Solving Equations Rigorously, without Circularity

Question

Suppose that I am asked to find the value of $h$ that makes the matrix $A = \begin{bmatrix}1 & 5 & 8 \\ 2 & 0 & 1 \\ 0 & 0 & h\end{bmatrix}$ singular.

Here, I can see that $h = 0$ works. However, what I'd typically do is this:

1. Set the determinant of the matrix to be $0.$
2. Find the value of $h$ that satisfies $\det(A) = 0.$ That is, we seek the value $c$ for which $$h = c \implies \det(A) = 0.$$

My issue here is that the step 2 seems to be committing some sort of a circular/invalid argument: instead of proving the above statement, it is proving its converse $$\det(A) = 0 \implies h = c.$$ Shouldn't we also set $h = c$ to calculate the determinant of the matrix $A,$ thereby actually proving (through calculations) the statement that we were originally trying to prove?

I've done such questions many times, but only recently has this thought occurred to me, and I haven't been able to formulate a satisfactory response. Am I needlessly confusing myself here? I always seem to get stuck on trivial issues no matter how much maths I study.

Answer 1

In principle, this has very little to do with linear algebra or determinants. In algebra, we often ask for values of $x$ (it could be numbers, tuples of numbers, matrices, etc.) such that $f(x) = 0$.

To do this, we set $f(x) = 0$ and then [hopefully] use logically valid steps to solve for $x$, which might have more than one value. Without more context, we cannot guarantee these $x$ actually satisfy $f(x)=0$, but we know that, if there is a solution, it's found in the list of values we found for $x$.

This is why some textbooks have a validation step at the end of the solution. This is where we take the values from the list and evaluate $f(x)$. If it comes out to be $0$, then it was an honest solution to the original equation $f(x)=0$. If not, it was an extraneous solution, not an actual one. This happens, for instance, when solving $f(x)=0$ involves a step where we "square both sides of the equation".

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You might therefore ask why we don't always have to do this last step. The answer is that sometimes the steps showing $f(x) = 0 \implies x=c$ are all reversible. Dividing by a nonzero value is reversed by multiplying by that nonzero value. Replacing a function by an equivalent expression is also reversible (effectively by the symmetry and transitivity of $=$).

In your case, that's all that happened. The determinant of $A$, when $A$ has the form given, is equivalent to $-10h$, so replacing the determinant of $A$ with $-10h$ is reversible. Also, dividing by $-10$ is reversible, as it's reversed by multiplying by $-10$.

The reason why we don't go through all of these details each time is for readability and respect. We don't want to diminish the readability of a solution by including details whose inclusion might feel patronizing to the reader.

Answer 2

You ask a good question that many have pondered over!

1. The statement $$f(x)=0\implies\ldots\implies x=b\;\text{ or }\;x=c\;\text{ or }\;x=d$$ says that the solution set of $f(x)=0$ is a (possibly improper or the empty) subset of $\{b,c,d\},$ in other words, that $\boldsymbol{b,c,d}$ are its only candidate solutions.

2. The statement $$x=c\implies\ldots\implies f(x)=0$$ says that the solution set of $f(x)=0$ is a (possibly improper) superset of $\{c\},$ in other words, that $\boldsymbol c$ is one of its solution(s).

3. The statement $$f(x)=0\iff\ldots\iff x=c\;\text{ or }\;x=d$$ says that the solution set of $f(x)=0$ is precisely $\{c,d\},$ in other words, that $\boldsymbol {c,d}$ are its only solutions.

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Examples:

1. To find all the values of $x$ that satisfy the equation $f(x)=0,$ by systematically verifying the candidate solutions:

   $f(x)=0\implies \ldots\implies x=b\;\text{ or }\;x=c\;\text{ or }\;x=d \\x=b\implies f(x)=7\ne0 \\x=c\implies f(x)=0 \\x=d\implies f(x)=0 \\\text{Hence, the required solution set is }\{c,d\}.\tag*{}$

2. I am asked to find the value of $h$ that makes the matrix $A = \begin{bmatrix}1 & 5 & 8 \\ 2 & 0 & 1 \\ 0 & 0 & h\end{bmatrix}$ singular.

   Here, I can see that $h = 0$ works.

   The exercise implicitly tells us that exactly one value of $h$ satisfies the given condition, and you've found it by inspection. In this case, this short presentation is perfectly valid and contains no circular reasoning: The required value is $0,$ since $$h=0\implies\ldots\implies A\text{ is singular.}$$

Answer 3

TLDR There is no circular argument: to find the necessary conditions for $P$ to occur, you supposed $P$, then found out that $P$ implies $det(A)=0$ which in turn implies $h=0$ (a necessary condition!) conversely you easily showed that $h=0$ is a sufficient condition for $A$ to be singular, so you can conclude that you found all values of $h$ for which the matrix is singular.

Note however that it is setting det(A) to be 0, is just implicitly using a known equivalence, and so is not source of a circular argument:

• The matrix $A$ is a variable (linked to the other variable $h$)
• det(A) is a variable (linked to the variable $A$ and the variable $h$
• The goal is to complete: $A$ singular $\iff det(A)=0 \iff h\in\ ?$

Strategy: Suppose $P$ then deduce $Q$ to show $P\implies Q$ then do the other way around to show $P\impliedby Q$ and conclude $P\iff Q$.

$\implies$: When there are variables in play what you implicitly do before supposing $A$ is fixing the variable: "Let $h$ be a real number, suppose $A$ is singular, then $det(A)=0$, so $h$ equals 0".

edited, I was wrong