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The assignment:

a) Prove that square-matrix A is orthogonal if and only if A has orthonormal columns.

b) Prove that square-matrix A is orthogonal if and only if A has orthonormal rows.

So I know that A matrix has orthonormal columns if and if only $A^TA=I$.

But how about orthonormal rows? Should I use $AA^T=I$ ?


b) For example, can I prove like this (?) :

Let be $A=\begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix}$

$AA^T=I$

$AA^T=\begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix} \times \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}^T = \begin{bmatrix} a_1a_1^T & a_2a_2^T & a_3a_3^T \end{bmatrix}$

$a_1a_1^T=1 \quad\quad a_2a_2^T=1 \quad\quad a_3a_3^T=1$

So $A$ is orthogonal, because rows of matrix A are orthonormal. $\Box$


a) I did it like this which I think is correct:

$A^TA=I$

$A^TA=\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}^T \times \begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix} = \begin{bmatrix} a_1^Ta_1 & a_1^Ta_2 & a_1^Ta_3 \\ a_2^Ta_1 & a_2^Ta_2 & a_2^Ta_3 \\ a_3^Ta_1 & a_3^Ta_2 & a_3^Ta_3 \end{bmatrix}$

$a_1^Ta_2=0 \quad a_1^Ta_3=0$

$a_2^Ta_1=0 \quad a_2^Ta_3=0$

$a_3^Ta_1=0 \quad a_3^Ta_2=0$

$a_1^Ta_1=1 \quad\quad a_2^Ta_2=1 \quad\quad a_3^Ta_3=1$

So $A$ is orthogonal, because columns of matrix A are orthonormal. $\Box$

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  • $\begingroup$ What definition of orthogonal matrix are you given? Is it '$A$ is orthogonal matrix if $A^{-1} = A^T$'? $\endgroup$
    – Student
    Commented Mar 19, 2017 at 16:06
  • $\begingroup$ Yes it is $A^T=A^{-1}$ $\endgroup$
    – RedRose
    Commented Mar 19, 2017 at 16:08
  • $\begingroup$ Your solution for (a) looks fine, up to some small details: 1) you denote $a_1$ for the first column of $A$, but then your notation for $A^T$ is not correct: the transpose should be at each column $a_i$ instead of how it is now. 2) You have shown this for $n = 3$, but you can do this (completely the same way) for general $n$. $\endgroup$
    – Student
    Commented Mar 19, 2017 at 16:11
  • $\begingroup$ Mostly my problem is with the "b)"...But what $n=3$ ? $\endgroup$
    – RedRose
    Commented Mar 19, 2017 at 16:13
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    $\begingroup$ For (a) $A^TA$ is easier and for the second one the other alternative is easier. $\endgroup$
    – Student
    Commented Mar 19, 2017 at 19:19

1 Answer 1

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For (b): Let me denote the matrix $A$ as follows: $$\begin{pmatrix} - & a_1 & -\\ - & a_2 & -\\ & \vdots & \\ - & a_n & - \end{pmatrix}$$ where the $a_i$ are row vectors and I emphasised this by adding '-'. We know that a matrix $A$ is orthogonal if $AA^T = I$. We want to show that the rows of $A$ form an orthonormal set, so let us take two arbitrary rows, $a_j$ and $a_k$, with $1 \leq j,k \leq n$. Note that we have that $$AA^T = \begin{pmatrix} - & a_1 & -\\ - & a_2 & -\\ & \vdots & \\ - & a_n & - \end{pmatrix}\begin{pmatrix} | & | & & | \\ a_1^T & a_2^T & \ldots & a_n^T\\ | & | & & | \end{pmatrix} = I$$ so if we compute $a_ja_k^T$, this corresponds to the entry in row $j$, column $k$ of the identity matrix. This entry is equal to $0$ if $j \neq k$ and equal to $1$ if $j = k$. This shows that the rows of $A$ form an orthonormal set. The other implication (orhtonormal rows implies $A$ orthogonal) follows in the same way.

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  • $\begingroup$ Okey thanks :) So as I have understood the a and b goes in same way as your answer...only in a) contents are horizontally and in b) vertically? $\endgroup$
    – RedRose
    Commented Mar 19, 2017 at 16:51
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    $\begingroup$ Yes, since we can easily express the product of two matrices by multiplying rows of the first matrix with columns of the second matrices, we like to have a row notation in the first matrix. And depending on what statement we want to prove, we can denote $A$ using rows (part b) or columns (part a). $\endgroup$
    – Student
    Commented Mar 19, 2017 at 16:53
  • $\begingroup$ One more thing, as I noticed it in textbook, that isn't $a_ka_k^T$ a matrix and only $a_k^Ta_k=1$ ? $\endgroup$
    – RedRose
    Commented Mar 19, 2017 at 20:48
  • $\begingroup$ Here is have denoted a rowvector by $a_k$, hence $a_ka_j^T$ is a number. If $a_k$ would have been a columnvector, then it would have been a matrix. If you are not sure about the dimensions of a product of matrices, look at the dimension of each matrix and you have the result :) $\endgroup$
    – Student
    Commented Mar 19, 2017 at 21:10
  • $\begingroup$ Hmm...Is there online some very detailed article or something about that rule? I have searched it online and in textbook but not much info. :/ $\endgroup$
    – RedRose
    Commented Mar 19, 2017 at 21:15

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