Homogeneous Systems of Equations

For each archetype that is a system of equations, we have formulated a similar, yet different, homogeneous system of equations by replacing each equation's constant term with a zero. To wit, for , we can convert the original system of equations into the homogeneous system,

Can you quickly find a solution to this system without row-reducing the augmented matrix?

Archetype B can be converted to the homogeneous system, whose augmented matrix row-reduces to

By , the system is consistent, and so the computation $n-r=3-3=0$ means the solution set contains just a single solution. Then, this lone solution must be the trivial solution.

can be converted to the homogeneous system, whose augmented matrix row-reduces to

By , the system is consistent, and so the computation $n-r=3-2=1$ means the solution set contains one free variable by , and hence has infinitely many solutions. We can describe this solution set using the free variable $x_3$, S=\setparts{\colvector{x_1\\x_2\\x_3}}{x_1=-x_3,\,x_2=x_3} =\setparts{\colvector{-x_3\\x_3\\x_3}}{x_3\in\complex{\null}}

Geometrically, these are points in three dimensions that lie on a line through the origin.

The write-up for lists several solutions of the corresponding homogeneous system. Here are two, written as solution vectors. We can say that they are in the null space of the coefficient matrix for the system of equations in . \vect{x}=\colvector{3\\0\\-5\\-6\\0\\0\\1}&& \vect{y}=\colvector{-4\\1\\-3\\-2\\1\\1\\1}

However, the vector \vect{z}=\colvector{1\\0\\0\\0\\0\\0\\2} is not in the null space, since it is not a solution to the homogeneous system. For example, it fails to even make the first equation true.

Let us compute the null space of A=\begin{bmatrix} 2 & -1 & 7 & -3 & -8 \\ 1 & 0 & 2 & 4 & 9 \\ 2 & 2 & -2 & -1 & 8 \end{bmatrix} which we write as $\nsp{A}$. Translating , we simply desire to solve the homogeneous system $\homosystem{A}$. So we row-reduce the augmented matrix to obtain \begin{bmatrix} \leading{1} & 0 & 2 & 0 & 1 & 0 \\ 0 & \leading{1} & -3 & 0 & 4 & 0 \\ 0 & 0 & 0 & \leading{1} & 2 & 0 \end{bmatrix}

The variables (of the homogeneous system) $x_3$ and $x_5$ are free (since columns 1, 2 and 4 are pivot columns), so we arrange the equations represented by the matrix in reduced row-echelon form to x_1&=-2x_3-x_5\\ x_2&=3x_3-4x_5\\ x_4&=-2x_5\\

So we can write the infinite solution set as sets using column vectors, \nsp{A}=\setparts{ \colvector{-2x_3-x_5\\3x_3-4x_5\\x_3\\-2x_5\\x_5} }{ x_3,\,x_5\in\complex{\null} }