Give an example of a linear map $T$ such that $\dim(\operatorname{null}T) = 3$ and $\dim(\operatorname{range}T) = 2$

Question

Can I get a verification if this is the right way to approach this problem?

Give an example of a linear map $T$ such that $\dim(\operatorname{null}T) = 3$ and $\dim(\operatorname{range}T) = 2$.

By the fundamental theorem of linear maps, $$\dim V = \dim \operatorname{range}T + \dim\operatorname{null}T,$$ thus $\dim V=5$. Let $e_1,e_2,e_3,e_4,e_5$ be a basis for $\mathbb{R}^5$. Let $f_1,f_2$ be a basis for $\mathbb{R}^2$. Define a linear map $T \in \mathcal{L}(\mathbb{R}^5,\mathbb{R}^2)$ by $$T(a_1e_1+a_2e_2+a_3e_3+a_4e_4+a_5e_5)=a_1f_1+a_2f_2.$$

Thus $\dim(\operatorname{null}T) = 3$ and $\dim(\operatorname{range}T) = 2$.

Answer 1

This is completely correct. This will give a linear map with the properties you're asked for.

I think that it is a bit too general to actually be "an example". I think it would be better if you actually pick a concrete basis. But that's a personal aesthetic belief, and one would have to be pretty pedantic about it to say that that makes you wrong.

One objection with a bit more substance is that you haven't actually proven that your claims about the kernel and the image actually holds. You don't need much, but if this were on a test or an assignment and I was correcting it, I would want you to spend a sentence or two on each of them. For instance

$\dim \operatorname{im} T=2$ because $T$ is clearly surjective and $\dim \Bbb R^2=2$. Then by the rank nullity theorem, we also get $\dim\ker T=3$.

Answer 2

Yes, your example is correct (though like the other answerer, I would tend to prefer something more specific).

Based on your previous post and your use of the word "thus", I assume that you're also trying to prove that the range and nullspace have the dimension you claim (even though such a step is typically considered unnecessary). I would say that you have not done so in the answer you have presented. One proof would be as follows:

We see that the range is given by $\{a_1 f_1 + a_2 f_2 : a_1,a_2 \in \Bbb R\}$. This is the span of the linearly independent set $\{f_1,f_2\}$. Thus, the dimension of the range is $2$.

On the other hand, we note that $$ T(a_1e_1+a_2e_2+a_3e_3+a_4e_4+a_5e_5) = 0 \iff\\ a_1f_1+a_2f_2 = 0 \iff\\ a_1 = 0\text{ and } a_2 = 0. $$ Thus, the kernel of $T$ is given by $\{0e_1 + 0e_2 + a_3e_3+a_4e_4+a_5e_5 : a_3,a_4,a_5 \in \Bbb R\}$. This is the span of the linearly independent set $\{e_3,e_4,e_5\}$. Thus, the dimension of the kernel is $3$.

Answer 3

Let's do an easy variation on your example, where $V=\Bbb R^5$ and $W=\Bbb R^2$.

Define $T$ by $T(x_1,x_2, x_3, x_4, x_5)=(x_1,x_3)$..

Then clearly $T$ is surjective. Hence $\operatorname {rank}T=2$. This forces, by the "Fundamental theorem", as you call it, that $\operatorname {null}T=3$.

It is easy to come up with other such variations. Here are some others, in a slightly different spirit: $T(v_1,v_2, v_3, v_4, v_5)=(av_1+bv_2, cv_3)$, for any $a,b, c\ne0$ Actually, we could let one of $a$ and $b$ be zero.

Actually, the collection of all such maps is in $1-1$ correspondence with the set of rank $2, 2×5$ matrices.

I think your approach was fine, and by coincidence, I defined the same map, essentially, without having read what you did carefully yet. Though I used, implicitly, the standard basis, and you, as @Arthur pointed out, did it in a more general setting.

Answer 4

I was also attempting to solve this question but your approach seemed abstract to me, and upon little reflection, I was able to come up with the following more concrete example which helped increase my understanding of the concept.

The key understanding I got, which prompted me to write this answer, is as follows : since any linear map $T \in \mathcal{L}(\mathbb{R}^5,\mathbb{R}^2)$, which has dim range $T = 2$ will definitely have dim null $T = 3$ (according to Fundamental theorem of Linear Maps), all we need to do is to look for a map $T \in \mathcal{L}(\mathbb{R}^5,\mathbb{R}^2)$ for which dim range $T = 2$.

Define $T \in \mathcal{L}(\mathbb{R}^5,\mathbb{R}^2)$ by : $$T(x_1, x_2, x_3, x_4, x_5) = (x_1 + x_2 +x_3 +x_4+x_5, x_1 + x_2)$$

The range has a basis $(1, 0),(0,1)$ and hence dim range $T = 2$ and null $T$ has basis $(1, -1, 0, 0, 0), (0, 0, 1, -1, 0), (0, 0, 1, 0, -1)$ and hence dim null $T =3$.