The Unapologetic Mathematician

Unsolvable Inhomogenous Systems

We know that when an inhomogenous system has a solution, it has a whole family of them. Given a particular solution, it defines a coset of the subspace of the solutions to the associated homogenous system. And that subspace is the kernel of a certain linear map.

But must there always be a particular solution to begin with? Clearly not. When we first talked about linear systems we mentioned the example

$3x^1+4x^2=12$
$x^1-x^2=11$
$x^1+x^2=10$

In our matrix notation, this reads

$\displaystyle\begin{pmatrix}3&4\\1&-1\\1&1\end{pmatrix}\begin{pmatrix}x^1\\x^2\end{pmatrix}=\begin{pmatrix}12\\11\\10\end{pmatrix}$

or $Ax=b$ in purely algebraic notation.

We saw then that this system has no solutions at all. What’s the problem? Well, we’ve got a linear map $A:\mathbb{F}^2\rightarrow\mathbb{F}^3$. The rank-nullity theorem tells us that the dimension of the image (the rank) plus the dimension of the kernel (the nullity) must equal the dimension of the source. But here this dimension is $2$, and so the rank can be at most $2$, which means there must be some vectors $b\in\mathbb{F}^3$ which can’t be written as $b=Ax$ no matter what vector $x$ we pick. And the vector in the example is just such a vector outside the image of $A$.

The upshot is that we can only solve the system $Ax=b$ if the vector $b$ lies in the image of the linear map $A$, and it might be less than obvious what vectors satisfy this requirement. Notice that this is more complicated than the old situation for single equations of single variables. In that case, the target only has one dimension, and the linear transformation “multiply by the number $A$” only misses this dimension if $A=0$, which is easy to recognize.

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Posted by John Armstrong | Algebra, Linear Algebra

3 Comments	»

1. [...] the system is inhomogenous in general, and as such it might not have any solutions. Since every short exact sequence of vector spaces splits we can write . Then the vector will have [...]

Pingback by | July 22, 2008 | Reply

2. [...] there’s no possible way to satisfy the equation. That is, we can quickly see if the system is unsolvable. On the other hand, if lies completely within the nonzero rows of , it’s straightforward to [...]

Pingback by | August 18, 2009 | Reply

3. [...] echelon form of a system of equations has a leading coefficient in the last column, that system is unsolvable. We don’t even need to climb the ladder to know [...]

Pingback by | September 2, 2009 | Reply

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