The Unapologetic Mathematician

Mathematics for the interested outsider

Unitary and Orthogonal Matrices

Let’s see what happens when we take a unitary or orthogonal transformation and turn it into a matrix by picking a basis for our vector space.

First a unitary transformation on a complex vector space. We pick a basis \left\{\lvert i\rangle\right\}_{i=1}^n and set up the matrix

\displaystyle u_{ij}=\langle i\rvert U\lvert j\rangle

We can also set up the matrix for the adjoint

\displaystyle {u^*}_{ij}=\langle i\rvert U^*\lvert j\rangle=\overline{\langle j\rvert U\lvert i\rangle}=\overline{u_{ji}}

That is, the adjoint matrix is the conjugate transpose. This isn’t really anything new, since we essentially saw it when we considered Hermitian matrices.

But now we want to apply the unitarity condition that U^*U=I_V. It will make our lives easier here to just write out the sum over the basis in the middle and find


Now, this isn’t particularly useful on its face. I mean, what does that mess even mean? But if nothing else it tells us that we can describe unitary matrices in terms of (a lot of) equations involving only complex numbers. We can then pick out all the n\times n complex matrices which represent unitary transformations. They form the “unitary group” \mathrm{U}(n).

What about orthogonal matrices? Again, we pick a basis \left\{\lvert i\rangle\right\}_{i=1}^n to get a matrix

\displaystyle o_{ij}=\langle i\rvert O\lvert j\rangle

and also a matrix for the adjoint

\displaystyle {o^*}_{ij}=\langle i\rvert O^*\lvert j\rangle=\langle j\rvert O\lvert i\rangle=o_{ji}

Here the adjoint matrix is just the transpose, not the conjugate transpose, since we’re working over a real inner product space. Then we can write down the orthogonality condition


Again, this doesn’t really seem to tell us much, but we can use these equations to cut out the matrices which represent orthogonal transformations from all n\times n real matrices. They form the “orthogonal group” \mathrm{O}(n).

But there’s something else we should notice here. The equations for the unitary group involved complex conjugation, so we need some structure like that to talk sensibly about unitarity. However, the orthogonality equations only involve basic field operations like addition and multiplication, and so these equations make sense over any field whatsoever. That is, given a field \mathbb{F} we can consider the collection of all n\times n matrices with entries in \mathbb{F}, and then impose the above orthogonality condition to cut out the matrices in the orthogonal group \mathrm{O}(n,\mathbb{F}), while the first orthogonal group is \mathrm{O}(n)=\mathrm{O}(n,\mathbb{R}).

One useful orthogonal group is \mathrm{O}(n,\mathbb{C}). This is not the same as the unitary group \mathrm{U}(n), though it can be confusing to keep the two separate at first. The unitary group consists of matrices whose inverses are their conjugate transposes, instead of just their transposes for the complex orthogonal group. The unitary group preserves a sesquilinear inner product, which has a clear geometric interpretation we’ve been talking about. The orthogonal group preserves a bilinear form, which doesn’t have such a clear visual interpretation. They are related in a way, but we’ll be coming back to that subject much later on.

July 29, 2009 Posted by | Algebra, Linear Algebra | 4 Comments