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 $V$, $<Tx,Ty>=<x,y>$. 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?

  • $\begingroup$ How can $a_{ij}$ be factored out if it depends on $i$? Same for $a_{lk}$. $\endgroup$ – Daniel Apr 7 '20 at 15:07
  • $\begingroup$ 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} $\endgroup$ – lonza leggiera Apr 7 '20 at 23:27

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