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

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 $$\dim(\operatorname{range}T) = 2$$.

• Note that $imT=span(T(e_1).,,,,,T(e_n))$ where $e_i$ is a basis vector.
– user643073
Dec 31, 2019 at 20:13
• What you've done seems like the most straightforward way to answer the question as it is phrased. Jan 1, 2020 at 4:06

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 $$3$$.

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.

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 $$T =3$$.

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 $$\dim\ker T=3$$.