I'm doing well with linear algebra, but I keep coming across AX=0. I don't really get it's significance. I'm getting in the habit of solving most of the questions I'm asked by pure mechanics and that's when I know I need better understanding.

I think the most recent time I came across it was when we were talking about basis of matrices. There are a lot of other times when that's come up. Can anyone give me a quick explanation of what the homogenous equation AX=0 means and maybe a hint as to how that relates to linear algebra?

By default when I see that I know I end up doing row reductions or augmenting a matrix, depending on the context, but I haven't figure out what it means yet.

  • $\begingroup$ Have you studied the kernel/null space? $\endgroup$ Mar 21, 2012 at 20:27
  • $\begingroup$ We just learned null, column, and row space. We had one lecture on the rank of a matrix. $\endgroup$
    – brandon
    Mar 21, 2012 at 20:28
  • $\begingroup$ I suspect that there are interesting results (isomorphisms) when considering quotient spaces, but that is above my pay grade! Sure wish I had been asking these questions when I first studied Linear Algebra... $\endgroup$ Mar 21, 2012 at 20:31

1 Answer 1


By asking "What is important about homogeneous equations?" You come pretty close to asking "Why is linear algebra important?" Most every problem in linear algebra no matter how abstract at some point boils down to solving linear systems.

In general a linear system can be written in the form $A{\bf x} = {\bf b}$. If you take any two solutions ${\bf x}_1$ and ${\bf x}_2$ then ${\bf x}_1-{\bf x}_2$ is a solution of the corresponding homogeneous system $A{\bf x}={\bf 0}$. This is turn implies that if you find one solution ${\bf x}_p$ (a particular solution) of $A{\bf x} = {\bf b}$ and then find the general solution of the corresponding homogeneous system $A{\bf x}={\bf 0}$, say ${\bf x}_h$, then ${\bf x}={\bf x}_p+{\bf x}_h$ is the general solution of $A{\bf x}={\bf b}$. So in some sense the homogeneous solutions account for all of the redundant solutions of $A{\bf x}={\bf b}$ once you've found a particular solution.

If you have a linear transformation, say $T:V \to W$, then the kernel (or nullspace) of $T$ is the subspace $\mathrm{Ker}(T)=\{ v \in V \;|\; T(v)={\bf 0} \}$ (everything in $V$ that maps to the zero vector in $W$). If your linear transformation is $T(v)=Av$ for some matrix $A$, then the kernel of $T$ is nothing more than the null space of the matrix $A$. The range of $T$ is all of the vectors of $W$ that get mapped to: $\mathrm{Range}(T) = T(V) = \{ T(v) \;|\; v\in V\}$. Again if your transformation is $T(v)=Av$, then the range of $T$ is nothing more than the column space of the matrix $A$.

Again the kernel (a set of solutions of a homogeneous linear system) accounts for redundancies. If $T(v_1)=w=T(v_2)$ (i.e. $v_1$ and $v_2$ both map to the same output $w$), then $v_1-v_2 \in \mathrm{Ker}(T)$. So if $w \in T(V)$ (the range of $T$) and $v_p \in V$ is a vector such that $T(v_p)=w$, then $T(v_p+k)=w$ for any $k \in \mathrm{Ker}(T)$. Briefly, let $K=\mathrm{Ker}(T)$ and $v_p+K=\{v_p+k\;|\; k\in K\}$. Then $v_p+K$ is the set of all vectors which map to $w$.

In general, each element in the range of $T$ corresponds to a set of the form $v+K=v+\mathrm{Ker}(T)$ (these are called cosets of the kernel). So if we take $V$ and quotient out $K$ (whatever that means), denoted $V/K$, then we are left with a collection of sets which exactly correspond (i.e. isomorphic) with the range $T(V)$. This is written: $V/\mathrm{Ker}(T) = \mathrm{Range}(T)$. This result is known as the first ismorphism theorem. A quick consequence is that "rank plus nullity equals the dimension of the domain".

I know that doesn't complete answer your question, but maybe it'll get you started.

  • $\begingroup$ Bill, I am grateful for your insight here. $\endgroup$ Mar 22, 2012 at 0:57
  • $\begingroup$ I didn't realize there was so much overlap between group theory and linear algebra. Good post! $\endgroup$
    – user94492
    Sep 12, 2013 at 12:31
  • $\begingroup$ @leojanssens Thanks. Yes there's a lot of overlap when one is looking at the beginnings of the theories of vector spaces (an Abelian group + something), rings (an Abelian group + something), and groups. Subobjects, direct sums, homomorphisms, kernels, isomorphism theorems...all this stuff looks about the same at first glance. $\endgroup$
    – Bill Cook
    Sep 12, 2013 at 18:21

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