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I am a high school student and my formal Linear Algebra education consisted merely of the definition of matrices as a list of numbers and then some random properties. While reading the BetterExplained article on Linear Algebra and some of 3Blue1Brown's videos on the same did help, I still often face difficulties while solving problems and when I look up the solutions, all I'm thinking is "Wait. You're allowed to do that to matrices too?"

So my question is - what can we do to these list of numbers that we can do to individual real and complex numbers?

To clarify, I do know that we can add matrices by adding their individual elements together and multiply matrices in a row-column order and stuff like that. My doubts are along the lines of the ones listed below:

  1. Does $AB = CD$ imply that $B^{-1}A^{-1}=D^{-1}C^{-1}$ ? (where A, B, C and D are 4 non-singular matrices of appropriate orders)
  2. Is $A \times A^n = A^n \times A$ valid? (where A is a matrix and n is a natural number)
  3. Is multiplication of matrices commutative only when a matrix is being multiplied by a null matrix or unit matrix of appropriate order?
  4. Is the inverse of the matrix $A^n$ the same as the inverse of $A$ multiplied $n$ times to itself?

I am not asking for proofs of the problems listed above - instead, I would greatly appreciate it if someone who has noticed some "patterns" in the doubts listed above would point me to some resource that I can study to clear all such doubts in my conceptual understanding. Alternatively, how should I approach matrix multiplication and their inverses in general that would solve these and other similar problems that I may be having?

Thank you.


Edit: These issues have been solved adequately and then some by Dave L. Renfro's comments on this question.

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    $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ Sep 9, 2021 at 10:37
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    $\begingroup$ 1. yes, 2.yes, 3. no, 4. yes. Please have a look at texts on linear algebra. And try out everything with $2\times 2$ matrices yourself. This gives already a concrete understanding and you don't have to ask then all of it. $\endgroup$ Sep 9, 2021 at 10:37
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    $\begingroup$ Concerning your second question, can't you see that both $A^n.A$ and $A.A^n$ are equal to$$\overbrace{A.A\cdots A}^{n+1\text{ times}}?$$ $\endgroup$ Sep 9, 2021 at 10:44
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    $\begingroup$ Yes, I have many texts in mind. This site has a huge amount of book recommendations for linear algebra, see for example here. Also useful is the term group, in your case $GL_n(K)$. In a group always $a^n\cdot a=a\cdot a^n$ and $(a^n)^{-1}=(a^{-1})^n$ and so on. $\endgroup$ Sep 9, 2021 at 10:44
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    $\begingroup$ @JoséCarlosSantos Indeed. I had doubts about whether pre-multiplying and post-multiplying would yield the same result or not. $\endgroup$
    – sonicsid
    Sep 9, 2021 at 10:46

1 Answer 1

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Everything that follows is under that assumption that we work over square matrices only. For non-square matrices some things are no longer well defined, e.g. there are no two-sided inverses (and one-sided are never unique), you can't multiply a matrix by itself, and so on.

  1. Does $AB = CD$ imply that $B^{-1}A^{-1}=D^{-1}C^{-1}$ ? (where A, B, C and D are 4 non-singular matrices of appropriate orders)

Yes. This follows from two general rules (taken from group theory): (1) that $(gh)^{-1}=h^{-1}g^{-1}$ and (2) that inverses are unique.

  1. Is $A \times A^n = A^n \times A$ valid? (where A is a matrix and n is a natural number)

I assume that by "$\times$" you mean matrix multiplication. The answer is yes, these are the same. This follows from the fact that matrix multiplication is associative. By applying induction on $n$. Also to solve that you need a proper definition of $A^n$ which formally is defined recursively as: $A^0:=I$ is the identity matrix (diagonal matrix with $1$s on the diagonal) and $A^n:=AA^{n-1}$ for $n\geq 1$. With that the induction step is quite simple:

$$AA^n=A(AA^{n-1})=A(A^{n-1}A)=(AA^{n-1})A=A^nA$$

The initial step for $n=0$ is the trivial $AI=IA$ equation.

  1. Is multiplication of matrices commutative only when a matrix is being multiplied by a null matrix or unit matrix of appropriate order?

Clearly $AB=BA$ whenever $A=B$, regardless of what matrix you pick.

This is however true under some stronger assumption, namely: if $A$ is a square matrix such that $AB=BA$ for any other square matrix $B$ then $A=\lambda I$ where $I$ the identity matrix and $\lambda$ is a scalar (at least over fields). Or in other words when $A$ is a diagonal matrix with a single fixed value on the diagonal. This is not a trivial result though, requires some work.

  1. Is the inverse of the matrix $A^n$ the same as the inverse of $A$ multiplied $n$ times to itself?

Yes. By induction on $n$ and the fact that inverses are unique. Indeed, write down that $B:=(A^n)^{-1}$. Then $I=A^nB=A^{n-1}AB$ and thus $AB$ is the inverse for $A^{n-1}$. By the induction step and uniqueness of inverses this means that $AB=(A^{-1})^{n-1}$ and thus by multiplying by $A^{-1}$ on the left (note that $A$ is invertible) we get $B=(A^{-1})^{n}$.

The case $n=0$ is trivial since $I^{-1}=I$.

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    $\begingroup$ Note that you can only multiply a SQUARE matrix by itself. (2) is not valid if $A$ is not square, because then you can't multiply A by itself. $\endgroup$ Sep 9, 2021 at 15:31

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