Let \(V\) be the set of complex numbers regarded as a vector space over the field of real numbers (Exercise 1). We define a function \(T\) from \(V\) into the space of \(2 \times 2\) real matrices, as follows. If \(z = x + iy\) with \(x\) and \(y\) real numbers, then \[ T(z) = \begin{bmatrix} x + 7y & 5y \\ -10y & x - 7y \end{bmatrix}. \]
Verify that \(T\) is a one-one (real) linear transformation of \(V\) into the space of \(2 \times 2\) real matrices.
Verify that \(T(z_1 z_2) = T(z_1) T(z_2)\).
How would you describe the range of \(T\)?
Solution
Write \[ J=\begin{bmatrix}7&5\\-10&-7\end{bmatrix}. \] Then \[ T(x+iy)=xI_2+yJ. \] This immediately shows real-linearity. If \(T(x+iy)=0\), the \((1,2)\) entry gives \(5y=0\), hence \(y=0\), and then the diagonal entries give \(x=0\). Thus \(T\) is one-to-one.
A direct multiplication gives \[ J^2=-I_2. \] Therefore, for \(z_1=x+iy\) and \(z_2=u+iv\), \[ T(z_1)T(z_2) =(xI+yJ)(uI+vJ) =(xu-yv)I+(xv+yu)J =T(z_1z_2). \]
The range is the two-dimensional real subspace \[ \operatorname{range}T=\{xI_2+yJ:x,y\in\mathbb R\}. \] Equivalently, it consists of the matrices \[ \begin{bmatrix}a&b\\c&d\end{bmatrix} \] satisfying \[ c=-2b,\qquad 5(a-d)=14b. \] Under matrix addition and multiplication this range is a subfield of \(M_2(\mathbb R)\) isomorphic to \(\mathbb C\).