Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thank you for your time!

share|improve this question
    
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

2 Answers 2

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.

share|improve this answer
    
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.