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#### baseball3030

##### New member

- Sep 18, 2013

- 9

**proof onto**

Prove: A linear Map T:Rn->Rm is an onto function :

The only way I have thought about doing this problem is by proving the contrapositive:

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- Thread starter
- #1

- Sep 18, 2013

- 9

Prove: A linear Map T:Rn->Rm is an onto function :

The only way I have thought about doing this problem is by proving the contrapositive:

Last edited:

- Jun 26, 2012

- 45

The counterpositive is "If T is onto, then m<=n" It's easier to do it by the theorem that says that the dimensions of the kernel and of the image of T add to n.Prove: A linear Map T:Rn->Rm is not onto if m>n.

The only way I have thought about doing this problem is by proving the contrapositive:

If m<=n then T:Rn->Rm is onto.

I would start by letting there be a transformation

matrix with dimension mxn.

Then the only thing I can think of doing is using the rank nullity thm to show that the dimension of the range=dimension of v. Does anyone know of any other ways to go about this or if my way would be correct? Thank you so much

- Feb 15, 2012

- 1,967

let $B = \{v_1,\dots,v_n\}$ be a basis for $\Bbb R^n$.

Consider the set $T(B) \subset \Bbb R^m$.

If $u \in T(\Bbb R^n) = \text{im}(T)$ we have:

$u = T(x)$ for some $x = c_1v_1 + \cdots + c_nv_n \in \Bbb R^n$.

Thus:

$u = c_1T(v_1) + \cdots c_nT(v_n)$, which shows $T(B)$ spans $\text{im}(T)$.

Since $|T(B)| \leq n < m$, we have that the dimension of $\text{im}(T) \leq n < m$.

However, if $\text{im}(T) = \Bbb R^m$, we have that $T(B)$ spans $\Bbb R^m$ leading to:

$m \leq |T(B)| \leq n < m$, a contradiction.

As a general rule, functions can, at best, only "preserve" the "size" of their domain, they cannot enlarge it. Dimension, for vector spaces, is one way of measuring "size".

While it is technically possible to have a function from $\Bbb R^n \to \Bbb R^m$ where $m > n$ (so called "space-filling functions") that is onto, such functions turn out to be rather bizarre and cannot be linear (they do not preserve linear combinations). Linear maps cannot "grow in dimension", for the same reason the column rank cannot exceed the number of rows in a matrix (even if we have more columns than rows).

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- Sep 18, 2013

- 9

Since |T(B)|≤n<m , we have that the dimension of im(T)≤n<m ?

I understand the T(B)≤n but don't know how you know that n<m?

I understand everything up to that point. Thank you very much, I appreciate it.

- Feb 15, 2012

- 1,967

$n < m$ is given by the problem, yes?

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- #6

- Sep 18, 2013

- 9

- Feb 15, 2012

- 1,967

Yes, I am doing a proof by contradiction (or *reductio ad absurdum).*