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Elementary Linear Algebra: An Active Learning Approach

Section A.4 Non-Uniqueness of the Expression for the General Solution of \(\A\x=\b\)

For any given wide or singular system \(\A\x=\b,\) the expression for the general solution is highly non-unique. The following discussion actually applies generally to any given affine set.

Remark A.4.1. It Does Not Matter What Names We Use for Variables.

One source of non-uniqueness is that of variable names. In basic algebra we are probably most comfortable with \(x\) as the independent variable and \(y\) as the dependent variable, but we could just as well use any other pair of variable names like \(\a\) as the independent variable and \(\beta\) as the dependent variable.
As long as we use the horizontal axis for the independent variable and the vertical axis for the dependent variable, the graph of \(y=x^2\) will look exactly like the graph of \(\beta=\a^2.\)
The same holds true in parametrized sets. Indeed,
\begin{align*} \amp\left\{\left(\left.\begin{array}{c}5/2\\0\\-2\,\,\end{array}\right)+x_2\,\left(\begin{array}{c}3/2\\1\\0\end{array}\right)\right|\,x_2\in\R\right\}\\ =\amp\left\{\left(\left.\begin{array}{c}5/2\\0\\-2\,\,\end{array}\right)+y\,\left(\begin{array}{c}3/2\\1\\0\end{array}\right)\right|\,y\in\R\right\}\\ =\amp\left\{\left(\left.\begin{array}{c}5/2\\0\\-2\,\,\end{array}\right)+\sigma\,\left(\begin{array}{c}3/2\\1\\0\end{array}\right)\right|\,\sigma\in\R\right\} \end{align*}
Essentially, any parameter in the element form of a set is a dummy variable in much the same way that \(x\) is a dummy variable in the definite integral \(\displaystyle{\int_0^1x^2\,dx.}\)
We do wish to be consistent about naming our variables to minimize confusion. We will typically use \(s\) when we need a single parameter, \(s,t\) when we need two parameters, \(r,s,t\) if three are needed, and \(s_1,s_2,\ldots,s_k\) when more than three parameter variables are present.
An observation about notation: If we were not using the notation \(\x\) for the unknown in \(\A\x=\b,\) we would be perfectly happy using \(\left(\begin{array}{c}x\\y\end{array}\right),\left(\begin{array}{c}x\\y\\z\end{array}\right),\) or \(\left(\begin{array}{c}x\\y\\z\\w\end{array}\right)\) to denote variable-component vectors. But the minor notation collision in for example \(\x=\left(\begin{array}{c}x\\y\\z\end{array}\right)\) exhibits enough of an ambiguity that we choose to use \(x_1,x_2,\ldots,x_n\) to denote component variables.

Remark A.4.2.

A more fundamental source of non-uniqueness is that we are not actually required to select free variables in the way we have done so far. Although we organize our solutions essentially from right to left, this organizational principle is just a way to standardize our process. There are other possibilities.
Indeed, consider the consistent system Example 3.4.9 with solution set
\begin{equation*} \left\{\left(\left.\begin{array}{c}5/2\\0\\-2\,\,\end{array}\right)+s\,\left(\begin{array}{c}3/2\\1\\0\end{array}\right)\right|\,s\in\R\right\}\text{.} \end{equation*}
We could just as well call \(x_1\) a free variable and solve \(2x_1-3x_2=5\) for \(x_2,\) leading to
\begin{equation*} x_2=-\frac{5}{3}+\frac{2}{3}x_1 \end{equation*}
yielding
\begin{equation*} \x=\left(\begin{array}{c}x_1\\-\frac{5}{3}+\frac{2}{3}x_1\\-2\,\,\,\end{array}\right)=\left(\begin{array}{c}0\\-5/3\\-2\,\,\,\end{array}\right)+x_1\,\left(\begin{array}{c}1\\2/3\\0\end{array}\right)\text{.} \end{equation*}
After replacing the component variable \(x_1\) with the parameter variable \(s\) we obtain the solution set
\begin{equation*} \left\{\left.\left(\begin{array}{c}0\\-5/3\\-2\,\,\,\end{array}\right)+s\,\left(\begin{array}{c}1\\2/3\\0\end{array}\right)\,\right|\,s\in\R\right\}. \end{equation*}
To reinforce our point, we have
\begin{equation*} \left\{\left(\left.\begin{array}{c}5/2\\0\\-2\,\,\end{array}\right)+s\,\left(\begin{array}{c}3/2\\1\\0\end{array}\right)\right|\,s\in\R\right\}=\left\{\left.\left(\begin{array}{c}0\\-5/3\\-2\,\,\,\end{array}\right)+s\,\left(\begin{array}{c}1\\2/3\\0\end{array}\right)\,\right|\,s\in\R\right\}. \end{equation*}
We will continue to organize our solutions essentially from right to left based on pivot variables and free variables but you should be aware that there are always equivalent re-expressions as above.

Remark A.4.3.

