# Determine whether linear mapping is invertible

Determine whether linear mapping $$~T:R^3 \rightarrow R^3$$ defined by $$T(a_1,a_2,a_3)=(3a_1-2a_3,a_2,3a_1+4a_2)$$ is invertible or not.

For $$T$$ to be invertible, it must be injective, so in this case set $$3a_1-2a_3=0, a_2=0 ,3a_1+4a_2=0$$, we get $$a_1=a_2=a_3=0$$. Hence $$\dim (\ker(T))=0$$.

Checking surjective: since it is $$R^3$$, $$\dim(\text{Im}(T))=3$$

By rank-nullity theorem, $$\dim(V)=3=\dim(\ker(T))=\dim(\text{Im}(T))=3$$

Hence it is invertible. Is that a right justification?

Also to generalize, for any finite dimension $$T:V \rightarrow W$$, is it always true that if $$\dim(V)=\dim(W)$$, then always invertible?

• For finite dimension if $T: V \to W$ is a injective linear map with $\dim V = \dim W$, then $T$ is invertible. The proof of this fact follows from the ideia used by you: just use the injectivity and the rank-nullity theorem. Feb 9, 2020 at 5:04
• Also, your attempt to this case looks good to me. Feb 9, 2020 at 5:04
• You just need to correct the equality, because $\dim \ker T = 0$. Feb 9, 2020 at 5:06

If $$Ta = b$$ then $$a_2 = b_2$$ and since $$3a_1 + 4 a_2 = b_3$$ we have $$a_1 = {1 \over 3} (b_3-4 b_2)$$ and similarly for $$a_3$$. Hence $$T$$ is invertible. Sometimes an equation is just an equation.