1), B = (0, 1)

Solution: Write -

a (1,1) + b(0, 1) = (1, 0)

a + b(0) = 1

a + b = 0

Thus: a = 1 and b = -a = -1

So the coordinates are: (1, -1)

b) X = (1,1), A = (2, 1), B = (-1, 0)

Solution:

Write the linear combination:

a (2,1) + b(-1, 0) = (1, 1)

2a - b = 1

a + 0 = 1

Therefore: a = 1 and b = 2a - 1 = 2(1) -1 = 1

The coordinates are (1,1)

Next problem set from the end of:

http://brane-space.blogspot.com/2011/12/more-linear-algebra-finding-orthonormal.html

(1) Find the orthonormal basis for the subspaces of R^3 generated by the vectors:
A = (2, 1, 1) and B = (1, 3, -1)

Solution:

The orthonormal basis for A is just:

A/ ‖A ‖ = (2, 1, 1)/ {2^2 + 1^2 + 1^2} = (2, 1, 1)/ [6]^½

The orthonormal basis for B is:

B/ ‖B ‖ = (1, 3, -1)/ {1^2 + 3^2 + (-1)^2} = (1, 3, -1)/ [11]^½

(2) Find the orthonormal basis for the subspaces of R^4 generated by the vectors:

A = (1, 1, 0, 0)B = (1, -1, 1, 1)C = (-1, 0, 2, 1)

Solution:

For A we have:

A/ ‖A‖ = (1, 1, 0, 0)/ [1^2 + 1^2 ]^½ = (1, 1, 0, 0)/ [2 ]^½

For B we have:

B/ ‖B‖ = (1, -1, 1, 1)/ [1^2 + (-1)^2 +1^2 + 1^2]^½

= (1, -1, 1, 1)/ [4 ]^½ = (1, -1, 1, 1)/ 2

For C we have:

C/ ‖C‖ = (-1, 0, 2, 1)/ [(-1)^2 + (2)^2 +1^2 + 1^2]^½

so:

C/ ‖C‖ = (-1, 0, 2, 1)/ [6 ]^½

(3) Find an orthonormal basis for the subspace of the complex space C^3 generated by the vectors: A = (1, -1, i) and B = (i, 1, 2)

For A we have:

A/ ‖A‖ = (1, -1, -i)/ [1^2 + (-1)^2 + (-i)]^½

But i is a complex number, i = [-1]^½ and therefore

(-i)^2 = -[-1]^2 = 1

So:

A/ ‖A‖ = (1, -1, i)/ [1^2 + (-1)^2 + 1]^½ = (1, -1, i)/ [3]^½

For B we have:

B/ ‖B‖ = (i, 1, 2)/ [(i)^2 + 1^2 + 2^2]^½=

(i, 1, 2)/ [-1 + 1 + 4]^½ = (i, 1, 2)/ [4]^½ = (i, 1, 2)/ 2