# Linear transformation of orthonormal vectors and orthogonality of transformation matrix

Question:

Let $v_1, v_2,\cdots, v_n$ be orthonormal vectors in $\mathbb{R^n}$. Show that $Av_1.Av_2,\cdots,Av_n$ are also orthonormal if and only if $A\in \mathbb{R^n}$ is orthogonal.

What I have Done:

$P: Av_1.Av_2,\cdots,Av_n$ are orthonormal

$Q: A \in \mathbb{R^n}$ is orthogonal

$Q \Rightarrow P$ is quite straightforward. But what confuses me is the proof of $P \Rightarrow Q$.

If $P$, then we could write $$(Av_i)^T(Av_j)=v_i^TA^TAv_j= \begin{cases} 1,i=j\\ 0,i\neq j \end{cases}$$ But what is next. Even though we could say $A^TA = I$, which I do not think I could conclude directly from the formulas we have, how should I prove $AA^T = I$, which is the premise to define a orthogonal matrix.

Thank you in advance. Any help would be much appreciated.

• You are almost there. Just write the matrix $A^{T}A$ in the basis $v_1, v_2, \ldots, v_n$ (you have done it already, in fact, just haven't realised it). – Petr Naryshkin Oct 14 '17 at 19:24
• Since you know that $A$ is full rank, its left inverse is equal to its right inverse, so if $A^TA=I$, $AA^T=I$. – Paul Oct 14 '17 at 19:28
• @ПетяНарышкин Thank you. But I do not really understand what you say. Can you provide more details? – Mr.Robot Oct 14 '17 at 19:39
• @Paul I just realize $A$ is full rank. But could I directly conclude $A^TA =I$ without necessary reasoning? – Mr.Robot Oct 14 '17 at 19:42
• @Mr.Robot, yes, you can conclude it directly. Notice that $v_1, \ldots, v_n$ is an orthonormal basis so every vector can be written as $\sum c_iv_i$. The formulas for $v_iA^{T}Av_j$ (you made a mistake, by the way, they are actually formulas for $v_i^{T}A^{T}Av_j$, which is just an inner product of $v_i$ and $A^{T}Av_j$) allow you to compute $A^{T}A(\sum c_iv_i)$. Turns out it is equal to $\sum c_iv_i$ which means $A^{T}A=I$ by definition of the identity matrix. – Petr Naryshkin Oct 14 '17 at 19:45

Firstly, since $\{v_i\}$ is an orthonormal system of $n$ vectors in $\mathbb{R}^n$, they form an orthonormal basis. Therefore, each vector $v \in \mathbb{R}^n$ can be written as the sum $$v = \sum\limits_{i=1}^n (v, v_i)v_i,$$ where $(\cdot, \cdot)$ is the standard inner product on $\mathbb{R}^n$. This gives us \begin{multline*} A^{T}A(v) = A^{T}A(\sum\limits_{i=1}^n (v, v_i)v_i) = \sum\limits_{i=1}^n (v, v_i)A^{T}A(v_i) = \sum\limits_{i=1}^n (v, v_i)\sum\limits_{j=1}^n(A^{T}A(v_i), v_j)v_j \\= \sum\limits_{i=1}^n (v, v_i)\sum\limits_{j=1}^nv_j^{T}A^{T}A(v_i)v_j = \sum\limits_{i=1}^n\sum\limits_{j=1}^n (v, v_i)\delta_{ij}v_j = \sum\limits_{i=1}^n (v, v_i)v_i = v \end{multline*} where $$\delta_{ij}=\begin{cases}1 & i=j \\ 0 & i \ne j\end{cases}.$$ Therefore for every $v \in \mathbb{R}^n$ $$A^{T}Av = v$$ which means $$A^{T}A = I.$$
Like @Paul mentioned in the comments, $A^{T}A = I$ means $A^{T} = A^{-1}$ and thus $$AA^{T} = AA^{-1} = I.$$