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

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?

• Yes that's correct. Aug 5 at 23:03

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