Answer to Question #259060 in Linear Algebra for Waweru

Question

(a) Let A=   be a matrix representing the linear transformation T
 i. Find the image of V = [−1, 2, 5] under T(2marks)
 ii. Determine the kernel or nullspaceof T(5marks)
Find the range and rank of T (3marks) 


(b) let T: R3 ... > R3 be defined by T (x, y, z) = (x + y, x − y, y + z)
Find the matrix representing T with respect to standard basis(3marks)
Find the kernel and nullity of T (4marks)
find Range and Rank of T(3marks) 


c) Let A=      be a matrix representing A respect to the standard under basis R3 
Find the image of V = [−2, 1, 2] under T (4marks)
Determine the kernel of T (3marks)
Find the rank of T (3marks) 

Answer 1

In part (a) and (c) of the question, the respective matrix denoted as A are ommited.

Adjusting the conditions of the question

thus taking:

Part (a) matrix as

$A=\begin{pmatrix} 1&0& -1\\ 1&2&-1\\1&0&-1 \end{pmatrix}$

and part(c) matrix as:

$A=\begin{pmatrix} 3&-3&0\\ 1&-1&0\\2&1&1 \end{pmatrix}$

a) (i) image of v

$\begin{pmatrix} 1&0&-1\\ 1&2&-1\\1&0&-1 \end{pmatrix}\begin{pmatrix} -1\\2\\5 \end{pmatrix}=\begin{pmatrix} -6\\-2\\ -6 \end{pmatrix}$

(ii) $ker(A)=[x\in\Reals^3|Ax=0]$

$\begin{pmatrix} 1&0&-1\\1&2&-1\\1&0&-1 \end{pmatrix}\begin{pmatrix} a\\b\\c \end{pmatrix}=[0]$

$\implies a=c\quad b=0\>and\> let \>c=1$

$ker\>A=[\>c\begin{pmatrix} 1\\0\\1 \end{pmatrix}]$

(iii) Range = columns of A corresponding to pivot column in rref (A)

$Range =$ {$\space\begin{pmatrix} 1\\1\\1 \end{pmatrix}\>,\>\begin{pmatrix}0\\2\\0\space \end{pmatrix}$ }

$Rank=$ dim( column space A)

$=2$

b) (i) $T=\begin{pmatrix} x\\y\\z \end{pmatrix}=\begin{pmatrix} x+y\\x-y\\ y+z \end{pmatrix}$

$=x\begin{pmatrix}1\\1\\0 \end {pmatrix}+\begin{pmatrix}1\\-1\\1 \end{pmatrix}+z\begin{pmatrix} 0\\0\\1 \end{pmatrix}$

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

(ii)rref of T$=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1 \end{pmatrix}$

The rank of T is therefore 3

$\implies$ the nullity is zero

Ker T is a zero vector [0]

(iii) Range of T= span( column space (T))

$= [\> \begin{pmatrix}1\\1\\0 \end{pmatrix}\>,\>\begin{pmatrix}1\\-1\\1 \end{pmatrix}\>,\>\begin{pmatrix}0\\0\\1 \end{pmatrix}$ ]

Rank of T= dim (volume space)

=3

C) (i) Taking the omitted matrix as

$A=\begin{pmatrix}3&-3&0\\1&-1&0\\2&1&1 \end{pmatrix}$

Image of v is

$\begin{pmatrix} 3&-3&0\\1&-1&0\\2&1&1 \end{pmatrix}\begin{pmatrix} -2\\1\\2 \end{pmatrix}=\begin{pmatrix} -9\\-3\\-1 \end{pmatrix}$

Im (v) = (-9,-3,-1)

iii) rref $A=\begin{pmatrix}1&0&\frac{1}{3}\\0&1&\frac{1}{3}\\0&0&0 \end{pmatrix}$

$\begin{pmatrix} 1&0&\frac{1}{3}\\ 0&1&\frac{1}{3}\\0&0&0 \end{pmatrix}\begin{pmatrix}a\\ b\\c \end{pmatrix}=[0]$

$a=-\frac{1}{3}c\quad b=-\frac{1}{3}c$

Let $c=t$

Ker $A=[\begin{bmatrix} -\frac{1}{3}\\ -\frac{1}{3}\\1 \end{bmatrix}t ]\quad t\in\reals$

iii). Rank of T= dim(column space (T)

=2