The Principle of Superposition

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The principle of superposition is just a restatement of the fact that matrix mappings are linear. Nevertheless, this restatement is helpful when trying to understand the structure of solutions to systems of linear equations.

Homogeneous Equations

A system of linear equations is homogeneous if it has the form

$\begin{equation} \label{homosys} Ax=0, \end{equation}$ where $A$ is an $m\times n$ matrix and $x\in \R ^n$ . Note that homogeneous systems are consistent since $0\in \R ^n$ is always a solution, that is, $A(0)=0$ .

The principle of superposition makes two assertions:

Suppose that $y$ and $z$ in $\R ^n$ are solutions to (??) (that is, suppose that $Ay=0$ and $Az=0$ ); then $y+z$ is a solution to (??).
Suppose that $c$ is a scalar; then $cy$ is a solution to (??).

The principle of superposition is proved using the linearity of matrix multiplication. Calculate $A(y+z) = Ay + Az = 0+0=0$ to verify that $y+z$ is a solution, and calculate $A(cy) = c(Ay) = c\cdot 0 = 0$ to verify that $cy$ is a solution.

We see that solutions to homogeneous systems of linear equations always satisfy the general property of superposition: sums of solutions are solutions and scalar multiples of solutions are solutions.

We illustrate this principle by explicitly solving the system of equations $\left (\begin {array}{rrrr} 1 & 2 & -1 & 1\\ 2 & 5 & -4 & -1 \end {array}\right )\left (\begin {array}{c} x_1\\x_2\\x_3\\x_4 \end {array}\right ) = \left (\begin {array}{c} 0\\0 \end {array}\right ).$ Use row reduction to show that the matrix $\left (\begin {array}{rrrr} 1 & 2 & -1 & 1\\ 2 & 5 & -4 & -1 \end {array}\right )$ is row equivalent to $\left (\begin {array}{rrrr} 1 & 0 & 3 & 7\\ 0 & 1 & -2 & -3 \end {array}\right )$ which is in reduced echelon form. Recall, using the methods of Section ??, that every solution to this linear system has the form $\left (\begin {array}{c} -3x_3-7x_4\\ 2x_3+3x_4\\ x_3\\ x_4\end {array}\right ) = x_3\left (\begin {array}{r}-3\\2\\1\\0\end {array}\right ) + x_4\left (\begin {array}{r}-7\\3\\0\\1\end {array}\right ).$ Superposition is verified again by observing that the form of the solutions is preserved under vector addition and scalar multiplication. For instance, suppose that $\alpha _1 \left (\begin {array}{r}-3\\2\\1\\0\end {array}\right ) + \alpha _2 \left (\begin {array}{r}-7\\3\\0\\1\end {array}\right ) \AND \beta _1 \left (\begin {array}{r}-3\\2\\1\\0\end {array}\right ) + \beta _2 \left (\begin {array}{r}-7\\3\\0\\1\end {array}\right )$ are two solutions. Then the sum has the form $\gamma _1 \left (\begin {array}{r}-3\\2\\1\\0\end {array}\right ) + \gamma _2 \left (\begin {array}{r}-7\\3\\0\\1\end {array}\right )$ where $\gamma _j = \alpha _j + \beta _j$ .

We have actually proved more than superposition. We have shown in this example that every solution is a superposition of just two solutions $\left (\begin {array}{r}-3\\2\\1\\0\end {array}\right ) \AND \left (\begin {array}{r}-7\\3\\0\\1\end {array}\right ).$

Inhomogeneous Equations

The linear system of $m$ equations in $n$ unknowns is written as $Ax=b$ where $A$ is an $m\times n$ matrix, $x\in \R ^n$ , and $b\in \R ^m$ . This system is inhomogeneous when the vector $b$ is nonzero. Note that if $y,z\in \R ^n$ are solutions to the inhomogeneous equation (that is, $Ay=b$ and $Az=b$ ), then $y-z$ is a solution to the homogeneous equation. That is, $A(y-z) = Ay - Az = b - b = 0.$ For example, let $A = \left (\begin {array}{rrr} 1 & 2 & 0 \\ -2 & 0 & 1 \end {array}\right ) \AND b = \vectwo {3}{-1}.$ Then $y = \left (\begin {array}{c} 1\\1\\1\end {array}\right ) \AND z = \left (\begin {array}{c} 3\\0\\5\end {array}\right )$ are both solutions to the linear system $Ax=b$ . It follows that $y-z = \left (\begin {array}{r} -2\\1\\-4\end {array}\right )$ is a solution to the homogeneous system $Ax=0$ , which can be checked by direct calculation.

Thus we can completely solve the inhomogeneous equation by finding one solution to the inhomogeneous equation and then adding to that solution every solution of the homogeneous equation. More precisely, suppose that we know all of the solutions $w$ to the homogeneous equation $Ax=0$ and one solution $y$ to the inhomogeneous equation $Ax=b$ . Then $y+w$ is another solution to the inhomogeneous equation and every solution to the inhomogeneous equation has this form.

An Example of an Inhomogeneous Equation

Suppose that we want to find all solutions of $Ax=b$ where $A = \left ( \begin {array}{rrr} 3 & 2 & 1 \\ 0 & 1 & -2 \\ 3 & 3 & -1 \end {array} \right )\quad \mbox {and}\quad b=\left ( \begin {array}{r} -2 \\ 4 \\ 2 \end {array} \right ).$ Suppose that you are told that $y=(-5,6,1)^t$ is a solution of the inhomogeneous equation. (This fact can be verified by a short calculation — just multiply $Ay$ and see that the result equals $b$ .) Next find all solutions to the homogeneous equation $Ax=0$ by putting $A$ into reduced echelon form. The resulting row echelon form matrix is $\left ( \begin {array}{rrr} 1 & 0 & \frac {5}{3} \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end {array} \right ).$ Hence we see that the solutions of the homogeneous equation $Ax=0$ are $\left (\begin {array}{r} -\frac {5}{3}s\\ 2s\\ s\end {array}\right ) = s\left (\begin {array}{r}-\frac {5}{3}\\2\\1\end {array}\right ).$ Combining these results, we conclude that all the solutions of $Ax=b$ are given by $\left (\begin {array}{r}-5\\6\\1\end {array}\right )+ s\left (\begin {array}{r}-\frac {5}{3}\\2\\1\end {array}\right ).$

Exercises

Consider the homogeneous linear equation $x+y+z = 0$

(a): Write all solutions to this equation as a general superposition of a pair of vectors $v_1$ and $v_2$ .
(b): Write all solutions as a general superposition of a second pair of vectors $w_1$ and $w_2$ .

Write all solutions to the homogeneous system of linear equations

$\begin{eqnarray*} x_1+2x_2+x_4-x_5 = 0\\ x_3-2x_4+x_5 = 0 \end{eqnarray*}$

as the general superposition of three vectors.

Find all solutions to the homogeneous equation $Ax=0$ where $A = \left (\begin {array}{ccc} 2 & 3 & 1 \\ 1 & 1 & 4 \end {array} \right ).$
Find a single solution to the inhomogeneous equation $\begin{equation} \label{E:inhom} Ax =\vectwo{6}{6}. \end{equation}$
Use your answers in (a) and (b) to find all solutions to (??).