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Why can I use equations with algebraic structures (like vector spaces, matrices) to prove theorems? Why can we "manipulate" equations of algebraic structures to prove things?

Example: $A\in \mathbb{F}^{n\times n}$, $A$ is an invertible matrix. $$ Ax=b\\ (A^{-1}A)x=A^{-1}b \\ Ix=A^{-1}b \\ x=A^{-1}b $$

Why is it correct to multiply both side of the equation by a matrix? aren't equations only true for number? Are all the rules related to equations (with numbers only) also true for equations without numbers but matrices or vector spaces? Can every operation used with equations involving numbers be used with other algebraic structures?

mawaior
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  • Because you have defined what matrix multiplication ( or whatever operation you are applying) means already. – Paul Dec 09 '23 at 13:37

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The concept of equation is not limited to numbers, it extends to all mathematical objects. In mathematical discourse, to say that one mathematical object is equal to another mathematical object means that they are the same object. Equality has its own laws, although these laws are sometimes regarded as laws of logic, as opposed to laws of mathematics. One important example is the transitive law of equality: $A=B$ and $B=C$ implies $A=C$.

As the notation for each new kind of mathematical object is introduced, it is common to state exactly what "equality" means in that new context.

For example, in the Cartesian coordinate plane, given two points $P=(a,b)$ and $Q=(c,d)$, one defines equality of points $P=Q$ to mean: $a=c$ and $b=d$.

For another example relevant to your post, given two matrices $A,B \in \mathbb F^{m \times n}$ one defines equality of matrices $A=B$ to mean that $A_{ij}=B_{ij}$ for all $i \in \{1,\ldots,m\}$ and $j \in \{1,\ldots,n\}$.

In your discussion of matrix multiplication, first one proves that the matrix multiplication formula defines a binary operation, meaning a function which inputs a pair of matrices $A \in \mathbb F^{k \times l}$, $B \in \mathbb F^{l \times m}$ and outputs their matrix product $AB \in \mathbb F^{k \times m}$. To be a function means that equal inputs implies equal outputs, and therefore $$Ax=B \implies A^{-1}(Ax) = A^{-1} b $$ Also, the associative law of matrix multiplication is a true theorem, that one must prove in a linear algebra course: for all $A \in F^{k \times l}$, $B \in \mathbb F^{l \times m}$, $C \in \mathbb F^{m \times n}$ we have $(AB)C=A(BC)$. It follows that $$A^{-1}(Ax) = (A^{-1} A) x $$ Putting the last two equations together, and using the transitive law of equality, it follows that $$Ax=B \implies (A^{-1} A) = A^{-1} b $$ which is the first two lines of your calculation.

Perhaps you know now how to continue with the remaining lines?

On the other hand, as for your final paragraph asking whether all rules for manipulating numerical equations apply to all other equations ... well, of course not. For example, I suspect you know that the commutative law of multiplication is true for pairs of numbers, namely $ab=ba$ for all real numbers, but that there exists $n \times n$ matrices $A,B$ such that $AB \ne BA$. But if you don't know that, well, play around with some simple matrices, perhaps you might discover a counterexample.

Lee Mosher
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  • In my one of my previous questions I asked why we sometimes lose solutions when solving equations, I provided this "rule" $\forall f: (a=b \implies f(a)=f(b))$ which means that we can do every manipulation that we want and get the solutions to the original equation and the solutions for the manipulated equation (extraneous). Some answers said that I am wrong and that not every "manipulation" is a function. Am I correct that we can do any operation"manipulation" that we want and get an equation which the solution to it are also solutions for the original equation? – mawaior Dec 09 '23 at 14:04
  • The question is in my profile, it is titled "Why do we sometimes lose solutions when solving equations?". – mawaior Dec 09 '23 at 14:07
  • The implication $a=b \implies f(a)=f(b)$ is true when $f$ is a function, as defined in set theory. Your concept of an operation"manipulation" is quite vague, so all that I can really say is this: if it's a function as defined in set theory then that implication is guaranteed to be true; whereas if it's .... something else? ... then there are no guarantees. – Lee Mosher Dec 09 '23 at 14:44
  • Every operation can be considered a function. Can you give me an example of an operation which is not a function? – mawaior Dec 09 '23 at 15:09
  • The discourse has now gone beyond your current post, so I'll just repeat myself and sign off: My answer refers only to functions in the sense of set theory, not to your vaguer concept of operation"manipulation" (define that concept, and you could perhaps have a real question). In this specific question, the functions considered are the binary operations of matrix multiplication. – Lee Mosher Dec 09 '23 at 16:22