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Proof of a linear transformation not being onto

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:
 
Last edited:

ModusPonens

Well-known member
Jun 26, 2012
45
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
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.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
You can do a direct proof, and I do not think one needs to invoke rank-nullity.

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).
 

baseball3030

New member
Sep 18, 2013
9
How are you able to say that
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.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
$n < m$ is given by the problem, yes?
 

baseball3030

New member
Sep 18, 2013
9
Yeah but the problem states that the linear map is not onto if m<n so are you assuming that it is onto and then arriving at a contradiction? Thanks,
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Yes, I am doing a proof by contradiction (or reductio ad absurdum).