Let $A$ be a positive definite symmetric matrix. Show that there exists an invertible matrix $B$ such that $A=B^TB$. [Hint: Use the Spectral Theorem to write $A = QDQ^T$. Then show that D can be factored as $C^TC$ for some invertible matrix $C$.]
I can't seem to get to the correct answer. I'm not entirely sure what is meant by the last line of the hint too. Could anyone please help me out?
Best Answer
$A$ is symmetric, so it can be written as $A=QDQ^T$ (Eigenvalue decomposition of a symmetric matrix), where $D$ is a diagonal matrix with real elements on main diagonal.
$A$ is PSD, so elements on main diagonal of $D$ are nonnegative, so they have square root. Let $D= D_1D_1^T$, where $D_1$ is also diagonal.
You can write $C^T= QD_1$. You will have $A=C^TC$.