Is it possible for these to be onto?

Question

Is it possible for a linear map to be onto if:

I know to be onto $\operatorname{rank} T = \dim W$ for $T\colon V \to W$.

My thoughts:

No, because $5 + \operatorname{nullity}T = 4$, thus $\operatorname{nullity}T = -1$ which isn't possible?
No, because 5 and $\dim M = 8$ are not equal?
Not sure about this one.

Any help would be appreciated, thanks.

Your question asks 'one-to-one' and your text asks 'onto'. These are two separate things. Which do you want to know? — Ian Coley, Mar 22 '13 at 02:56
Well for a map to be onto, the dimension of the codomain (range) cannot exceed the dimension of the domain. Does that help? — Ian Coley, Mar 22 '13 at 03:00
Ahh, so 1. is Yes? Is the dimension of range of M(4,4) = 8? And as for 3. I'm not sure what the dimension of F(R) is. — Goose, Mar 22 '13 at 03:05
@FrankMcGovern , what about http://math.stackexchange.com/questions/123326/a-simple-bijection-between-mathbbr-and-mathbbr4-or-mathbbrn ? — tacos_tacos_tacos, Mar 22 '13 at 03:14
Also, in latex, the $\to$ symbol is \to. And to get operators such as $\operatorname{nullity}$ and $\operatorname{rank}$, type \operatorname{nullity} and \operatorname{rank}, though $\dim$ is built-in as \dim. — Avi Steiner, Mar 22 '13 at 03:15
@tacos_tacos_tacos That post is looking at arbitrary maps, while this one is restricting to linear maps. — Avi Steiner, Mar 22 '13 at 03:15
If $M(4,4)$ is the vector space of $4\times4$ matrices, its dimension is not $8$. If $M(4,4)$ is not the vector space of $4\times4$ matrices, what is it? — Gerry Myerson, Mar 22 '13 at 06:49

DonAntonio · Answer 1 · 2013-03-22T06:50:02.313

2

I think you may want to concentrate in one single lemma: let $\,V\,,\,W\,$ be two linear spaces over the same field, then

Claim: There exists an onto linear map $\,V\to W\,$ iff $\,\dim V\ge\dim W\,$

The above is true even if the dimensions are infinite (at least with the aid of AC).

edited Mar 22 '13 at 06:50

answered Mar 22 '13 at 06:36

DonAntonio

Make that $\gt$ into $\ge$, and I'll buy it. – Gerry Myerson Mar 22 '13 at 06:47
Ok, done. It's $0.05 Klingonian dollars, please...and thanks, of course. – DonAntonio Mar 22 '13 at 06:50
1

Sorry, all I have is upvotes. – Gerry Myerson Mar 22 '13 at 06:58
I understand that seeing as how Frank posted something similar above: "Well for a map to be onto, the dimension of the codomain (range) cannot exceed the dimension of the domain." I've figured out the first two, but I can't figure out the dimension of F(R). If you could help me out on that I'll accept your answer. – Goose Mar 22 '13 at 15:26
I apologize if I sounded rude, that was not my intent. As far as what F(R) is, this problem I posted comes straight from my book, I suppose it assumes you should already know what F(R) is like you know what R4 and P5 are.. – Goose Mar 22 '13 at 21:46
Surely, somewhere the book explains what $F(R)$ is. – Gerry Myerson Mar 24 '13 at 05:56

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