Othorgonal matrices and full rank

Question

I have a question regarding orthogonal matrices. Is it the case that orthogonal matrices always have full rank?

I tried to illustrate a $2\times 2$ orthogonal matrix with $\det=-1$ and come to the conclusion that it actually has full rank. However, I read in the text book that an orthogonal matrix with $\det=-1$ can be diagonizable or not. I do think this has something to do with the rank and hence invertibility.

Answer 1

The diagonalisability has nothing to do with invertibility. Orthogonal matrices always have full rank and so are always invertible. However they do not have to be diagonalisable. For instance, any $2\times 2$ rotation which is not $\pm \mathrm{Id}$ is represented by a $2\times 2$ orthogonal matrix which is invertible but not diagonalisable over $\mathbb{R}$.

Answer 2

Yes. Indeed an orthogonal matrix has full rank.
An Orthogonal matrix is a square matrix which has orthonormal columns. Consider an $n\times n$ orthogonal matrix with columns $q_1, q_2,...,q_n$. Since $q_i'$s are orthonormal, they are linearly independent (Can you prove it?).
Hence the matrix has $n$ linearly independent columns. Therefore, rank of the matrix = $n$ = dimension of column space.

Answer 3

As the previous replies have said, yes, they do! I think one could "remember" this fact geometrically by seeing that an orthogonal matrix is a rotation matrix, and so it won't ever zero out any vector; in other words, it doesn't have a null space.