# Tag Info

Accepted

### Recurrence relation for matrices

Another different way to obtain an equivalent formula: let $Y = A^{1/2}XA^{1/2}$. Then, multiplying your equation from both sides by $A^{1/2}$, we have $Y^2+Y = A^{1/2}BA^{1/2} = C$. The matrix $C$ ...
• 8,610