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I want to show that if there is an injective linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ then $n \leq m$. I approached it as follows. $T$ is injective implies that the standard matrix has nullity $0$ and so rank $n$. Hence, the standard matrix has $n$ linearly independent columns, which means that $T(e_1),T(e_2),\dots,T(e_n)$ are linearly independent vectors in $\mathbb{R}^m$. Since we can have at most $m$ linearly independent vectors in $\mathbb{R}^m$, $n \leq m$. Similarly I argue that if $T$ is surjective then the standard matrix $[T]$ has rank $m$ and so it has $m$ linearly independent columns. But the total number of columns is $n$, so $n\geq m$.

RGS
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user30523
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1 Answers1

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Your proof is perfectly fine. It can be proven for any $T: V \to U $ linear spaces of finite dimension, and thus if $T $ is bijective, the dimensions of $V $ and $U$ are the same.

Or, there is only a bijective linear transformation between two linear spaces if they have the same dimension. In that case, you can pick two bases and say that the image of one of the basis is the other and your $T$ is defined.

RGS
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