Exercise $1$, Section $3.B$ - Linear Algebra Done Right

Question

Exercise: Give an example of a linear map $T$ such that $\dim$ null $T = 3$ and $\dim$ range $T = 2$.

Solution: Define $T:R^5\to R^5$ by $T(x_1,x_2,x_3,x_4,x_5)=(x_1,x_2,0,0,0)$. This is indeed a linear map. Then we see that a basis of range$T$ consists of two vectors. Namely, $(1, 0, 0, 0, 0), (0, 1, 0, 0, 0)$. By the fundamental theorem of linear maps we have that $\dim $ null $T=3$.

Is this solution correct?

Answer 1

I think your solution is correct.

If $(y_1,y_2,y_3,y_4,y_5)\in\operatorname{range}T$, then $(y_1,y_2,y_3,y_4,y_5)=T(x_1,x_2,x_3,x_4,x_5)$ holds for some $(x_1,x_2,x_3,x_4,x_5)\in\mathbb{R}^5$.
So, $(y_1,y_2,y_3,y_4,y_5)=(x_1,x_2,0,0,0)=x_1(1,0,0,0,0)+x_2(0,1,0,0,0)$ holds for some $x_1,x_2\in\mathbb{R}$.
So, $(y_1,y_2,y_3,y_4,y_5)\in\operatorname{span}((1,0,0,0,0),(0,1,0,0,0))$.

Conversely, if $(y_1,y_2,y_3,y_4,y_5)\in\operatorname{span}((1,0,0,0,0),(0,1,0,0,0))$, then $(y_1,y_2,y_3,y_4,y_5)=z_1(1,0,0,0,0)+z_2(0,1,0,0,0)=(z_1,z_2,0,0,0)$ holds for some $z_1,z_2\in\mathbb{R}$.
And $(y_1,y_2,y_3,y_4,y_5)=(z_1,z_2,0,0,0)=T(z_1,z_2,0,0,0)$.
So, $(y_1,y_2,y_3,y_4,y_5)\in\operatorname{range}T$.

Hence, $\operatorname{range}T=\operatorname{span}((1,0,0,0,0),(0,1,0,0,0))$.

It is easy to check that $\dim\operatorname{span}((1,0,0,0,0),(0,1,0,0,0))=2$.

So, $\dim\operatorname{range}T=2$.

If $(x_1,x_2,x_3,x_4,x_5)\in\operatorname{null}T$, then $T(x_1,x_2,x_3,x_4,x_5)=(x_1,x_2,0,0,0)=(0,0,0,0,0,0)$.
So, $x_1=x_2=0$.
So, $(x_1,x_2,x_3,x_4,x_5)=(0,0,x_3,x_4,x_5)=x_3(0,0,1,0,0)+x_4(0,0,0,1,0)+x_5(0,0,0,0,1)$.
So, $(x_1,x_2,x_3,x_4,x_5)\in\operatorname{span}((0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1))$.

Conversely, if $(x_1,x_2,x_3,x_4,x_5)\in\operatorname{span}((0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1))$, then $(x_1,x_2,x_3,x_4,x_5)=z_3(0,0,1,0,0)+z_4(0,0,0,1,0)+z_5(0,0,0,0,1)=(0,0,z_3,z_4,z_5)$ holds for some $z_3,z_4,z_5\in\mathbb{R}$.
And $T(x_1,x_2,x_3,x_4,x_5)=T(0,0,z_3,z_4,z_5)=(0,0,0,0,0)$.
So, $(x_1,x_2,x_3,x_4,x_5)\in\operatorname{null}T$.

Hence, $\operatorname{null}T=\operatorname{span}((0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1))$.

It is easy to check that $\dim\operatorname{span}((0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1))=3$.

So, $\dim\operatorname{null}T=3$.