Let $T\colon C\to C$ be a linear transformation such that $T\begin{pmatrix}x\\y\end{pmatrix}= \begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$ Define $S(x+iy)= (ax+by)+i(cx+dy)$ Then is $S$ a linear transformation for all $a, b, c, d\in C$? It is a L.T. if $a = d$ and $b=-c$, but doubtful for arbitrary $a, b, c, d$. Please give your precious suggestions. Thank you.
S(x+iy)= (ax+by)+i(cx+dy)
-
2After asking over ten questions, please learn how to format your question using LaTeX/MathJax markup. – 2012-12-26
1 Answers
Note that $\mathbb{C}$ can be viewed as a vector space over various fields $\mathbb{F}$. So the answer really depends on $\mathbb{F}$.
By definition, $S$ is linear if $S(\alpha z_1 + z_2)=\alpha S(z_1)+S(z_2)$ for any $\alpha\in\mathbb{F}$ and $z_1,z_2\in\mathbb{C}$. When $\mathbb{F}=\mathbb{R}$ (so that $\mathbb{C}$ is a two-dimensional vector space over $\mathbb{F}=\mathbb{R}$ with basis $\{1,i\}$), you may verify that $S$ is linear.
If $\mathbb{F}=\mathbb{C}$ (so that $\mathbb{C}$ one-dimensional vector space over $\mathbb{F}$ with basis $\{1\}$) and the $a,b,c,d$ in your definition of $S$ are real numbers, then as you said, $S$ is linear iff $a=d$ and $b=-c$. However, if $a,b,c,d$ --- as you have stated in your question --- are complex numbers, then the necessary and sufficient condition for $S$ to be linear is $b+c=i(a-d)$.
-
0I was stuck at $(a+ic)(x-X)=(b+id)(Y-y)$. Now I understand. Thank you. – 2018-12-14