Solutions to Simple Linear Algebra Problems (1)

 1) Show the following vectors are linearly independent:

a) (π, 0) and (0, 1)

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

Solutions:

 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)

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)

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

Solutions:

Write:  a (1,1) + b(0, 1) = (1, 0)

Which leads to:

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)

Write the linear combination:

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

This leads to:

2a - b = 1

a + 0 = 1

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

The coordinates are (1,1)

3) Show the following vectors are linearly independent over C or R:

a)    (1, 1, 1) and (0, 1, -2)

b)    (-1, 1, 0)  and (0, 1, 2)

c)     (0, 1, 1),  (0, 2, 1) and (1, 5, 3)

Solutions:

To show linear independence we require the sum of the linear combination to be zero, and hence a = 0, b = 0, c=0 in each case 

a) Write out: a (1,1, 1) + b(0, 1, -2) = (0, 0, 0)

Then: a1 + b 0  =  0,  a1 + b1 = 0,  a1 + b2 = 0

Or in normal form:

a          = 0

a + b   = 0

a - 2b = 0

From which we find a = 0 and also b = 0. Hence, linear independence applies.

b) Write out: a (-1,1, 0) + b(0, 1, 2) = (0, 0, 0)

Then: - a1 + b 0  =  0,  a1 + b1 = 0,  a0 + b2 = 0

Or in normal form:

- a   + 0  = 0

a + b   = 0

0  - 2b = 0

From which we find a = 0 and also b = 0. Hence, linear independence applies.

c)Write out:

 a (0, 1, 1) + b(0, 2, 1) + c(1, 5 , 3 ) = (0, 0, 0)

Then: a0 + b0 + c1 = 0, a1 + b2 + c5 =0, a1 + b1 + c3 =0

Or in normal form:

0   + 0  + c   = 0

a + 2b + 5c = 0

a  + b  + 3c  = 0

From which we can see:  c= 0, a = 0,  b = 0

Hence, linear independence applies.