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?)
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