Finding the rank of a linear transformation

Question

I am stuck on the following problem:

Let $T$ be arbitrary linear transformation from $\Bbb R^n$ to $\Bbb R^n$ which is not one-one.Then I have show that Rank $(T)=n-1.$

I know that Rank$(T)$+ Nullity $(T)=n \implies$ Rank$(T)=n-$Nullity$(T)$. But what is the Nullity of $T?$ Can someone point me in the right direction?

EDIT: It was in fact a multiple choice question where the options were :
1. Rank$(T)>0 \space $,
2. Rank$(T)<n \space$,
3.Rank$(T)=n \space$ ,
4.Rank$(T)=n-1$.

The answer was given to be option 4 (which appears to be wrong from the responses).So,choice 2 appears to be correct one.

The nullity is the dimension of the kernel. If a linear transformation sends any nonzero vector to the zero, it has a nontrivial kernel, i.e. nullity $\ge 1$. — vadim123, Apr 23 '13 at 18:08
That statement is patently incorrect: the rank of the zero map for instance is always 0. Did you mean to write $\leq n-1$ instead of $=n-1$? — rschwieb, Apr 23 '13 at 18:08
I think it was a mistake. Because one could have a map from $\mathbb{R}^3$ to $\mathbb{R}^3$ which sends $(1,0,0)$ to itself and $(0,1,0)$ and $(0,0,1)$ to the zero vector. — Cameron Williams, Apr 23 '13 at 18:13
$2$ is of course a correct choice. I see no reason for $4$ to be correct. — Sugata Adhya, Apr 23 '13 at 18:18

rschwieb · Accepted Answer · 2013-04-23T18:25:31.777

4

Since the transformation is not one-to-one, $Null(T)>1$ so in terms of what you are given and the rank-nullity theorem, the dimensions satisfy

$n-\mathrm{Rank}(T)=\mathrm{Null}(T)>0$

Which is the same thing as saying

$n-\mathrm{Rank}(T)=\mathrm{Null}(T)\geq 1$ or more simply

$n-\mathrm{Rank}(T)\geq 1$

Rearrange the last equation to put $n-1$ alone on a side of the inequality.

edited Apr 23 '13 at 18:25

answered Apr 23 '13 at 18:11

rschwieb

153,510

How does $rank~T\leq n-1\implies rank~T=n-1?$ – Sugata Adhya Apr 23 '13 at 18:15
@SugataAdhya It doesn't. I gave a counterexample indicating as much in the comments above. If the downvote was given because you misread my solution, I would appreciate it if you reversed it: thanks! – rschwieb Apr 23 '13 at 18:24
1

rectified @rschwieb – Sugata Adhya Apr 23 '13 at 18:28

score 4 · Answer 2 · edited May 26 '13 at 11:36

4

Hint: Since $T$ is not one-one so there are vectors $v\neq w$ in $\mathbb R^n$ such that $T(u)=T(w)$ or $T(u-w)=0$. But $u-w\neq 0$ so $\dim(\operatorname{Ker}\, T)\neq 1$ and so...

edited May 26 '13 at 11:36

learner

6,726

answered Apr 23 '13 at 18:12

Mikasa

67,374

Great hint and lead, and so...$\quad\color{green}{\bf \Large \checkmark};$ from me, if I could... – amWhy Apr 24 '13 at 00:21
@amWhy: You are soooooo Welcome to me friend. – Mikasa Apr 24 '13 at 05:18
Hello, dear Babak! I will be heading to bed shortly, but so good to see you! – amWhy Apr 24 '13 at 05:20
@amWhy: If you let me driving here till you can have a deep rest. I hope I have the capability to perform good as you have done here. :-) – Mikasa Apr 24 '13 at 05:33
@BabakS.: I agree, excellent hint +1 – Amzoti Apr 24 '13 at 14:44

score 2 · Answer 3 · answered Nov 21 '13 at 22:52

2

I think the nullity which is Kernel of T has one element which is zero since the image of the zero under this map is zero and unique.

answered Nov 21 '13 at 22:52

MathGuy

43
7

Finding the rank of a linear transformation

3 Answers3