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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.

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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
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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
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$2$ is of course a correct choice. I see no reason for $4$ to be correct. –  Sugata Adhya Apr 23 '13 at 18:18
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@SugataAdhya yes,I agree... –  user52976 Apr 23 '13 at 18:24
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3 Answers 3

up vote 2 down vote accepted

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.

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

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Great hint and lead, and so...$\quad\color{green}{\bf \Large \checkmark}\;$ from me, if I could... –  amWhy Apr 24 '13 at 0:21
    
@amWhy: You are soooooo Welcome to me friend. –  B. S. Apr 24 '13 at 5:18
    
Hello, dear Babak! I will be heading to bed shortly, but so good to see you! –  amWhy Apr 24 '13 at 5:20
    
@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. :-) –  B. S. Apr 24 '13 at 5:33
    
@BabakS.: I agree, excellent hint +1 –  Amzoti Apr 24 '13 at 14:44
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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.

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