# Orthogonal Matrix

1. Dec 15, 2009

### sjeddie

Is the definition of an orthogonal matrix:

1. a matrix where all rows are orthonormal AND all columns are orthonormal

OR

2. a matrix where all rows are orthonormal OR all columns are orthonormal?

On my textbook it said it is AND (case 1), but if that is true, there's a problem:
Say we have a square matrix A, and we find its eigenvectors, they are all distinct so A is diagonalizable. We put the normalized eigenvectors of A as the columns of a matrix P, and (our prof told us) P becomes orthogonal and P^-1 = P^T. My question is how did P become orthogonal straight away? By only normalizing its columns how did we guarantee that its rows are also orthonormal?

2. Dec 15, 2009

### rochfor1

It turns out that the rows of a square matrix are orthonormal if and only if the columns are orthonormal. Another way to express that the condition that all columns are orthonormal is that $$A^T A = I$$ (think about why this is). Then we see that if $$v \in \mathbb{R}^n$$, $$\parallel x \parallel^2 = x^T x = x^T ( A^T A ) x = ( A x )^T ( A x ) = \parallel A x \parallel^2$$, and therefore A is injective. Since we are working with finite-dimensional spaces, A must also be surjective, so for $$v \in \mathbb{R}^n$$, there exists $$w \in \mathbb{R}^n$$ with v = Aw, and therefore $$A A^T v = A A^T A w = A w = v$$, so $$A A^T = I$$ as well. You can check this this implies that the rows of A are orthonormal. The proof of the converse is similar.

Note that this argument relies on the finite-dimensionality of our vector space. If you move up to infinite dimensional spaces, there may be transforms T with $$T^*T = I$$ but $$T T^* \neq I$$. This type of behavior is what makes functional analysis and operator algebras fun!

Last edited: Dec 15, 2009
3. Dec 15, 2009

### rochfor1

Theres actually an easier way to see that $$A^T A = I$$ implies A is injective, I just tend to think in terms of isometries like I wrote. If v is such that Av = 0, then $$0 = A^T 0 = A^T A v = v$$, so A is injective. Some may prefer this purely algebraic argument.

4. Dec 15, 2009

### sjeddie

Ah I see, thank you rochfor1, the (A^T)(A) = I thing makes a lot of sense :)