Question

(a) Let α : R
3 −→ R
3 be a linear transformation satisfying
α(1, 1, 0) = (1, 2, −1), α(1, 0, −1) = (0, 1, 1) and α(0, −1, 1) = (3, 3, 3).
i. Express (1, 0, 0) as a linear combination of (1, 2, −1), (0, 1, 1) and (3, 3, 3).

ii. Hence find v ∈ R
3
such that α(v) = (1, 0, 0).
(b) Let the map β : R
3 −→ R
3 be defined by
β((a, b, c)) = (a + b + c, −a − c, b)
for any (a, b, c) ∈ R
3
.
i. Show that β is a linear transformation.
ii. Find the kernel of β.

Answer 1

(a) i.

$$(1, 0, 0)=c_1(1, 2, −1)+c_2(0, 1, 1)+c_3(3, 3, 3)$$

$$c_1+3c_3=1$$ $$2c_1+c_2+3c_3=0$$ $$-c_1+c_2+3c_3=0$$

$$c_1+3c_3=1$$ $$3c_1=0$$ $$-c_1+c_2+3c_3=0$$

$$c_1=0$$ $$c_2=-1$$ $$c_3={1\over 3}$$

$$(1, 0, 0)=(0)(1, 2, −1)+(-1)(0, 1, 1)+({1\over 3})(3, 3, 3)$$

ii.

$$\text{v}=(0)(1, 1, 0)+(-1)(1, 0, -1)+({1\over 3})(0, -1, 1)$$

$$\text{v}=(-1, -{1\over 3}, {4\over 3})$$

(b) i.

$$β((a, b, c)) = (a + b + c, −a − c, b)$$

Let $\text{x}=(a_1, b_1,c_1), \text{y}=(a_2, b_2,c_2), a_1, b_1,c_1,a_2, b_2,c_2\in \R$

$$\beta(\text{x}+\text{y})=\beta((a_1, b_1,c_1)+(a_2, b_2,c_2))=$$

$$=\beta((a_1+a_2, b_1+b_2,c_1+c_2)=$$

$$=(a_1+a_2+b_1+b_2+c_1+c_2, -(a_1+a_2)-(c_1+c_2), b_1+b_2)=$$

$$=(a_1+b_1+c_1, -a_1-c_1,b_1)+(a_2+b_2+c_2,-a_2-c_2,b_2)=$$

$$=\beta((a_1, b_1, c_1))+\beta((a_2, b_2, c_2))=\beta(\text{x})+\beta(\text{y})$$

Let $r\in \R$

$$\beta(r\text{x})=\beta(r(a_1, b_1,c_1))=\beta((ra_1, rb_1,rc_1))=$$

$$=(ra_1+rb_1+rc_1, -ra_1-rc_1,rb_1)=$$

$$=r(a_1+b_1+c_1, -a_1-c_1,b_1)=r\beta(\text{x})$$

Therefore $\beta$ is a linear transformation.  

ii.

System of linear equations associated to the implicit equations of the kernel, resulting from equalling to zero the components of the linear transformation formula.

$$a+b+c=0$$ $$-a-c=0$$ $$b=0$$

The system has infinitely many solutions:

$$a=-c$$ $$b=0$$ $$c=arbitrary$$

$$A=\begin{pmatrix} 1 & 1 & 1 \\ -1 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}$$

$$\begin{pmatrix} 1 & 1 & 1 \\ -1 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}\xmapsto{R_2=R_2+R_1}\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{pmatrix}$$

$$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{pmatrix}\xmapsto{R_3=R_3-R_2}\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

$$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\xmapsto{R_1=R_1-R_2}\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

The solution can be written in the vector form:

$$\begin{pmatrix} a \\ b \\ c \end{pmatrix}=\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}c$$

Therefore the kernel of $\beta$ has a basis formed by the set

$$\bigg\{\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}\bigg\}$$