Question

Find a basis for the subspace W of R4, spanned by the set of vectors V1 {[1 1 0 -1]}, V2 {[0 1 2 1]}, V3 {[1 0 1 -1]}, V4 {[1 1-6 -3]}
and V5 {[-1 -5 1 0]}

What is the dimension of W?
Let W be the subspace of P3 spanned by:
{t^3 + t^2 -2t +1 , t^2 + 1 , t^3 - 2t , 2t^3 +3t^2 -4t +3}.
Find a basis for W. What is the dimension of w?

Answer 1

Answer on Question #56702 – Math – Linear Algebra

Question

Find a basis for the subspace W of R4, spanned by the set of vectors V1 {[1 1 0 -1]}, V2 {[0 1 2 1]}, V3 {[1 0 1 -1]}, V4 {[1 1-6 -3]} and V5 {[-1 -5 1 0]}.

Solution

Given the set

W = \left\{ \begin{array}{c} 1 \\ 1 \\ 0 \\ -1 \end{array} \right., \quad \begin{array}{c} 0 \\ 1 \\ 2 \\ 1 \end{array}, \quad \begin{array}{c} 1 \\ 0 \\ 1 \\ -1 \end{array}, \quad \begin{array}{c} 1 \\ 1 \\ -6 \\ -3 \end{array}, \quad \begin{array}{c} -1 \\ -5 \\ 1 \\ 0 \end{array} \right\}

of vectors in the vector space $\mathbb{R}^4$, find a basis for span S.

The set $W = \{ \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4, \mathbf{v}_5 \}$ of vectors in $\mathbb{R}^4$ is **linearly independent** if the only solution of

$$(*) \quad c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 + c_4 \mathbf{v}_4 + c_5 \mathbf{v}_5 = 0$$ $$c_1 \begin{bmatrix} 1 \\ 1 \\ 0 \\ -1 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \\ 2 \\ 1 \end{bmatrix} + c_3 \begin{bmatrix} 1 \\ 0 \\ 1 \\ -1 \end{bmatrix} + c_4 \begin{bmatrix} 1 \\ 1 \\ -6 \\ -3 \end{bmatrix} + c_5 \begin{bmatrix} -1 \\ -5 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$ $$c_1 = c_2 = c_3 = c_4 = c_5 = 0.$$

Rearranging the left hand side yields

\begin{array}{l} \left\{ \begin{array}{l} 1 \ c_1 + 0 \ c_2 + 1 \ c_3 + 1 \ c_4 - 1 \ c_5 \\ 1 \ c_1 + 1 \ c_2 + 0 \ c_3 + 1 \ c_4 - 5 \ c_5 \\ 0 \ c_1 + 2 \ c_2 + 1 \ c_3 - 6 \ c_4 + 1 \ c_5 \\ -1 \ c_1 + 1 \ c_2 - 1 \ c_3 - 3 \ c_4 + 0 \ c_5 \end{array} \right. = \begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \end{array}

The matrix equation above is equivalent to the following **homogeneous system of equations**

can be transformed by a sequence of elementary row operations to the matrix

The reduced row echelon form of the coefficient matrix of the homogeneous system $(^{**})$ is

which corresponds to the system

The leading entries have been highlighted in yellow.

Those columns in the matrix, that do not contain leading entries, correspond to unknowns that will be arbitrary. The system has infinitely many solutions:

$$c_1 = -3 c_4 +4 c_5$$ $$c_2 = 2 c_4 +1 c_5$$ $$c_3 = 2 c_4 -3 c_5$$ $$c_4 = \text{arbitrary}$$ $$c_5 = \text{arbitrary}$$

Since the variables $c_4$, $c_5$ are arbitrary, then each of the vectors $\mathbf{v}_4$, $\mathbf{v}_5$ can be expressed as a linear combination of vectors in the set $T = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$.

