Answer to Question #153651 in Linear Algebra for pallavi

Question

Let F be the field of complex numbers and let T be the function from F

3

into F

3 defined by

T(x1, x2, x3) = (x1 − x2 + 2x3 ,2x1 + x2 , − x1 − 2 x2 + 2x3 )

i. Verify that T is a linear transformation.

ii. If (a, b, c) is a vector in F

3, what are the conditions on a, b and c that the vector bein the range of T? What is the rank of T?

iii. What are the conditions on a, b and c that the vector (a, b, c) be in the null space of T? What is the nullity of T?

Answer 1

$T:F_3\rightarrow F_3\\ (x_1,x_2,x_3)\rightarrow (x_1 − x_2 + 2x_3 ,2x_1 + x_2 , − x_1 − 2 x_2 + 2x_3 )$

$=\begin{bmatrix}1&-1&2\\2&1&0\\-1&-2&2\end{bmatrix}\times\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=AX \;[let]$

where $A=\begin{bmatrix}1&-1&2\\2&1&0\\-1&-2&2\end{bmatrix},X=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$

i.e. $T(X)=AX$

(i)

Let $X,Y\in F_3, c\in \mathbb{C}$

$\therefore T(aX+Y)=A(aX+Y)=aAX+AY=aT(X)+T(Y)$

Hence $T$ is a linear transformation.

(ii)

Claim:- $\{A_1=\begin{bmatrix}1\\2\\-1\end{bmatrix},A_2=\begin{bmatrix}-1\\1\\-2\end{bmatrix}\}$ is a Basis of $Im(T)$ .

Proof:-

Any element of $Im(T)$ can be expressed by columns of $A \;\;i.e. \;A_1, A_2,A_3$

$\therefore Im(T)=span(\{A_1, A_2,A_3\})$

$A_3=\frac{2}{3}A_1-\frac{4}{3}A_2$

$\therefore A_3\in span(\{A_1,A_2\})$

$\Rightarrow span(\{A_1,A_2\})= span(\{A_1,A_2,A_3\})=Im(T)$ ...(1)

Now $A_1,A_2$ are independent [ as $\exists no\;c\in \mathbb{C}\ni A_1=cA_2$ ] ...(2)

So, from above two facts, we can conclude that ,

$\{\begin{bmatrix}1\\2\\-1\end{bmatrix},\begin{bmatrix}-1\\1\\-2\end{bmatrix}\}$ is a Basis of $Im(T)$ ...(Proved)

Therefore $\pmb{rank(T)=dim(Im(T))=2}$

(iii)

Null Space of $\;T=Null(T)=\{X\in F_3 | AX=0\}$

So,

$\;\;\;\;\;AX=0\\ \Rightarrow \begin{bmatrix}1&-1&2\\2&1&0\\-1&-2&2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=0\\ \Rightarrow \begin{bmatrix}1&0&2/3\\0&1&-4/3\\0&0&0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=0\\ \Rightarrow x_1+\frac{2x_3}{3}=0, \;x_2-\frac{4x_3}{3}=0$

[let $x_3=k$]

$\Rightarrow X\in\{k\begin{bmatrix}-2/3\\4/3\\1\end{bmatrix}: k\in \mathbb{C}\}$

$\therefore Null(T)=\{k\begin{bmatrix}-2/3\\4/3\\1\end{bmatrix}: k\in \mathbb{C}\}$

Clearly Null(T) is generated by 1 element.

Hence, $\pmb{Nullity\;\;of \;\;T=dim(Null(T))=1}$.