18.06 Problem Set #3 Solutions

1. The set in (a) can’t be a basis because two vectors can span an at most 2 dimensional vector space, while ℝ^{3} is 3 dimensional. The sets in (b) and (d) can’t be bases since these sets cannot be linearly independent. (If the vectors are the columns of a matrix, then the matrix is 3 × 4, so has a non trivial null space, and the vectors have a non-trivial dependence relation.) For (c), we put the vectors in a matrix and row reduce.

A = [ \begin{matrix}3 & −1 & 0 \\ 2 & 2 & 1 \\ 2 & 1 & 0 \end{matrix} ] → ⋯ → [ \begin{matrix}3 & −1 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{matrix} ] .

Since the matrix is invertible, the columns of the original matrix form a basis. (A has full column rank so the columns span ℝ^{3}, and A has a trivial nullspace so the columns are linearly independent.)

2. We show that the set is a basis by showing that it spans M_{2} and is linearly independent. Denote the vectors in this set m_{1}, m_{2}, m_{3}, m_{4}.

Spanning: Let [ \begin{matrix}x & y \\ z & w \end{matrix} ] be any vector in M_{2}, and we look for constants a, b, c, d so that

am_{1} + bm_{2} + cm_{3} + dm_{4} = [ \begin{matrix}x & y \\ z & w \end{matrix} ] .

This gives

[ \begin{matrix}a + c & a + d \\ b + d & b + c + d \end{matrix} ] = [ \begin{matrix}x & y \\ z & w \end{matrix} ] .

This gives rise to a system of four equations in the four unknowns a, b, c, d. We get the augmented matrix

[ \begin{matrix}1 & 0 & 1 & 0 & x \\ 1 & 0 & 0 & 1 & y \\ 0 & 1 & 0 & 1 & z \\ 0 & 1 & 1 & 1 & w \end{matrix} ] .

Row reducing, we get the unique solution

a = x − w + z, b = 2z − y + x − w, c = w − z, d = y − x + w − z.

We conclude that the set above spans M_{2}.

Linear Indepencence: Suppose am_{1} + bm_{2} + cm_{3} + dm_{4} = 0, where 0 = [ \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ]. We get the same system of equations

as above with x, y, z, w all zero. But the coefficient matrix has full rank, so the nullspace is the zero vector. Hence am_{1} + bm_{2} + cm_{3} + dm_{4} = 0 only when a = b = c = d = 0. We conclude that the set is linearly independent.

3. (a) We have {(a, b, c, a + b)} = span{(1, 0, 0, 1), (0, 1, 0, 0), (0, 0, 1, 0), (0, 1, 0, 1)}. Since the last vector in this set is a linear combination of the other three, we can remove it. The resulting set is a basis, so the dimension is 3.

(b) The set {(a, b, a − b, a + b)} is the span of the linearly independent set {(1, 0, 1, 1), (0, 1, −1, 1)}. Hence the dimension is 2.

(c) The set {(a, a, c, d)} is the span of the linearly independent set {(1, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)}. Hence the dimension is 3.

(d) The set {(a+c, a−b, b+c, −a+b)} is the span of the linearly independent set {(0, −1, 1, 1), (1, 0, 1, 0)}. Hence the dimension is 2.

4. Row reducing A gives

[ \begin{matrix}0 & 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 4 & 6 \\ 0 & 0 & 0 & 1 & 2 \end{matrix} ] → ⋯ → [ \begin{matrix}0 & 1 & 2 & 0 & −2 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{matrix} ] .

The rank is 2. A basis for the row space consists of the non-zero rows of the echelon matrix, so B_{row} = {(0, 1, 2, 0, −2), (0, 0, 0, 1, 2)}. The nullspace has dimension 5−2 = 3. Assigning the value one or zero to the free variables x_{1}, x_{3}, x_{5} gives our basis B_{null} = {(1, 0, 0, 0, 0), (0, −2, 1, 0, 0), (0, 2, 0, −2, 1)}. A basis for the column space (which has dimension 2) is given by the columns of the original matrix that correspond to pivot columns in the echelon matrix. Hence B_{col} = {(1, 1, 0), (3, 4, 1)}. Finally, for the left nullspace, N(A^{T}), we can use the fact that N(A^{T}) is the orthogonal complement of the column space in ℝ^{3}. It’s easy to see that the (one dimensional) space of vectors orthogonal to the column space of A has basis B_{N(A^{T})} = {(1, −1, 1)}.

5. We show that this set spans ℝ^{n} and is linearly independent.Since A is invertible, given any vector b in ℝ^{n}, there is exactly one vector x in ℝ^{n} with Ax = b. But x can be written as a linear combination of the basis vectors v_{1}, …, v_{n}. That is, x = c_{1}v_{1} + ⋯ + c_{n}v_{n} for some constants c_{1}, …, c_{n}. But then we have

b = Ax = A(c_{1}v_{1} + ⋯ + c_{n}v_{n}) = c_{1}Av_{1} + ⋯ + c_{n}Av_{n}.

Hence, we have written b as a linear combination of the vectors Av_{1}, …, Av_{n}. Since b was an arbitrary vector in ℝ^{n}, The set {Av_{1}, …, Av_{n}} spans ℝ^{n}.

Now, suppose that there are constants d_{1}, …, d_{n} so that d_{1}Av_{1} + ⋯ + d_{n}Av_{n} = 0. Then

0 = d_{1}Av_{1} + ⋯ + d_{n}Av_{n} = A(d_{1}v_{1} + ⋯ + d_{n}v_{n}).

Since A is invertible, it has trivial nullspace, and we conclude that d_{1}v_{1} + ⋯ + d_{n}v_{n} = 0. But the set {v_{1}, …, v_{n}} is linearly independent, so we must have d_{1} = d_{2} = ⋯ = d_{n} = 0. But that says that if d_{1}Av_{1} + ⋯ + d_{n}Av_{n} = 0 then all the constants are zero. That is, the set {Av_{1}, …, Av_{n}} is linearly independent. Since this set is a linearly independent spanning set in ℝ^{n}, it is a basis.