[Math] Proof that the change of basis matrix is invertible

change-of-basislinear algebraproof-verificationvector-spaces

I'm going to briefly describe my idea and solution, which I don't know if it's correct, but I think it should be.

Suppose we have two basis $f$ and $e$ for a vector space $V$. We know
that the vectors of $e$ can be represented as a linear combination of
the vectors in $f$, since, by definition, $f$ spans $V$.

The change of basis matrix from $F = \begin{pmatrix} f_1 & \cdots &
f_n \end{pmatrix}$ to $E = \begin{pmatrix} e_1 & \cdots & e_n
\end{pmatrix}$ is $C$ in the following formula (see here for more details)

$$E = F \cdot C$$

Since the vectors of $F$ are linearly independent, then $F$ is
invertible, therefore we have

$$F^{-1}E = C$$

$E$ is also invertible for the same reason.

We know that a matrix $X$ is invertible if and only if $\det(X)
\neq 0$.

Thus we have

$$det(C) = det(F^{-1} \cdot E) = det(F^{-1}) \cdot det(E) \neq 0$$

Since $det(C) \neq 0$, then $C$ is invertible.

Best Answer

If I have understood your question properly, the argument should be the following. First of all $$ E = F \cdot C, $$ as you said. But then since $E$ is also a basis, there is a matrix $D$ such that $$ F = E \cdot D. $$ Therefore $$ E = F \cdot C = (E \cdot D) \cdot C = E \cdot (D \cdot C). $$ Since $E$ is a linearly independent, this yields that $D \cdot C$ is the identity matrix. You have proved that $C$ is invertible.

Clearly you can do the same starting with $F = E \cdot D = \dots = F \cdot (C \cdot D)$ to see that also $C \cdot D$ is the identity, but the above suffices anyway.

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