Question

Is it true that the difference of two matrices is equal to a squared (multiplied by itself) matrice subtracted from another squared matrice? Why?

Is it true that the difference of two matrices is equal to a squared (multiplied by itself) matrice subtracted from another squared matrice? Why, can you prove? (Is there a "the difference of two matrices" rule in matrice algebra?)

Answer 1

It depends...

Explanation:

I'm not sure exactly what you are asking, but let's take a look...

First note that the determinant preserves multiplication. That is, if `A. B` are any matrices then:

`abs(A B) = abs(A) * abs(B)`

In particular:

`abs(A^2) = abs(A)^2`

In particular note that if `A` has real coefficients then `abs(A) in RR` and `abs(A)^2 >= 0`.

We can deduce that it is not possible for the determinant of a squared matrix over `RR` to be negative.

We can deduce that a matrix such as `((0, 1), (1, 0))` (which has negative determinant) is not the square of any matrix over `RR`.

Can it be the difference of two squared matrices?

Yes - for example:

`((1/sqrt(2), 1/sqrt(2)),(1/sqrt(2), 1/sqrt(2)))^2 - ((1,0),(0,1))^2 = ((1,1),(1,1))-((1,0),(0,1)) = ((0,1),(1,0))`

Is this possible for all square real matrices?

I don't know, but note that not every matrix over integers is the difference of two squared matrices over integers. For example, the `1xx1` matrix `(2)` is not the difference of two squared integer `1xx1` matrices.

Notes

I did wonder if you actually intended to ask whether the difference of squares identity holds for matrices - i.e.:

`A^2-B^2 = (A-B)(A+B)`

This does not hold in general due to matrix multiplication being non-commutative in general.

We have:

`(A-B)(A+B) = A^2+AB-BA-B^2`

So if `AB != BA` then `(A-B)(A+B) != A^2-B^2`