Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.