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Can someone please explain the logic behind this solution simply?

I don't understand how the statements are equivalent (must be some formula I've never seen before), or how the invertibility proves a basis. Thanks!

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$$ Q = (t_1, \dots, t_n) $$ we have $$ [t_i]_T = e_i = A t_i $$ where the $e_i$ are the canonical base vectors. So we need a matrix $A$ with $$ A Q = I \iff \\ A = Q^{-1} $$ That is why $$ [u]_T = Q^{-1} u $$ The $n$ vectors $Q^{-1} v_i$ form a basis if they are linear independent, thus if the system $$ Q^{-1}V x = (Q^{-1} v_1, \dotsc, Q^{-1} v_n) x = 0 $$ has the only solution vector $x = 0$. This is the case because $$ \det(Q^{-1} V) = \det(Q^{-1}) \det(V) \ne 0 $$ because $Q$ is invertible and $V$ is a matrix of basis vectors.

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    Why must AQ = I? I don't really understand how it's necessary to have the identity matrix be involved here – 2017-02-15
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    A basis vector $b_i$ regarding its own basis always has the coordinates of the canonical vector $e_i$, and $I = (e_1, \dotsc, e_n)$. – 2017-02-15
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    What exactly is a canonical vector? Is it like the corresponding rref column? Your edited answer makes perfect sense though – 2017-02-15
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    $e_i$ has the $i$-th component $1$ and all other components $0$. – 2017-02-15