是。因为\(H\)是\(\C^{2\times 2}\)的子集，且对矩阵加法和实数数乘封闭： 设\(A = \begin{pmatrix}\alpha_{1}&\beta_{1} \\ -\overline{\beta_1}&\overline{\alpha_1}\end{pmatrix}, B = \begin{pmatrix}\alpha_{2}&\beta_{2} \\ -\overline{\beta_2}&\overline{\alpha_2}\end{pmatrix} \in H\)，则
\begin{align*}
A + B \amp = \begin{pmatrix}\alpha_{1}+\alpha_{2}&\beta_{1}+\beta_{2} \\ -(\overline{\beta_1}+\overline{\beta_2})&\overline{\alpha_1}+\overline{\alpha_2}\end{pmatrix} \\
\amp = \begin{pmatrix}\alpha_{1}+\alpha_{2}&\beta_{1}+\beta_{2} \\ -\overline{\beta_1+\beta_2}&\overline{\alpha_1+\alpha_2}\end{pmatrix} \in H,
\end{align*}
加法关于\(H\)封闭。对任意\(c\in\R\)，\(cA = \begin{pmatrix}c\alpha_{1}&c\beta_{1} \\ -c\overline{\beta_1}&c\overline{\alpha_1}\end{pmatrix} = \begin{pmatrix}c\alpha_{1}&c\beta_{1} \\ -\overline{c\beta_1}&\overline{c\alpha_1}\end{pmatrix}\)，因为\(c\)是实数，所以\(\overline{c\beta_1}=c\overline{\beta_1}\)，\(\overline{c\alpha_1}=c\overline{\alpha_1}\)。因此\(cA\in H\)。实数数乘关于\(H\)也封闭。同时矩阵的加法和数乘满足线性空间要求的其他性质，所以\(H\)构成\(\R\)上的线性空间。下面求维数与基。设\(\alpha = a+b{\rm i}, \beta = c+d{\rm i}\)，其中\(a,b,c,d\in\R\)。则
\begin{align*}
\begin{pmatrix}\alpha & \beta \\ -\overline{\beta} & \overline{\alpha}\end{pmatrix}= \amp \begin{pmatrix}a+b{\rm i}&c+d{\rm i}\\ -c+d{\rm i}&a-b{\rm i}\end{pmatrix}\\
=\amp \phantom{+} a \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} + b \begin{pmatrix}{\rm i}& 0 \\ 0 & -{\rm i}\end{pmatrix} \\
\amp + c \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix} + d \begin{pmatrix}0 & {\rm i}\\ {\rm i}& 0\end{pmatrix}.
\end{align*}
记
\begin{equation*}
I= \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix},\quad J = \begin{pmatrix}{\rm i}& 0 \\ 0 & -{\rm i}\end{pmatrix},
\end{equation*}
\begin{equation*}
K = \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix},\quad L = \begin{pmatrix}0 & {\rm i}\\ {\rm i}& 0\end{pmatrix}.
\end{equation*}
则显然这四个矩阵是实数域上线性无关的（因为系数\(a,b,c,d\)唯一确定）。且任意\(H\)中矩阵可由它们线性表示，所以\(\dim_{\R}H = 4\)，它的一个基为\((I, J, K, L)\)。