For example, set $c_4 = 1$, $c_5 = 0$, then

$$c_1 = -3, \quad c_2 = 2, \quad c_3 = 2.$$

Use the equation (*) to express $\mathbf{v}_4$ as a linear combination of the remaining vectors in the set $S$.

Let

$$c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 + c_4 \mathbf{v}_4 + c_5 \mathbf{v}_5 = 0,$$ $$(-3 c_4 + 4 c_5) \mathbf{v}_1 + (2 c_4 + 1 c_5) \mathbf{v}_2 + (2 c_4 - 3 c_5) \mathbf{v}_3 + 1 \mathbf{v}_4 + 0 \mathbf{v}_5 = 0.$$

Put $c_4 = 1$, $c_5 = 0$, then

$$-3 \mathbf{v}_1 + 2 \mathbf{v}_2 + 2 \mathbf{v}_3 + \mathbf{v}_4 + 0 \mathbf{v}_5 = 0,$$

hence

$$\mathbf{v}_4 = 3 \mathbf{v}_1 - 2 \mathbf{v}_2 - 2 \mathbf{v}_3.$$

If $c_4 = 0$, $c_5 = 1$, then

$$c_1 = 4, \quad c_2 = 1, \quad c_3 = -3.$$

Let

$$c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 + c_4 \mathbf{v}_4 + c_5 \mathbf{v}_5 = 0,$$ $$(-3 c_4 + 4 c_5) \mathbf{v}_1 + (2 c_4 + 1 c_5) \mathbf{v}_2 + (2 c_4 - 3 c_5) \mathbf{v}_3 + 1 \mathbf{v}_4 + 0 \mathbf{v}_5 = 0.$$

Put $c_4 = 0$, $c_5 = 1$, then

$$4 \mathbf {v} _ {1} + \mathbf {v} _ {2} + - 3 \mathbf {v} _ {3} + 0 \mathbf {v} _ {4} + \mathbf {v} _ {5} = 0,$$

hence

$$\mathbf {v} _ {5} = - 4 \mathbf {v} _ {1} - \mathbf {v} _ {2} + 3 \mathbf {v} _ {3}.$$

Since the set $\mathbf{T} = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly independent and it spans span $\mathbf{W}$ , then the set

$$\mathbf {T} = \left\{\left[ \begin{array}{c c c} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \\ - 1 & 1 & - 1 \end{array} \right] \right\}$$

forms a basis for span W.

Answer:

$$\mathbf {T} = \left\{\left[ \begin{array}{c c c} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \\ - 1 & 1 & - 1 \end{array} \right] \right\}$$

Question

What is the dimension of W?

Solution

The dimension of $\mathbf{W}$ is 3 (number of nonzero rows).

Answer: 3.

Question

Let W be the subspace of P3 spanned by:

$$\{t ^ {\wedge} 3 + t ^ {\wedge} 2 - 2 t + 1, t ^ {\wedge} 2 + 1, t ^ {\wedge} 3 - 2 t, 2 t ^ {\wedge} 3 + 3 t ^ {\wedge} 2 - 4 t + 3 \}.$$

(i) Find a basis for W. (ii) What is the dimension of W?

Solution

1 0 1 2

1 1 0 3

-2 0 -2 -4

1 1 0 3

1 0 1 2

0 1 -1 1

0 0 0 0

0 1 -1 1

1 0 1 2

0 1 -1 1

0 0 0 0

0 0 0 0

As the columns of matrix containing leading entries are the first and the second columns, the first and the second polynomial form a basis for W. That is, the set

$$\{t ^ {\wedge} 3 + t ^ {\wedge} 2 - 2 t + 1, t ^ {\wedge} 2 + 1 \} \text{ is a basis for W. Thus the dimension of W is 2.}$$

Answer: (i) $\{t^{\wedge}3 + t^{\wedge}2 - 2t + 1, t^{\wedge}2 + 1\}$; (ii) 2.

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