Prove Linear Transformation And Find the Nullity

Question

Let $T: R^3 \rightarrow R^3$ be defined by $T(x_1,x_2,x_3)=(x_1-x_2+2x_3,2x_1+x_2,-x_1−2x_2+2x_3)$ then,
1. Show that T is a linear transformation
2. What are the conditions on a, b, c such that (a, b, c) is in the null space of T? Specifically, find the nullity of T.

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Solution

Rewriting the transformation function T for convenience:

$\begin{align*} \left T(u,v,w)=(u-v+2w,2u+v,-u-2v+2w) \end{align*}$

Part 1 – Show that T is a linear transformation

We say that T is a linear transformation if it satisfies the below conditions:

(1) $\begin{equation*} For \ all \ x, y \ \epsilon \ V,T(x+y)=T(x)+T(y) \end{equation*}$

(2) $\begin{equation*} \left For \ all \ x \ \epsilon \ V, c \ \epsilon \ R, T(c.x)=c.T(x) \end{equation*}$

Proving #1

(3) $\begin{equation*} \left Let \ x = (u_1,v_1,w_1), y = (u_2,v_2,w_2) \end{equation*}$

Applying (3) in (1),

$\begin{align*} \left T(x+y) = T((u_1,v_1,w_1)+(u_2,v_2,w_2)) \left = T(u_1+u_2,v_1+v_2,w_1+w_2) \end{align*}$

(4) $\begin{equation*} \left =(u_1+u_2-v_1-v_2+2w_1+2w_2,2u_1+2u_2+v_1+v_2,-u_1-u_2-2v_1-2v_2+2w_1+2w_2) \end{equation*}$

Similarly,

$\begin{align*} \left T(x)+T(y)=T(u_1,v_1,w_1)+T(u_2,v_2,w_2) \end{align*}$

(5) $\begin{equation*} \left =(u_1+u_2-v_1-v_2+2w_1+2w_2,2u_1+2u_2+v_1+v_2,-u_1-u_2-2v_1-2v_2+2w_1+2w_2) \end{equation*}$

Since (4) = (5), T satisfies the first condition.

Proving #2

(6) $\begin{equation*} \left Let \ x = (u_1,v_1,w_1), c \ be \ a \ scalar \ from \ R \end{equation*}$

$\begin{align*} \left T(c.x) = T(c.(u_1,v_1,w_1)) = T(c.u_1,c.v_1,c.w_1) \end{align*}$

(7) $\begin{equation*} \left = (c.u_1-c.v_1+2c.w_1,2c.u_1+c.v_1,-c.u_1+2c.v_1+2c.w_1) \end{equation*}$

Similarly,

$\begin{align*} \left c.T(x)=c.T(u_1,v_1,w_1) \end{align*}$

(8) $\begin{equation*} \left = (c.u_1-c.v_1+2c.w_1,2c.u_1+c.v_1,-c.u_1+2c.v_1+2c.w_1) \end{equation*}$

Since (7) = (8), T satisfies the second condition as well.

Hence, T(u,v,w)=(u-v+2w,2u+v,-u-2v+2w) is a linear transformation.

Part 2 – Finding Nullspace & Nullity

Convert the linear transformation into the form, $A_x=0$ . Thus,

$A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 1 & 0 \\ -1 & -2 & 2 \end{bmatrix}, b = \begin{bmatrix} a \\ b \\ c \end{bmatrix}, 0 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

To find the nullspace, N(A), we need to apply Gauss Elimination to convert the matrix into its row-echelon form.

This can be done by applying in sequence: $R_1 \leftrightarrow R_2, R_2 \rightarrow R_2-R_1/2, R_3 \rightarrow R_3+R_1/2, R_3 \rightarrow R_3-R_2$

Thus, the REF form of A,

$R = \begin{bmatrix} 2 & 1 & 0 \\ 0 & -3/2 & 2 \\ 0 & 0 & 0 \end{bmatrix}$

$Ax=0$ & $Rx=0$ are equivalent systems.

Hence,

$R = \begin{bmatrix} 2 & 1 & 0 \\ 0 & -3/2 & 2 \\ 0 & 0 & 0 \end{bmatrix}, b = \begin{bmatrix} a \\ b \\ c \end{bmatrix}, 0 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Observations from the above system:

c is the free variable
rank of the system is 2
number of columns in the system is 3

$2a + b = 0$

$-3/2b + 2c = 0$

Applying back-substitution $\Rightarrow b=4/3c, a=−2/3c$

$x = \begin{bmatrix} 4/3c & -2/3c & c \end{bmatrix} = t_1 \begin{bmatrix} 4 \\ -2 \\ 3 \end{bmatrix} \ where \ t_1=c/3, \ c \ \epsilon \ R$

$Therefore, \ Nullspace \ N(A) = span \begin{bmatrix} 4 \\ -2 \\ 3 \end{bmatrix}$

Applying Rank Nullity Theorem:

Rank(A) + Nullity(A) = No. of Columns = n

Nullity(A) = 3-2 = 1

Question

Solution

Part 1 – Show that T is a linear transformation

Part 2 – Finding Nullspace & Nullity

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