Are linear maps defined in a 1-1 manner by setting their kernel and image? In other words, If I have a vector space, and I define a set to be the kernel of my would-be linear map, and another set to be it's image. Would I get a well defined, one linear map? (Given that my Ker and Im are okay, e.g. the dimensions are fine, etc.) I have the following task: Let T be a linear map in R^4. Given that orthogonal_space_of(k e rT) = { (1, 2, 0, 4) , ( 1, 0,1, 0) }, and T(1,0,1,1)=(1,2,1,1), give an example of such T ("you don't have to explicitly find T(x1,x2,x3,x4).") Any ideas on how to do that (how can I characterize a linear map T without finding an explicit formula for it?) I thought about finding it's Ker and Im, but that doesn't work. Any thoughts?

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    $\begingroup$ Any two automorphisms of a given space, for example, have the same kernel/image data. $\endgroup$ – Dylan Moreland Jun 16 '12 at 17:32

No. A linear map $f : V \to W$ is uniquely determined by the images of a basis of $V$. Consider given $\mathrm{Ker} \, f$ and $\mathrm{Im} \,f$. When you define $\mathrm{Ker} \, f$, you are specifying the images of some elements of a basis. (Suppose $V$ is finite dimensional, $\mathrm{Ker} \, f = \langle v_1, \cdots, v_k \rangle$, with $\{v_1, \cdots, v_k\}$ linearly independent, and extend this set of vectors to a basis $\{v_1, \cdots, v_n\}$). If $\mathrm{Ker} \, f \neq V$, i.e., if $f \neq 0$, you still have some elements of the basis that you have to map somewhere to determine the linear transformation: $v_{k+1}, \dots, v_n$.

Suppose you have one such linear map, with $f(v_{i}) = y_i$, for all $i \in {k+1, \dots, n}$, and $\mathrm{Im} \,f = \langle y_{k+1},...y_n \rangle$. Then you have a couple of ways in which you could get a different linear map while keeping $\mathrm{Ker} \, f$ and $\mathrm{Im} \,f$: for any $i \in {k+1, \dots, n}$, you could redefine $f(v_{i}) = k\cdot y_i$, with $k$ any scalar, or $f(v_{i}) = y_j$, with $j \in {k+1, \dots, n}$ such that $y_j \neq y_i$, or even define $f(v_{i})$ to be some linear combination of the vectors in the image. That is, it's very easy to change a linear map and keep the image and the kernel.

A very simple example: consider $f(x) = 2x$ and $g(x) = 3x$, two linear maps $\mathbb{R} \to \mathbb{R}$ with $\mathrm{Ker} \,f = \mathrm{Ker} \,g = {0}$ and $\mathrm{Im} \,f = \mathrm{Im} \,g = \mathbb{R}$.

To answer your second question: You cannot uniquely determine such a map, but there are lots of them. Since you have $\mathrm{Ker} \,f$ (first you have to calculate the orthogonal complement of the given subspace), pick a basis of it, add the other element whose image you know ($(1,0,1,1)$) and add vectors in order to get a basis of $\mathbb{R}^4$. Then choose any images for the vectors on which you don't have any other conditions. The important thing is: a linear map is uniquely determined by the images of the elements of a basis of the domain, so you could give an answer in the following form, without computing an explicit equation for $T(x_1,x_2,x_3,x_4)$:

$\begin{cases} T(v_1) = y_1\\ T(v_2) = y_2\\ T(v_3) = y_3\\ T(v_4) = y_4 \end{cases}$, where $\{v_1, v_2, v_3, v_4\}$ is a basis of $\mathbb{R}^4$.

  • $\begingroup$ So can I do that with any given dimension of the image (in my case 2)? (see my updated question) and by so define a linear map? $\endgroup$ – idan Jun 16 '12 at 17:40

No. As for example for $a \in \mathbb R \setminus \{0\}$ all the maps $f_a\colon \mathbb R^2 \to \mathbb R^2$, $f_a(x) = (0,ax_1)$ have kernel and image $\{0\} \times \mathbb R$.


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