Question

Show that T:R^3 →R^2: T(x,y,z)= (2x+y-z,x+z) is a linear transformation. Verify that T satisfies the Rank-nullity theorem

Answer 1

Let $\alpha=(x_1,y_1,z_1)\isin R^3, \beta=(x_2,y_2,z_2)\isin R^3$

Then $T(\alpha)=(2x_1+y_1-z_1,x_1+z_1)$

$T(\beta)=(2x_2+y_2-z_2,x_2+z_2)$

$\alpha+\beta=(x_1+x_2,y_1+y_2,z_1+z_2)$

$T(\alpha+\beta)=(2(x_1+x_2)+(y_1+y_2)-(z_1+z_2),(x_1+x_2)+(z_1+z_2))$

$=((2x_1+y_1-z_1)+(2x_2+y_2-z_2),(x_1+z_1)+(x_2+z_2))$

$=T(\alpha)+T(\beta)$

Also let $c\isin R.$ 

Then

 $c\alpha=(cx_1,cy_1,cz_1)$

$T(c\alpha)=(2cx_1+cy_1-cz_1,cx_1+cz_1)$

$=c(2x_1+y_1-z_1,x_1+z_1)$

$=cT(\alpha)$

Thus 

$T(\alpha+\beta)=T(\alpha)+T(\beta)$ 

for all $\alpha,\beta\isin R^3$

and 

$T(c\alpha)=cT(\alpha)$ for all $c\isin R$ and $\alpha\isin R^3.$

Hence $T$ is a linear transformation.

$K$$er$ $T=$ {$(x,y,z)\isin R^3:T(x,y,z)=(0,0)$}

Let $(x_1,y_1,z_1)\isin$ $Ker$ $T$

Then $2x_1+y_1-z_1=0,x_1+z_1=0.$

From second equation we get $x_1=-z_1$

Let $x_1=k$ (say) , then $z_1=-k$

Now from first equation we get , $y_1=-3k$

Therefore $(x_1,y_1,z_1)=k(1,-3,-1)=c(-1,3,1)$, $c\isin R$

Let $\alpha=(-1,3,1)$. Then $Ker$ $T=$linear span of $\alpha=L${$\alpha$ } and $dim$$Ker$ $T=1.$

$ImT$ is the linear span of the vectors $T(\alpha_1),T(\alpha_2),T(\alpha_3)$ where {$\alpha_1,\alpha_2,\alpha_3$} is any basis of $R^3$.

{$\epsilon_1=(1,0,0),\epsilon_2=(0,1,0),\epsilon_3=(0,0,1)$} is a basis of $R^3$.

Now $T(\epsilon_1)=(2,1),T(\epsilon_2)=(1,0),T(\epsilon_3)=(-1,1).$

$ImT=L${$(2,1),(1,0),(-1,1)$ }

These vectors are linearly dependent in $R^2.$

But the subset {$(2,1),(1,0)$} is linearly independent in $R^2$.

$\therefore ImT=L${$(2,1),(1,0)$ }

Therefore $dim$ $ImT=2.$

Now according to the Rank- Nullity theorem we know that

$dim$ $KerT+$ $dim$ $ImT$ $=dim$ $R^3$

Here, $dim$ $KerT+dim$ $ImT=1+2=3$.

$dim$ $R^3=3$

Therefore $T$ satisfies the Rank-Nullity theorem.