# Can a solution set be an Image of a linear transformation?

Given a solution set of a system of linear equations with dim(solution_set) = dim(SolSet) = 1, could it be the Image of a linear equation?

What we know is that the solution set is equal to the Kernel of a linear transformation, right?

So using the rank-nullity theorem we get this:

$\dim V = \dim\mathrm{Im}f + \dim\mathrm{Ker}f$
($\dim\mathrm{Ker}f = \dim(\text{SolSet})$ )

$\dim V = \dim\mathrm{Im}f + \dim(\text{SolSet}) \Rightarrow \dim\mathrm{Im}f = \dim V - \dim(\text{SolSet})$

For the $\dim\mathrm{Im}f$ to be equal to the of $\dim(\text{SolSet})$, $\dim V$ must be 2 times $\dim(\text{SolSet})$. (maybe using the 1st isomorphism theorem?) Is that possible?

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The solution set of a system of linear equations is the image of a linear transformation if and only if it is a subspace, if and only if the system of equations is homogeneous. –  Arturo Magidin Apr 30 '12 at 17:17

Yes, it is related to what you said.

For vector spaces, if $W\oplus U=V$ where $W$and $U$ are two subspaces, then there is always the projection map $\pi_W:V\rightarrow W$ which just deletes $U$. Since every subspace $W$ can be expressed as a summand this way, there will always be a linear projection for which $W$ is the image.

The rank-nullity theorem uses the isomorphism theorem to decompose the space into two pieces like this, and then makes an observation about their dimensions.

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Could you please explain it with a more simpler? (we haven't worked with projection maps at all - studying linear algebra for university) –  Chris May 1 '12 at 23:43
Sure :) As you probably know, if you have written $W\oplus U=V$, then each element $v$ has a unique representation $v=w_v+u_v\in W\oplus U$. This ensures that the map $\pi(w_v+u_v)=w_v$ for all $v\in V$ is well-defined. (I dropped the subscript on $\pi$ just to simplify things.) You can check the axioms to see that it is indeed a linear map. Try that and then let me know if more hints are necessary. –  rschwieb May 2 '12 at 0:54
What you meant to write here is this: "$\pi(w_v+u_v)=w_v$ " or this "$\pi(w_v+u_v)=v_v$" ? (since $v=w_v+u_v$) –  Chris May 2 '12 at 12:59
The first one is the one I mean. So for example, $V=\mathbb{R}\oplus\mathbb{R}=\{(a,0)\mid a\in \mathbb{R}\}\oplus \{(0,b)\mid b\in \mathbb{R}\}$ where the left summand is named $W$ and the right summand $U$. To project onto $W$, the map would look like: $\pi(a,b)=\pi((a,0)+(0,b))=(a,0)\in W$. –  rschwieb May 2 '12 at 13:05
1. The image of a linear transformation is always a subspace.

2. Suppose we have a system of linear equations $A\mathbf{x}=\mathbf{b}$. If the solution set is nonempty, when is it a subspace? Well, to be a subspace we need $\mathbf{0}$ to be in the set, so we need $A\mathbf{0}=\mathbf{b}$. So, a necessary condition for the solution set of a system of linear equations to be a subspace is that the system be homogeneous.

3. Conversely, the solution set of a homogeneous system of linear equations $A\mathbf{x}=\mathbf{0}$ is a subspace: it contains $\mathbf{0}$, and if $\mathbf{a}$ and $\mathbf{b}$ are solutions and $\alpha$ is a scalar, then $A(\mathbf{a}+\alpha\mathbf{b}) = A\mathbf{a}+\alpha A\mathbf{b}=\mathbf{0}+\alpha\mathbf{0}=\mathbf{0}$.

4. Given any subspace $W$ of $V$, there is always a linear transformation $T\colon V\to V$ with image $W$.

Proof. Let $\beta$ be a basis for $W$, and let $\gamma$ be a subset of $V$ such that $\beta\cup\gamma$ is a basis for $V$. Then we can define $T\colon V\to V$ by specifying what it does on $\beta\cup \gamma$. Let $$T\mathbf{u}=\left\{\begin{array}{ll} \mathbf{u}&\text{if }\mathbf{u}\in\beta\\ \mathbf{0}&\textbf{if }\mathbf{u}\in\gamma. \end{array}\right.$$ Then $\mathrm{Im}(T) = \mathrm{span}(\beta) = W$.

(There are many other linear transformations with image $W$)

In summary, the solution set of a system of linear equations is the image of a linear transformation if and only if the system is homogeneous.

A bit more generally, given an arbitrary system of linear equations $A\mathbf{x}=\mathbf{b}$ that has at least one solution $\mathbf{s}$ (that is, the system is consistent). Then you should verify that the set of all solutions is equal to $$\Bigl\{\mathbf{s}+\mathbf{s}_0\mid \mathbf{s}_0\text{ is a solution of }A\mathbf{x}=\mathbf{0}\Bigr\},$$ that is, the solution set is a translate of the solution set of a homogeneous system, hence the translate of the image of a linear transformation.

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Thank you for your answer! Could you explain how we got this result the solution set of a system of linear equations is the image of a linear transformation if and only if the system is homogeneous? (It was difficult for me to fully understand the whole process(/answer)) Thank you! –  Chris May 1 '12 at 23:41
@Chris: In other words, "Thank you for your answer, but I didn't understand it..." Well, that means I wasn't clear. Essentially: the image of a linear transformation is always a subspace, and subspaces are always images of (lots and lots of) linear transformations. So a subset is the image of a linear transformation if and only if it is a subspace. So your question is equivalent to "When is the solution set of a system of linear equations a subspace?" The answer is that it is a subspace if and only if the system is homogeneous. Putting the two "if and only if"s together gives the statement –  Arturo Magidin May 2 '12 at 1:38
Well, I didn't want to be rude or sth, but you still got the point :-) and now I got it! Thank you very much! –  Chris May 2 '12 at 12:53