Answer to Question #271357 in Linear Algebra for Gestavo

Question

It is known that the vectors in the vector space R³:

⃗v₁ = (1, 1, 1), v₂ = (2, -1, 1), v₃ = (0, 2, 1)

and

⃗w₁ = (2, -1, 3), w₂ = (3, -1, 7), w₃ = (-1, 1, 1)

The vectors v₁, v₂, v₃ are basis in R³. Transformation T : R³ → R³ is a linear transformation defined by:

 T( ⃗v_i) = ⃗w_i

Define:

1. Matrix transformation of T

2. Basis of Ker(T) and Range(T)

Answer 1

1.

$$\vec e_1=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}=a\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}+b\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}+c\begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}$$

$$\begin{matrix} a+2b=1 \\ a-b+2c=0 \\ a+b+c=0 \end{matrix}=>\begin{matrix} a=3 \\ b=-1 \\ c=-2 \end{matrix}$$

$$\vec e_1=3\vec v_1-\vec v_2-2\vec v_3$$

$$T\begin{pmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \end{pmatrix}= 3\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}-\begin{bmatrix} 3 \\ -1 \\ 7 \end{bmatrix}-2\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}$$

$$=\begin{bmatrix} 5 \\ -4 \\ 0 \end{bmatrix}$$

$$\vec e_2=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}=d\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}+f\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}+g\begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}$$

$$\begin{matrix} d+2f=0 \\ d-f+2g=1 \\ d+f+g=0 \end{matrix}=>\begin{matrix} d=2 \\ f=-1 \\ g=-1 \end{matrix}$$

$$\vec e_2=2\vec v_1-\vec v_2-\vec v_3$$

$$T\begin{pmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \end{pmatrix} = 2\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}-\begin{bmatrix} 3 \\ -1 \\ 7 \end{bmatrix}-\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}$$

$$=\begin{bmatrix} 2 \\ -2 \\ -2 \end{bmatrix}$$

$$\vec e_3=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}=k\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}+m\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}+n\begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}$$

$$\begin{matrix} k+2m=0 \\ k-m+2n=0 \\ k+m+n=1 \end{matrix}=>\begin{matrix} k=-4 \\ m=2 \\ n=3 \end{matrix}$$

$$\vec e_3=-4\vec v_1+2\vec v_2+3\vec v_3$$ $$T\begin{pmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \end{pmatrix} =-4\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}+2\begin{bmatrix} 3 \\ -1 \\ 7 \end{bmatrix}+3\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}$$

$$=\begin{bmatrix} -5\\ 5 \\ 5 \end{bmatrix}$$

$$T(\vec x)=A\vec x$$

$$A=\begin{bmatrix} 5 & 2 & -5 \\ -4 & -2 & 5 \\ 0 & -2 & 5 \end{bmatrix}$$

2.

$$A=\begin{bmatrix} 5 & 2 & -5 \\ -4 & -2 & 5 \\ 0 & -2 & 5 \end{bmatrix}$$

$R_1=R_1/5$

$$\begin{bmatrix} 1 & 2/5 & -1 \\ -4 & -2 & 5 \\ 0 & -2 & 5 \end{bmatrix}$$

$R_2=R_2+4R_1$

$$\begin{bmatrix} 1 & 2/5 & -1 \\ 0 & -2/5 & 1 \\ 0 & -2 & 5 \end{bmatrix}$$

$R_2=-(5R_2)/2$

$$\begin{bmatrix} 1 & 2/5 & -1 \\ 0 & 1 & -5/2 \\ 0 & -2 & 5 \end{bmatrix}$$

$R_1=R_1-(2R_2)/5$

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -5/2 \\ 0 & -2 & 5 \end{bmatrix}$$

$R_3=R_3+2R_2$

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix}$$

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$$

If we take $x_3=t,$ then $x_1=0, x_2=(5/2)t$

Therefore the kernel of $T$ has a basis formed by the set

$$\begin{bmatrix} 0\\ 5/2 \\ 1 \end{bmatrix}$$

$$\vec u_3=0\vec u_1+(-5/2)\vec u_2$$

The set $\bigg\{\begin{bmatrix} 5\\ -4 \\ 0 \end{bmatrix}, \begin{bmatrix} 2\\ -2\\ -2 \end{bmatrix}\bigg\}$ forms a basis for the range of $T.$