 Rectangular kernel and solution-fibre example

Consider the linear map $A:{\mathbb{R}}^3\to {\mathbb{R}}^2$ represented by the rectangular matrixTo compute its kernel, solve the homogeneous linear system $x+2y=0$, $y+z=0$. Taking $y=t$ gives $x=-2t$, $z=-t$, soThus $v$ represents a null direction of a linear map: decreasing the first input by two, increasing the second by one, and decreasing the third by one changes neither output. Solving the homogeneous linear system is the concrete computation behind a kernel; the column space instead decides which outputs are attainable.^[ref]

Now prescribe the nonzero output $b=(3,2{)}^T$. The linear system $Au=b$ becomes $x+2y=3$, $y+z=2$. Again take $y=t$:Direct substitution gives $A{u}_0=(3,2{)}^T$, and $A(u_0+tv)=A{u}_0+tAv=b$. Conversely, the two linear equations force exactly this form, so no solutions have been omitted. This fiber is an affine subspace of dimension one in the three-dimensional input vector space. It is parallel to the kernel, but does not pass through the origin. For instance, $u_0=(3,0,2{)}^T$ and $u_1=(1,1,1{)}^T$ both give $b$, while their difference $u_1-u_0=v$ gives zero. Adding $u_0$ and $u_1$ gives output $2b$, which is another way to see that this nonzero-output fiber is not a vector subspace.

The first and third columns of $A$ are $(1,0{)}^T$ and $(0,1{)}^T$, a basis of the output vector space. Hence $A$ has matrix rank two and reaches every output. The rank-nullity theorem predicts nullity $3-2=1$ before we solve any linear system. It provides a useful completeness check: one nonzero kernel vector already forms a basis here, so a second independent null direction of a linear map would signal a calculation error. In general, matrix rank $r$ for an $m\times n$ matrix leaves $n-r$ independent kernel directions, rather than $m-r$.^[ref]