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?
Best Answer
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.