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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?

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    $\begingroup$ 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. $\endgroup$
    – Lucas
    Feb 9, 2020 at 5:04
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    $\begingroup$ Also, your attempt to this case looks good to me. $\endgroup$
    – Lucas
    Feb 9, 2020 at 5:04
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    $\begingroup$ You just need to correct the equality, because $\dim \ker T = 0$. $\endgroup$
    – Lucas
    Feb 9, 2020 at 5:06

1 Answer 1

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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.

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