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Not One to One


New member
Sep 18, 2013
One to One

Prove that A linear map T: Rn->Rm is one to one:

Again, the only thing I can think of doing is possibly using rank nullity theorem but then again I think this can be proved by using independence assumption.
Last edited:


Well-known member
MHB Math Helper
Jan 26, 2012
[JUSTIFY]Hint: find a basis of $\mathbb{R}^n$ (the columns of $I_n$ will do) and show that the image of this basis under $T$ cannot be linearly independent in $\mathbb{R}^m$. Use this to show that $\mathbf{0} \in \mathrm{R}^m$ has two preimages under $T$.[/JUSTIFY]


Well-known member
MHB Math Scholar
Feb 15, 2012
From a previous topic, if $B$ is any basis for $\Bbb R^n$, then $T(B)$ is linearly independent if $T$ is 1-1.

Thus $\text{dim}(\text{im}(T)) = |T(B)| = n$.

Since any basis $C$ of $\Bbb R^m$ is a MAXIMAL linearly independent set, we have $n \leq m$, contradicting $m < n$.

Our only assumption was that $T$ was 1-1, so this cannot be the case.

(Bacterius' approach is good, too :))