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I have an orthogonal matrix $A^{n\times n}$, that is $A^TA=I$. What is the number of equations $A$ has to satisfy?

It seems to me that the answer is $\frac{n}{2}(n+1)$. My reasoning as follows. Each element of $A^TA$ corresponds to an equation. The diagonal elements are not repeated but the off diagonal elements in the upper triangular matrix are repeated also in the lower triangular matrix. So we need half the elements + the rest of the diagonal matrix that is $n^2/2 + n/2$. Is this correct?

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  • $\begingroup$ Your title and your question is different. $\endgroup$ – Dog_69 May 12 '18 at 16:55
  • $\begingroup$ When are "equations" linearly independent? The matrix equality $AA^T=I$ corresponds to $n^2$ equations in the coefficients. $\endgroup$ – Dietrich Burde May 12 '18 at 16:57
  • $\begingroup$ Sorry, I don't know why I put that in the title. I just mean the number of equations that arise from $A^T A = I$ by equating the elements of the right hand side to the elements on the left hand side. Some of those are the same of course. $\endgroup$ – berrygreen May 12 '18 at 17:14
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    $\begingroup$ In formal language, you are asking about the dimension of the orthogonal group $\operatorname{O}_n$. Yes, dimension is a subtle notion that isn't easy to define; but it's still one of the easiest ways to make "number of independent equations" rigorous. If you prefer algebra, you can ask for the maximum number of algebraically independent elements in the coordinate ring of $\operatorname{O}_n$ instead. $\endgroup$ – darij grinberg May 12 '18 at 18:12
  • $\begingroup$ To add slightly to @darijgrinberg's comment; the two equations $x+y = 1, x-y = 1$ are equivalent to the single equation $(x-1)^2 + y^2 = 0$, so the notion of equation in the context of degrees of freedom needs more care than just equation counting. $\endgroup$ – copper.hat May 12 '18 at 22:38
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As noted in the comments $$ A^T A = I $$ is equivalent to $n^2$ equations in the matrix components: $$ \sum_j a_{ij}^T a_{jk} = \sum_j a_{ji} a_{jk} = \delta_{ik} \quad\quad (i, k \in \{1,\dotsc, n\}) $$ On the main diagonal we have $i=k$ and up to $n$ different equations $$ \sum_j a_{ji}^2 = 1 $$ Otherwise we have $$ 0 = \sum_j a_{ji} a_{jk} = \sum_j a_{jk} a_{ji} $$ which gives up to $(n-1) + (n-2) + \dotsb + 1 = (n-1)n/2$ different equations.

So we have up to $n(n+1)/2$ different equations.

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  • $\begingroup$ yes but some of the equations are redundant, I only want to count equations that tell me something new $\endgroup$ – berrygreen May 12 '18 at 17:37
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Since $A^TA$ is symmetric, it is not difficult to see that $A^TA=I$ can be written in the form of $n(n+1)/2$ "different" equations. Unfortunately, the concept of "different equations" has no geometric interpretation. That is important is the concept of algebraically independent relations between the $(a_{i,j})$ (cf. the Darij Grinberg's post); here our $n(n+1)/2$ relations are independent but we must prove that. To do that, we can use the @DisintegratingByParts 's post; yet, is not easy because behind, there is a fibration...

The easiest and best method is to calculate the dimension $d$ of the kernel of $Df_A$ where $f:A\rightarrow A^TA-I$. Then the number of degrees of freedom of $O(n)$ is $n^2-d$. On this website, I wrote the associated proof several times; unfortunately, when talking about derivatives, most students throw themselves on the "matrix cookbook"; then, of course, they do not understand anything about the proof they read.

In short, I am discouraged and I will not write the same proof again (it suffices to use the search button). When I see that the OP gave the green chevron to @mvw (he himself knows that his answer is just smoke), I tell myself that there is still a long way ahead of us.

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The matrix equation $A^{T}A=I$ is equivalent to saying that the columns of $A$ form an orthonormal basis. There are $n$ length equations. After that, there are $n-1$ equations stating that column $1$ is orthogonal to columns $2$ to $n$. And there are $n-2$ equations stating that column $2$ is orthogonal to columns $3$ to $n$. So the total number of equations is $$ n+(n-1)+(n-2)+\cdots+1=\frac{n(n+1)}{2} $$

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