# Let $Q$ is the matrix with the column vectors from orthogonal basis $\beta$

Let $Q$ be a matrix such that the column vectors form an orthogonal basis $\beta$={$v_1,\dots,v_n$} of $V$.
Let $\alpha$ be the standard ordered basis of $V$.
Then $[I]_\alpha^\beta=Q$.
I think it is trivial (or not..?) but I would like to see a complete proof.
Does anyone know how to prove that?

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Let $Q$ be a matrix such that the column vectors form an orthogonal basis $\beta=\{v_1,\dots,v_n\}$ of $V$.
Let $\alpha=\{e_1,e_2,...,e_j\}$ be the standard ordered basis of $V$.

Claim: $[I]_\alpha^\beta=Q$.

The proof is trivial in the sense that it uses things you already know. However, I would argue that it is important to know what facts go behind what statements. For this statement, we would use the following facts (loosely stated):

Lemma: a linear transformation is completely determined by what it does to a basis of the domain.

Lemma: matrix multiplication on the left defines a linear transformation from column vectors to column vectors

By definition, $[I]_\alpha^\beta \,e_j=v_j$ for all $j$ from 1 to $n$

Claim: $Q\,e_j=v_j$

This follows from the nature of matrix multiplication.

By our lemmas, because $Q$ maps the standard basis to the vectors in $\beta$ and because $[I]_\alpha^\beta$ maps this basis in the same way, we may conclude that the two represent the same linear transformation.

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