Do exist an injective linear map from $\mathbb{R}^5\rightarrow\mathbb{R}^4$

Question

I already asked If there exists an injective function $f:\mathbb{R}^5\rightarrow\mathbb{R}^4$ and I found that it exists a such function: Do exist an injective function from $\mathbb{R}^5 \rightarrow \mathbb{R}^4$

Now, I would like to know If there exist such an injective linear map. I was trying to find out but I am unable. In think it does not exists because of the difference between dimmnsions but I am really unsure.

Please help me. Thank you!!!!

What do you mean by application? – Taufi Jun 19 '17 at 19:18 — Taufi, Jun 19 '17 at 19:18
@Taufi "application" is French for function/mapping. – Angina Seng Jun 19 '17 at 19:20 — Angina Seng, Jun 19 '17 at 19:20

score 6 · Accepted Answer · edited Jun 19 '17 at 19:30

6

The rank-nullity theorem tells us No. In fact

$$\dim\ker f=\dim\Bbb R^5-\operatorname{rank}f\ge 5-4=1$$ hence $\ker f\ne\{0\}$, and so $f$ isn't injective.

edited Jun 19 '17 at 19:30

Kyle Miller

19,353

answered Jun 19 '17 at 19:22

user296113

7,570

score 3 · Answer 2 · answered Jun 19 '17 at 19:26

No: if $f:V\rightarrow W$ is an injective linear map, then $\dim V \leq \dim W$. This is because if $v_1,\dots,v_n\in V$ are vectors such that $f(v_1),\dots,f(v_n)$ are dependent (i.e., there exist $c_1,\dots,c_n\in\mathbb{R}$ not all zero such that $c_1f(v_1)+\dots+c_nf(v_n)=0$), then $f(c_1v_1+\dots+c_nv_n)=0$, and injectivity implies $c_1v_1+\dots+c_nv_n=0$, hence the vectors are dependent. Dimension is the maximum number of linearly independent vectors, so $\dim V$ cannot possibly exceed $\dim W$.

In your case, $\dim \mathbb{R}^n=n$, and it is not true that $5\leq 4$.

Do exist an injective linear map from $\mathbb{R}^5\rightarrow\mathbb{R}^4$

2 Answers2