1) a) Find a basis for the solution space of:

b) What is the dimension of this basis?

(5)

2) State which of these sets of vectors are linearly dependent and say why.

a) u = v =	b) u = v =
c) u = v =	d) u = v =

(8)

3) If u, v, and w is a set of linearly independent vectors in vector space V,
prove that (u + v), (u – v), and (u – 2v + w) are also linearly independent vectors in V.

(3)

4) Given that S = {v₁ , v₂ , ......v_m} contains a linearly dependent subset R = {v₁ , v₂ , ......v_r}
where r m, prove that S is a set of dependent vectors.

(4)

5) Which of these sets of vectors form a basis for R³. If they don't, state why.

a) ,

b), ,

c), ,

(5)

TOTAL (25)

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Solutions

1) when we row reduce the homogeneous system of 3 equations in 5 variables -- we get:

We have 2 equations in 5 unknowns
so we need 3 parameters -- r, s, and t.

The solutions are:

The basis vectors for the nullspace are:

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2) b) and d) are dependent vector sets.
In b) v = 3u, and in d) v = – ½ u.

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3) Proof:

Since { u, v, w } is a linearly independent set, we know that if
k₁u + k₂v + k₃w = 0, then k₁ = 0, k₂ = 0, and k₃ = 0.

Now we consider the set of vectors
{ u + v, u – v, u – 2v + w }.

Assume that l₁ (u + v) + l₂ (u – v) + l₃ (u – 2v + w) = 0.

We remove brackets and rearrange the coefficients of u, v, and w to get:

(l₁ + l₂ + l₃) u + (l₁ – l₂ – 2l₃) v + l₃ w = 0

And since u, v, and w is linearly independent,

( l₁ + l₂ + l₃ ) = 0 ( l₁ – l₂ – 2l₃ ) = 0 and l₃ = 0
Solving this system gives l₁ = l₂ = l₃ = 0
therefore the vectors (u + v), (u – v) and (u – 2v + w)
are linearly independent.

4) Since R is a dependent set,
if k₁v₁ + k₂v₂ + k₃v₃ + ...... k_rv_r = 0, some of the k's are not = zero.
Now we consider l₁v₁ + l₂v₂ + l₃v₃ + ...... l_rv_r + l_{r +1}v_{r +1} + ..... + l _mv_m = 0
Since the first r vectors are from set R, some of the l 's are not = zero.
therefore S is linearly dependent.

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5) a) not a basis -- only 2 vectors.

b) yes, since det A = 5.

c) no, since det A = 0.

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