: Consider the special unitary Lie algebra
$$\mathfrak{su}(2) = \{A ∈ \mathfrak{gl}_2(\Bbb{C}) | A + A^∗ = 0, tr A = 0\} , A^∗= \overline{A}^t$$
over $\Bbb{R}$ and a bilinear form
$$(A, B) = \frac12tr(AB^∗), A, B ∈ \mathfrak{su}(2)$$.
Show that this form is symmetric. Show that $\mathfrak{su}(2)$ is 3-dimensional and has an orthonormal
basis $iσ_1, iσ_2, iσ_3$, where
$$σ_1 = \begin{pmatrix}0 & 1\\
1 & 0 \end{pmatrix}, σ_2 = \begin{pmatrix}0 & -i\\
i & 0 \end{pmatrix} σ_3 =
\begin{pmatrix}1 & 0\\
0 &-1 \end{pmatrix}$$
are called Pauli matrices.
I've gotten as far as proving its symmetric, by just using general examples for A and B, and equating them. I have no idea how to prove it is 3-dimensional, or that it has such a basis. My only background in this topic is in groups and fields, and some linear algebra, so any help would be greatly appreciated
Best Answer
For the question in the title you only need to use the definition of vector space dimension. The three matrices $i\sigma_1,i\sigma_2,i\sigma_3$ are a vector space basis of $\mathfrak{su}(2)$, so that $\dim \mathfrak{su}(2)=3$ over $\mathbb{R}$. To see this, note that they are linearly independent and span the vector space. Using a Lie algebra isomorphism, we see that $\mathfrak{su}(2)$ is isomorphic to $\mathfrak{so}(3)$, which consists of skew-symmetric $3\times 3$ matrices. In this case it is perhaps even more obvious to see that this vector space has real dimension $3$. Write the $3\times 3$ matrices as vectors in $9$-dimensional space and determine a basis.