Another source of non-uniqueness is that we may replace any parametric variable with a constant multiple of that variables without changing the solution set (though each individual vector in the set would require different parametric variables values in the re-expression).
This substitution, typically implemented to re-express vectors \(\vec{x_h}_i\) with fewer fractions, amounts to a dilation (compression or rarefaction) of the real axis associated with the replaced variable. For example, the solution set
\begin{equation} \left\{\left.\left(\begin{array}{c}\,\,2/3\\\,\,0\\-9/5\\\,\,0\end{array}\right) + s\left(\begin{array}{c}-2/3\\1\\0\\0\end{array}\right) + t\left(\begin{array}{c}2/3\\0\\6/5\\1\end{array}\right)\right|\,\,s,t\in\R\right\}\tag{A.4.1} \end{equation}
to the system in Example 3.4.12 may be re-expressed with fewer fractions as follows:
\begin{align} \amp\left\{\left.\left(\begin{array}{c}\,\,2/3\\\,\,0\\-9/5\\\,\,0\end{array}\right) + 3s\left(\begin{array}{c}-2/3\\1\\0\\0\end{array}\right) + 15t\left(\begin{array}{c}2/3\\0\\6/5\\1\end{array}\right)\right|\,\,s,t\in\R\right\}\notag\\ =\amp\left\{\left.\left(\begin{array}{c}\,\,2/3\\\,\,0\\-9/5\\\,\,0\end{array}\right) + s\left(\begin{array}{r}-2\\3\\0\\0\end{array}\right) + t\left(\begin{array}{c}10\\0\\18\\15\end{array}\right)\right|\,\,s,t\in\R\right\}\tag{A.4.2} \end{align}

Remark A.4.4.

A similar non-uniqueness is related to the choice of \(\vec{x_p}.\) For example, in (A.4.1) we may define new variables \(s_1,t_1\) by \(s=s_1+1\) and \(t=t_1+1,\) substitute them into (A.4.1), collect the constant terms into a new \(\vec{x_p},\) and rename the parameter variables as follows:
\begin{align} \amp\left\{\!\left.\left(\begin{array}{c}\,\,2/3\\\,\,0\\-9/5\\\,\,0\end{array}\right) + s\left(\begin{array}{r}-2\\3\\0\\0\end{array}\right) + t\left(\begin{array}{c}10\\0\\18\\15\end{array}\right)\right|\,s,t\in\R\!\right\}\notag\\ =\,\amp\left\{\!\left.\left(\begin{array}{c}\,\,2/3\\\,\,0\\-9/5\\\,\,0\end{array}\right) + (s_1\!+\!1)\!\left(\begin{array}{r}-2\\3\\0\\0\end{array}\right) + (t_1\!+\!1)\!\left(\begin{array}{r}10\\0\\18\\15\end{array}\right)\right|s,t\!\in\!\R\!\right\}\notag\\ =\,\amp\left\{\!\left.\left(\begin{array}{c}\,\,26/3\\\,\,3\\-81/5\\\,\,15\end{array}\right) + s_1\left(\begin{array}{r}-2\\3\\0\\0\end{array}\right) + t_1\left(\begin{array}{c}10\\0\\18\\15\end{array}\right)\right|\,s_1,t_1\in\R\right\}\notag\\ =\,\amp\left\{\!\left.\left(\begin{array}{c}\,\,26/3\\\,\,3\\-81/5\\\,\,15\end{array}\right) + s\left(\begin{array}{r}-2\\3\\0\\0\end{array}\right) + t\left(\begin{array}{c}10\\0\\18\\15\end{array}\right)\right|\,s,t\!\in\!\R\right\}.\tag{A.4.3} \end{align}
The latter expression does indeed represent the same set as (A.4.1), though computing any given vector in the solution set using (A.4.3) will require different \(\boldsymbol{s}\text{,}\)\(\boldsymbol{t}\) than computing the vector using (A.4.1).

Remark A.4.5.

In fact, the typical \(\vec{x_p}\) is given by substituting \(0\) for all parameters in the expression of the the general solution, but we may use any parameter values to create an \(\vec{x_p}.\)
Because of this, believe it or not one can always express the general solution to a consistent wide or singular system using only integers! All we need to do is find values of the parameters which yield a solution in \(\Z^n,\) and to use this solution as \(\vec{x_p}.\)
As an example, one may substitute \(s=1/3\) and \(t=3/5\) into (A.4.1) to obtain \(\vec{x_p}=\left(\begin{array}{c}6\\1\\9\\9\end{array}\right)\in\Z^4,\) leading to the all-integer representation
\begin{equation*} \left\{\!\left.\left(\begin{array}{c}6\\1\\9\\9\end{array}\right) + s\left(\begin{array}{r}-2\\3\\0\\0\end{array}\right) + t\left(\begin{array}{r}10\\0\\18\\15\end{array}\right)\right|\,s,t\in\R\right\}. \end{equation*}
For any given consistent wide or singular system the determination of such \(\vec{x_p}\in\Z^n\) is an interesting exercise, as is the formalization of the process including the determination of parameter values in the general case, but we will leave that to the interested reader and not require \(\vec{x_p}\in\Z^n.\) Your instructor may feel otherwise :)
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