Jtdylktuy

I was trying to find a quicker way to identify whether a set of vectors was an orthogonal set or not. I discovered something interesting, but I don't understand how to put this intuition into words.

Let S = a set of 3 vectors, which I combined into a $M_{33}$

$

S
=
begin{bmatrix}
3\1\1
end{bmatrix}
,
begin{bmatrix}
-1\2\1
end{bmatrix}
,
begin{bmatrix}
-.5\-2\3.5
end{bmatrix}
=
begin{bmatrix}
3&-1&-.5\1&2&-2\1&1&3.5
end{bmatrix}
\$

$S^T$*S = D, a diagonal matrix

D =
$begin{bmatrix}
11&0&0\0&6&0\0&0&16.5
end{bmatrix}$

I noticed that this is a pattern between all orthogonal square matrices. I also noticed that this may hold up for non-square matrices as well. It seems that this has something to do with eigenvalues and the spectral theorem, but I only briefly learned how to calculate eigenvalues and eigenvectors just yesterday yesterday, and we aren't going to touch this 'spectral theorem' thing until the end of the semester.

My questions are:

1) If you could sum up briefly, how would you describe what is happening?

2) Does my theory hold up for non-square matrices as well? My proofing skills are not strong enough to walk through this by myself.

Let’s go back to the definition of matrix multiplication. The elements of the product $C=AB$ are given by the formula $c_{ij}=sum_k a_{ik}b_{kj}$. That is, the $ij$-th element of the product of two matrices is the dot product of the $i$th row of the first matrix with the $j$th row of the second.

Now, the rows of $S^T$ are the columns of $S$, so $S^TS$ consists of all of the pairwise dot products of the columns of $S$. Thus, its diagonal elements are the squares of the norms of those columns, and the off-diagonal elements are zero iff the corresponding pair of columns is orthogonal. None of this requires that $S$ be square, but the product $S^TS$ will always be a square matrix.