Let $$ B = \begin{bmatrix} -2 & 0 & 0 \\ -1 & -4 & -1 \\ 2 & 4 & 0 \\ \end{bmatrix} $$ Then $$ B^2 = A = \begin{bmatrix} 4 & 0 & 0 \\ 4 & 12 & 4 \\ -8 & -16 & -4 \\ \end{bmatrix} $$
But the square of rcf of B is not equal to rcf of A.
Why is that so ?
Thank you .
Because matrix multiplication isn't transitive.
Meaning: Suppose there exists $A, B, C \in M_{n\times n}$ such that :$ C = A\cdot B$. Does not mean that $C = B \cdot A$ For that you need to demand more properties.
Let $\hat{B} = \begin{bmatrix} 2 & 0 & 0 \\ -1 & -4 & -1 \\ -2 & 4 & 0 \\ \end{bmatrix}$. notice that $B \neq \hat{B} \ $, but:
$$ B^2 = \left(\begin{bmatrix} -2 & 0 & 0 \\ -1 & -4 & -1 \\ 2 & 4 & 0 \\ \end{bmatrix}\right)^{2} = \left(\begin{bmatrix} 2 & 0 & 0 \\ -1 & -4 & -1 \\ -2 & 4 & 0 \\ \end{bmatrix}\right)^{2} = (\hat{B})^2 = \begin{bmatrix} 4 & 0 & 0 \\ 4 & 12 & 4 \\ -8 & -16 & -4 \\ \end{bmatrix} = A $$
If $B = \sqrt{A} \ \Longrightarrow \ B^2 = A $, but in order to prove the other part you'd have to demand more properties for $A$ and $B$. ;