00:02
We're told that two real matrices a and b are said to be orthogynally equivalent if there exists an orthogonal matrix p such that b is equal to p transpose ap.
00:14
We're asked to show that this relation is an equivalence relation on real matrices.
00:21
To prove it's an equivalence relation, we'll want to prove that it is a reflexive relation, a symmetric relation, and a transitive relation.
00:36
Prove it is reflexive.
00:39
Well, notice that if a is a real matrix, then it follows that a is equal to a, which is the same as i transpose a, moreover, i is in fact an orthogonal matrix.
01:15
Its rows are mutually orthogonal and are unit vectors.
01:26
Therefore, it follows that matrices a and a are orthogonally equivalent.
01:43
Therefore, it follows that the relation is reflexive.
01:51
Now we want to show the relation is symmetric.
01:54
Well, we'll suppose that a and b are real matrices, and we'll also suppose that a and b are orthogonally equivalent.
02:18
That means that there exists an orthogonal matrix p, such that b is equal to p transpose times a times p now this implies this implies that a is equal to well we know that because p is orthogonal p transpose equals p inverse solving for a we get a is equal to the inverse of p transpose times b times the inverse of p then using interchangeability of transpose and inverse, this is the inverse of the transpose of p times b times p transpose.
03:24
P transpose.
03:26
You know what? i'm sorry.
03:35
I'm not going to do that.
03:36
This is p inverse...