# 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?)

Jun 30, 2018

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:

$\left\mid A B \right\mid = \left\mid A \right\mid \cdot \left\mid B \right\mid$

In particular:

$\left\mid {A}^{2} \right\mid = {\left\mid A \right\mid}^{2}$

In particular note that if $A$ has real coefficients then $\left\mid A \right\mid \in \mathbb{R}$ and ${\left\mid A \right\mid}^{2} \ge 0$.

We can deduce that it is not possible for the determinant of a squared matrix over $\mathbb{R}$ to be negative.

We can deduce that a matrix such as $\left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right)$ (which has negative determinant) is not the square of any matrix over $\mathbb{R}$.

Can it be the difference of two squared matrices?

Yes - for example:

${\left(\begin{matrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{matrix}\right)}^{2} - {\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)}^{2} = \left(\begin{matrix}1 & 1 \\ 1 & 1\end{matrix}\right) - \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) = \left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right)$

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 $1 \times 1$ matrix $\left(2\right)$ is not the difference of two squared integer $1 \times 1$ matrices.

Notes

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

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

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

We have:

$\left(A - B\right) \left(A + B\right) = {A}^{2} + A B - B A - {B}^{2}$

So if $A B \ne B A$ then $\left(A - B\right) \left(A + B\right) \ne {A}^{2} - {B}^{2}$