# Prove the matrix of an orthogonal linear transformation relative to an orthonormal basis is orthogonal.

Let $$V$$ be an $$n$$-dimensional inner product space (not necessarily the standard inner product), and let $$L:V \rightarrow V$$ be an orthogonal linear transformation. Let $$B=\{v_1,...,v_n\}$$ be an orthonormal basis for $$V$$. Let $$A$$ be the matrix of $$L$$ relative to this basis. Prove that $$A$$ is orthogonal.

My attempt:

To prove that $$A$$ is orthogonal, I could prove that $$A^TA=I$$, or I could prove that the inner product of any two columns of $$A$$ is zero and the length of any column of $$A$$ is one. (I use standard the inner product for the first proof but not the second, right?)

I know how to prove that $$A$$ is orthogonal if $$V= \mathbb R^n$$, but I am not sure how to generalize to any orthogonal basis. This is how I would do it for $$V= \mathbb R^n$$:

1. Column $$i$$ of $$A$$ is simply $$T(e_i)$$, where $$e_i$$ is the $$i$$th vector in the $$n$$-dimensional standard basis.
2. Since $$T$$ is an orthogonal transformation and therefore preserves length, $$||T(e_i)||=||e_i||=1$$ since the basis is orthogonal.
3. Since $$T$$ is an orthogonal transformation, for all $$x,y$$ in $$V$$, $$=$$. Moreover, Since the basis $$B$$ is orthogonal, the inner product of any two basis vectors is $$0$$. Therefore, the inner product of any two columns of $$A$$ is $$0$$.
4. Since the length of every column of $$A$$ is $$1$$ and the inner product of any two columns of $$A$$ is $$0$$, $$A$$ is orthogonal.

My problem with generalizing to any $$V$$ is step 1: Specifically, what would $$A$$ look like for an arbitrary $$V$$?

Hint: For a general $$\ V\$$: $$T\left(v_j \right)= \sum_{i=1}^na_{ij}v_i\ ,$$ so $$\left\langle v_j, v_k\right\rangle= \left\langle T\left(v_j\right), T\left(v_k\right)\right\rangle = \left\langle \sum_{i=1}^na_{ij}v_i, \sum_{l=1}^na_{lk}v_l\right\rangle\ .$$ What happens if you use the bilinearity of the inner product to pull the coefficients of $$\ v_i\$$ and $$\ v_l\$$ in the rightmost term of these equations out of it?
• How can $a_{ij}$ be factored out if it depends on $i$? Same for $a_{lk}$. – Daniel Apr 7 '20 at 15:07
• Linearity of the inner product in its first argument implies that $$\left\langle \sum_{i=1}^na_{ij}v_i,\sum_{l=1}^na_{lk}v_l\right\rangle= \sum_{i=1}^na_{ij} \left\langle v_i,\sum_{l=1}^na_{lk}v_l\right\rangle\$$ and then linearity in its second argument implies that \begin{align} \left\langle \sum_{i=1}^na_{ij}v_i,\sum_{l=1}^na_{lk}v_l\right\rangle&= \sum_{i=1}^na_{ij} \sum_{l=1}^na_{lk} \left\langle v_i,v_l\right\rangle\\ &= \sum_{i=1}^n\sum_{l=1}^n a_{ij}a_{lk} \left\langle v_i,v_l\right\rangle\ . \end{align} – lonza leggiera Apr 7 '20 at 23:27