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I know that if $T: \mathbb{R}^m \to \mathbb{R}^n$ is invertible then it is also onto and 1-1. But is it equivalent? In other words, are all linear transformations that are bijective considered invertible?

Mike F
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merlin
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    In general, one proves a function $f:A\to B$ is bijective $\iff$ is it invertible $\iff$ it is one-one and onto. – Pedro Oct 18 '13 at 00:59
  • @merlin : The answer and above comment are correct. Note that the bijective linear map $T$ can exist only if $m=n$. Can you see why? – Stefan Smith Oct 18 '13 at 02:20

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By basic set theory, as alluded to in @Pedro's comment, a bijective linear transformation $T : \mathbb{R}^m \to \mathbb{R}^n$ will have an inverse function $S : \mathbb{R}^n \to \mathbb{R}^m$. The interesting question from a linear algebra standpoint is

Is the inverse $S$ also a linear transormation?

The answer to this question is yes. Given $x,y \in \mathbb{R}^n$ and $a,b \in \mathbb{R}$, we have $$ ax + by = a TSx + b TSy = T( aSx + bSy) $$ using $TS = I_n$ for the first equality and linearity of $T$ for the second. Applying $S$ to both sides of the above equation and using $ST = I_m$ gives $$ S(ax+by) = aSx + bSy$$ which says $S$ is linear.

Mike F
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