Question

1. Find a basis for the subspace of R4 spanned by the given vectors.

(1,1,-5,-6), (2,0,2,-2), (3,-1,0,8).

2. Find a subset of the vectors that forms a basis for the space spanned by the vectors, then express each vector that is not in the basis as a linear combination of the basis vectors.

v1=(1,0,1,1), v2 = (-4,4,-1,4), v3 = (2,4,5,10), v4 = (-10,4,-7,-2)

3. Find a subset of the vectors that forms a basis for the space spanned by the vectors, then express each vector that is not in the basis as a linear combination of the basis vectors.

v1=(1,-1,5,2), v2 = (-2,3,1,0), v3 = (5,-6,14,6), v4 = (0,4,2,-3), v5 = (3,15,45,2)

4. Determine whether b is in the column space of A, and if so, express b as a linear combination of the column vectors of A.

A = 3 4 B = -1
6 -9 15

Answer 1

1.As the set { $v_1,v_2,v_3$ } span a subspace of $R^{4}$

then, $c_1v_1+c_2v_2+c_3v_3=0$

$c_1(1,1,-5,-6)+c_2(2,0,2,-2)+c_3(3,-1,0,8)=0$

We get following equations from above equation

$c_1+2c_2+3c_3=0$

$c_1-c_3=0$

$-5c_1+2c_2=0$

$-6c_1-2c_2+8c_3=0$

From the above equation we can easily conclude that $c_1=c_2=c_3=0$

So,{ $v_1,v_2,v_3$ } are linearly independent.

Thus, the set is the basis for the subspace of R⁴.

2.As the set $v_1,v_2,v_3,v_4$ span R⁴.

then,$c_1v_1+c_2v_2+c_3v_3+c_4v_4=0$

$c_1(1,0,1,1)+c_2(-4,4,-1,4)+c_3(2,4,5,10)+c_4(-10,4,-7,-2)=0$

We get equations:

$c_1 -4 c_2 +2 c_3 -10 c_4 = 0$

$4 c_2 +4 c_3 +4 c_4 = 0$

$c_1 -c_2 +5 c_3 -7 c_4 = 0$

$c_1 +4 c_2 +10 c_3 -2 c_4 = 0$

Solving these equations we can easily get

$c_1 = -6 c_3 +6 c_4$

$c_2 = -c_3 -c_4$

$c_3 = arbitrary$

$c_4 = arbitrary$

{ $v_1,v_2$ } span { $v_3,v_4$ }

{$v_1,v_2$ } spans a subspace of R⁴ and vectors are linearly independent.Thus,vectors serve as the basis of a subspace of R⁴.

3.$c_1v_1+c_2v_2+c_3v_3+c_4v_4+c_5v_5=0$

$c_1(1,-1,5,2)+c_2(-2,3,1,0)+c_3(5,-6,14,6)+c_4(0,4,2,-3)+c_5(3,15,45,2)=0$

We get the following equations from the above equation.

$c_1 -2 c_2 +5 c_3 +3 c_5 = 0$

$- c_1 +3 c_2 -6 c_3 +4 c_4 +15 c_5 = 0$

$5 c_1 +c_2 +14 c_3 +2 c_4 +45 c_5 = 0$

$2 c_1+6 c_3 -3 c_4 +2 c_5 = 0$

Solving these equations we can get:

$c_1 = -3 c_3 -7 c_5$

$c_2 =c_3 -2 c_5$

$c_3 = arbitrary$

$c_4 = -4 c_5$

$c_5 = arbitrary$

{$v_1,v_2,v_4$ } span {$v_3,v_5$ }

{ $v_1,v_2,v_4$ } span a subspace of R⁴ and vectors are linearly independent.Thus, vectors serve as the basis of a subspace of R⁴.

4.$A=\begin{bmatrix} 3 & 4 \\ 6 & -9 \end{bmatrix}$

$B=\begin{bmatrix} -1 \\ 15 \end{bmatrix}$

$B=c_1\begin{pmatrix} 3 \\ 6 \end{pmatrix}+c_2\begin{pmatrix} 4 \\ -9 \end{pmatrix}$

$3c_1+4c_2=-1$

$6c_1-9c_2=15$

$c_1=1,c_2=-1$

Yes, B is in the column space of A as it can be expressed as a linear combination of column vectors of A.