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.

  • 1
    $\begingroup$ 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). $\endgroup$ Commented Oct 14, 2017 at 19:24
  • $\begingroup$ Since you know that $A$ is full rank, its left inverse is equal to its right inverse, so if $A^TA=I$, $AA^T=I$. $\endgroup$
    – Paul
    Commented Oct 14, 2017 at 19:28
  • $\begingroup$ @ПетяНарышкин Thank you. But I do not really understand what you say. Can you provide more details? $\endgroup$
    – Mr.Robot
    Commented Oct 14, 2017 at 19:39
  • $\begingroup$ @Paul I just realize $A$ is full rank. But could I directly conclude $A^TA =I$ without necessary reasoning? $\endgroup$
    – Mr.Robot
    Commented Oct 14, 2017 at 19:42
  • $\begingroup$ @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. $\endgroup$ Commented Oct 14, 2017 at 19:45

1 Answer 1


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 $$AA^{T} = AA^{-1} = I.$$

  • $\begingroup$ Now I finally understand. Thank you very much! $\endgroup$
    – Mr.Robot
    Commented Oct 14, 2017 at 20:52
  • $\begingroup$ @Mr.Robot, No problem $\endgroup$ Commented Oct 14, 2017 at 20:53
  • $\begingroup$ I am wondering how you could see it through without any difficulty and so fast. Is there a theorem or something related to this topic? $\endgroup$
    – Mr.Robot
    Commented Oct 14, 2017 at 21:19
  • $\begingroup$ @Mr.Robot Well, not really, just quite a good understanding of these concepts. The orthogonal matrices are actually pretty simple. Once you understand how they work everything becomes easy. I can't really explain it here (there are a lot of aspects involved) but I'm sure you will learn all of it soon. Just keep doing those excercises=) Also, uh, do you mind marking my answer as corrext and/or upvoting it? $\endgroup$ Commented Oct 14, 2017 at 21:35
  • $\begingroup$ I have already done it. Thank you again for your advice and solutions. $\endgroup$
    – Mr.Robot
    Commented Oct 14, 2017 at 22:08

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