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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. $\Box$

RedRose
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  • What definition of orthogonal matrix are you given? Is it '$A$ is orthogonal matrix if $A^{-1} = A^T$'? – Student Mar 19 '17 at 16:06
  • Yes it is $A^T=A^{-1}$ – RedRose Mar 19 '17 at 16:08
  • 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 $n$. – Student Mar 19 '17 at 16:11
  • Mostly my problem is with the "b)"...But what $n=3$ ? – RedRose Mar 19 '17 at 16:13
  • For (b), it might be usefull to denote $A$ using rows. This allows you to use the same reasoning you have used in (a). Also, note that you have shown only one implication in each point, so you still need to show that if $A$ is orhtogonal, then both rows and columns form orthonormal sets. – Student Mar 19 '17 at 16:14
  • My comment about $n = 3$ is that you did not prove this statement for an $n \times n$-matrix but for a $3 \times 3$ matrix. However, the result is proven in exactly the same way. – Student Mar 19 '17 at 16:15
  • In my assignment I used $n=3$ just as example – RedRose Mar 19 '17 at 16:23
  • oh, okey :) anyway, if you denote $A$ using row-notation, (b) should be exactly the same as (a), using $AA^T = I$ as you mentioned. – Student Mar 19 '17 at 16:25
  • Okey thanks, but I don't still get it...so the way I did the "b)" is correct? – RedRose Mar 19 '17 at 16:34
  • No, not really, your result seems to be a a vector having as only entries $1$'s... – Student Mar 19 '17 at 16:36
  • Oh :( Could you show short example of what should I do? If it is not too much to ask. – RedRose Mar 19 '17 at 16:38
  • Posted an answer for part (b). – Student Mar 19 '17 at 16:46
  • Btw I forgot to ask that is it same whether I use $AA^T$ or $A^TA$ for both a and b? – RedRose Mar 19 '17 at 19:17
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    For (a) $A^TA$ is easier and for the second one the other alternative is easier. – Student Mar 19 '17 at 19:19

1 Answers1

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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.

Student
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  • 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? – RedRose Mar 19 '17 at 16:51
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    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). – Student Mar 19 '17 at 16:53
  • One more thing, as I noticed it in textbook, that isn't $a_ka_k^T$ a matrix and only $a_k^Ta_k=1$ ? – RedRose Mar 19 '17 at 20:48
  • 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 :) – Student Mar 19 '17 at 21:10
  • 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. :/ – RedRose Mar 19 '17 at 21:15
  • You can only multiply an $m \times n$ -matrix with an $n \times k$ matrix... Otherwise multiplication is not defined :) it should be the first thing you see about matrixmultiplication :) – Student Mar 19 '17 at 21:17