I'm self studying Linear Algebra from Hoffman Kunze, and I've come upon this problem.
With complex number $z=x+iy$, $$T(z)=\begin{pmatrix} x-7y & 5y \\ -10y & x+7y \\ \end{pmatrix}$$ is a transformation from the set of complex numbers taken as a vectorspace over the field of real numbers into the set of $2\times2$ real matrices.
Show that $T$ is one-one real linear transformation and describe its range.
My attempt:
- $T(z)$ is one-one: Suppose $T(z)$ is one one. Then $T(z_1) = T(z_2) \implies z_1 = z_2$, that is, $T(z_1-z_2)=0\implies z_1-z_2=0$, that is only solution of $T(z)=0$ should be $z=0$.
Equating $$\begin{pmatrix}x-7y & 5y \\ -10y & x+7y \\ \end{pmatrix}=\begin{pmatrix}\ \ 0 & \ \ \ 0 \ \ \\ \ \ 0 &\ \ \ 0\ \ \\ \end{pmatrix}$$ we get $x=0,\ y=0 \implies z=0$
$T(z)$ is linear: $$T(cz_1+z_2) = \begin{bmatrix}cx_1+x_2 -7cy_1 -7y_2& 5cy_1+5y_2 \\ -10cy_1 -10y_2& cx_1+x_2+7cy_1+7y_2 \\ \end{bmatrix} \\ = cT(z_1)+T(z_2)$$ Hence $T$ is a linear one-one transform from vector space of complex numbers into $R^{2\times 2}$
Range of $T$ is a subspace of $R^{2\times 2}$. It can be written as $$T(z) = x\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}+y\begin{bmatrix} -7 & 5 \\ -10 & 7 \\ \end{bmatrix}$$ Since, $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$ and $\begin{bmatrix} -7 & 5 \\ -10 & 7 \\ \end{bmatrix}$ are linearly independent vectors, and span the range, we take them as a basis. Then the coordinate matrix of $T(x+iy)$ is $\left[T(x+iy)\right] = \begin{bmatrix} x \\ y \end{bmatrix}$.
However, I'm unsure as to how to describe the range of a transformation. What do I have to do/show?
$\endgroup$ 72 Answers
$\begingroup$So we have $\;T:\Bbb C_{\Bbb R}\to M_2(\Bbb R)\;$ and you proved this is an injective linear mapping, so $\;\dim\ker T=0\;$ , and by the dimensions theorem:
$$2=\dim_{\Bbb R}\Bbb C=\dim\ker T+\dim\,\text{Im}\,T=\dim\,\text{Im}\,T\implies\;T(\Bbb C)\;$$
is a plane in $\;M_2(\Bbb R)\;$, and it is thus spanned by both (linearly independent) columns of any matricial expression of $\;T\;$ . For example, taking the standard basis in both domain and codomain we get:
$$\begin{align*}&T1=\begin{pmatrix}1&0\\0&1\end{pmatrix}\implies 1 / 0 / 0/ 1\\{}\\&Ti=\begin{pmatrix}-7&5\\-10&7\end{pmatrix}\implies-7/5/-10/7\end{align*}\;\implies [T]=\begin{pmatrix}1&-7\\0&5\\0&-10\\1&7\end{pmatrix}$$
and it is now trivial that the image of $\;T\;$ is the subspace
$$\text{Im}\,T=\text{Span}\,\left\{\;\;\begin{pmatrix}1&0\\0&1\end{pmatrix}\;\;,\;\;\;\begin{pmatrix}-7&5\\-10&7\end{pmatrix}\;\;\right\}\le M_2(\Bbb R)$$
Observe that we didn't actually need the middle step ( with $\;T1,\,Ti\;$) , but I think it is always nice, and may even be important, to have it...and this proves you actually answered completely, or almost, your own question (maybe the above could help you mostly to understand the theoretical details). Good job
$\endgroup$ $\begingroup$when we know a generator system of the beginning space, so the image of this system is also a generator system of the image, and the theorem of the incomplete base allows us to ensure an extraction of a base starting the image of this system. so in your example
$\{(1,0),(0,1)\}$ is real base of $C$, so its image generate the range of $T$ that is $\{ \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) , \left( \begin{array}{cc} -7 & 5 \\ -10 & 7 \end{array} \right)\}$ a generator system of rage $T$, but you want to assure it is a base so you can check the linear independent of this two vectors or check that $T$ is injective and use the rang theorem, also you can prove this by use the rang of vectors notion.
the expression $T(x,y)=(x,y)$ is really not of $T$ but of the isomorphism ${\bar T}$ obtained by decomposition of $T$ respectivelly to basis mentienned
$\endgroup$
