Question

Show that the function T:R2-R2 defined by T(X1,X2)=(x1-X2, X1+X2) for (X1-X2) e R2 Is biyetive

Answer 1

Answer:

i don't know if I can I Will help

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

The linear mapping $T : \mathbb{ {R}^{2} } \mapsto\mathbb{ {R}^{2} }$

defined by

$T(x_1,x_2)= (x_1 - x_2, x_1 + x_2) \: \: for \: \: (x_1,x_2) \in \: \mathbb{ {R}^{2} }$

TO PROVE

T is bijective

FORMULA TO BE IMPLEMENTED

T is said to be bijective if T is one to one and onto

EVALUATION

CHECKING FOR ONE TO ONE

Let $(x_1,x_2) \in \: Ker T$

$\implies \: T(x_1,x_2) \: = (0,0)$

$\implies \: (x_1 - x_2 \: , \: x_1 + x_2) \: = (0,0)$

$\implies \: x_1 - x_2 = 0\: \: and \: \: \: x_1 + x_2 \: = 0$

On addition

$2x_1 = 0$

$\implies \: x_1 = 0$

From

$x_1 + x_2 \: = 0 \: \:$

We get

$x_2 \: = 0$

Hence

$x_1 = 0\: \: and \: \: \: x_2 \: = 0$

$Ker T \: = \{\theta \}$

Hence T is one to one

CHECKING FOR ONTO

We know that $\{(1,0) , \: (0,1) \} \: \: is \: a \: basis \: for \: \: \mathbb{ {R}^{2} }$

So Im T is a linear span of $T(1,0) ,T (0,1)$

Now

$T(1,0) = (1,1) \: \: and \: \: T(0,1) = ( - 1,1)$

Now we are proceeding to check whether

$\{(1,1) , ( - 1,1) \}$are linearly independent

Let

$c(1,1) + d( - 1,1) = (0,0) \: \: for \: some \: \: c \: , \: d \: \in \mathbb {R}$

$\implies \: (c - d,c + d) = (0,0)$

$\implies \: c - d = 0 \: \: and \: \: c + d = 0$

On addition

$2c = 0$

$\implies \:c = 0$

Now from $c - d = 0$We get

$d= 0$

So

$c = 0 \: \: and \: \: d = 0$

Therefore $\{(1,1) , ( - 1,1) \} \:$is linearly independent

So

$Im T = \mathbb{ {R}^{2} }$

Hence T is onto

RESULT

$\fbox{T \: IS \: BIJECTIVE}$