Is this a complete test to show that a matrix is orthogonal?

[tamtam402]

I used to test orthogonality by using the definition M^T = M^-1, which means I always calculated the inverse of the matrices. However, isn't it true that if M is orthogonal, then MM^T = I?

If we multiply both side by M^-1, we get M^T = M^-1.

Can I use this to proof the orthogonality of a matrix M, instead of calculating it's (often tedious) inverse?

 

[DonAntonio]

I used to test orthogonality by using the definition M^T = M^-1, which means I always calculated the inverse of the matrices. However, isn't it true that if M is orthogonal, then MM^T = I?

If we multiply both side by M^-1, we get M^T = M^-1.

Can I use this to proof the orthogonality of a matrix M, instead of calculating it's (often tedious) inverse?

Of course...it's exactly the same, right?!

DonAntonio

 

[tamtam402]

Of course...it's exactly the same, right?!

DonAntonio

This might be a dumb question, but would it be possible to have a matrix M where MM^T = I, yet M^TM ≠ I? I'm pretty sure that by definition, proving either of these 2 equalities is enough to know that M^-1 = M^T, but I wanted to make sure.

 

[micromass]

This might be a dumb question, but would it be possible to have a matrix M where MM^T = I, yet M^TM ≠ I? I'm pretty sure that by definition, proving either of these 2 equalities is enough to know that M^-1 = M^T, but I wanted to make sure.

That is actually a very good question. There is no reason why $MM^T=I$ should imply $M^TM=I$.
But this is actually quite a deep result in linear algebra. The result is that every left-invertible matrix is invertible. And likewise with right-invertible.
So if AB=I, then it holds that BA=I. This is a special result that only holds for matrices.

 

[DonAntonio]

That is actually a very good question. There is no reason why $MM^T=I$ should imply $M^TM=I$.
But this is actually quite a deep result in linear algebra. The result is that every left-invertible matrix is invertible. And likewise with right-invertible.
So if AB=I, then it holds that BA=I. This is a special result that only holds for matrices.

Well, that holds in any group and in any monoid where every element has a right inverse, since then ( with $\,a'=\,$ right inverse of $\,a$ ):
$$a'a=a'a(a'a'')=a'(aa')a''=a'ea''=a'a''=e$$
so the right inverse is also the left one.

DonAntonio