Suppose that $T : \Bbb R^n \to \Bbb R^m$ is a linear transformation. Prove that $T$ is injective if and only if $N(T) = \{0\}$.

So I know that if something is injective it has a one to one mapping and if $T(X)=T(X')$ then $X=X'$.

I thought initially that since $N(T)=\{x \in \Bbb R \mid T(x)=0\}$ I could multiply both sides by $T^{-1}$ but since it goes from $\Bbb R^n$ to $\Bbb R^m$ it's not a square matrix so it can't have an inverse so I'm stuck. Does anyone have sang ideas?


1 Answer 1


Since $T^{-1}$ doesn't necessarily exist, that's not going to be the correct approach. We're going to have to use only the properties we know -- such as the definitions of injectivity and nullspace.

"Only if":

Let $T: \Bbb R^n \to \Bbb R^m$ be an injective linear transformation. Then $T(x) = T(y) \implies x=y$. Let $a\in N(T)$. Then $$T(a) = 0 = T(0) \implies a = 0$$ Thus $N(T) = \{0\}$.


Let $T: \Bbb R^n \to \Bbb R^m$ be a linear transformation such that $N(T) = \{0\}$. Let $a,b\in\Bbb R^n$ such that $T(a) = T(b)$. Then $$T(a) - T(b) = 0 \implies T(a-b) = 0 \stackrel{(*)}\implies a-b = 0 \implies a = b$$

where $(*)$ is due to the fact that the only vector which maps to $0$ under $T$ is $0$. Thus $T$ is injective.$\ \ \ \square$


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