Representing a transformation from C to C with respect to the basis 1, i

I am having trouble understanding why the transformation: $T(z) = (3+4i)z$ from C to C can be represented by the matrix $\begin{bmatrix} 3, -4 \\ 4, 3 \end{bmatrix}$ with respect to the basis $\begin{bmatrix} 1 \\ i \end{bmatrix}$.

Isn't the following matrix the correct matrix for the transformation? I believe it returns the correct vector. $\begin{bmatrix} 3, 0 \\ 0, 4 \end{bmatrix}$

Multipy $\begin{bmatrix} 3 & -4\\ 4 & 3 \end{bmatrix} \begin{bmatrix} a\\ b \end{bmatrix} = \begin{bmatrix} 3a - 4b\\ 4a+3b \end{bmatrix}$

Now consider transformation: $T(a+bi) = (3+4i)(a+bi) = 3a-4b+(4a+3b)i$

Hence matrix $\begin{bmatrix} 3 & -4\\ 4 & 3 \end{bmatrix}$ is correct.

• @Timbuc no, there are exactly 4 Apr 23 '15 at 14:41
• Yes. I assume the OP is taking the transpose of the coefficients matrix, not the matrix itself, as he wrote $\;T(1)\;$ instead of $\;(1)T\;$ . Apr 23 '15 at 14:44
• I believe T represents transformation in this case. This is from page 348 of Otto Bretscher's Linear Algebra with Applications Apr 23 '15 at 14:48
• @120MinuteMan Of course it represents a (linear) transformation: it is written sepcifically in the question, and what else could it represent? Apr 23 '15 at 15:22

Apparently you're taking $\;\Bbb C\;$ as a vector space over $\;\Bbb R\;$ . Apply $\;T\;$ on the basis elements:

\begin{align}&T(1)=3\cdot1+4\cdot1i\\{}\\&T(i)=-4\cdot1+3\cdot i\end{align}\;\;\implies\;\;[T]=\begin{pmatrix}3&\!\!-4\\4&3\end{pmatrix}=\begin{pmatrix}3&4\\\!\!-4&3\end{pmatrix}^t

• If $T(1) = 3 + 4i$ the first column should be \begin{bmatrix} 3\\ 4 \end{bmatrix} Apr 23 '15 at 14:40
• Yep. I already edited and wrote both possibilities, one for when the function's applied on the left, the second (with the transpose) when the function's applied on the right. Apr 23 '15 at 14:46