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

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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?)

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?)

##### 1 Answer

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

#abs(A B) = abs(A) * abs(B)#

In particular:

#abs(A^2) = abs(A)^2#

In particular note that if

We can deduce that it is not possible for the determinant of a squared matrix over

We can deduce that a matrix such as

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

**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