Show that if an \(n \times n\) matrix \(A\) is positive definite, then there exists a positive definite matrix \(B\) such that \(A = {B^T}B\). (Hint: Write \(A = PD{P^T}\), with\({P^T} = {P^{ - 1}}\). Produce a diagonal matrix \(C\) such that \(D = {C^T}C\), and let \(B = PC{P^T}\). Show that \(B\) works.)

When any Symmetric Matrix \(A\) is diagonalized orthogonallyas\(PD{P^{ - 1}}\) we have:

\(\begin{aligned}{}{x^T}Ax = {y^T}Dy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{as }}x = Py} \right\}\\{\rm{and}}\\\left\| x \right\| = \left\| {Py} \right\| = \left\| y \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\left\{ {\forall y \in \mathbb{R}} \right\}\end{aligned}\)

As per the question, we have:

The positive\(n \times n\)matrix\(A\),Let we have:

\(A = PD{P^T},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{where }}{P^T} = {P^{ - 1}}} \right\}\)

For\(C\)being its diagonal matrix with all positive eigenvalues, we have:

\(\begin{aligned}{}D &= {C^2}\\ &= {C^T}C\end{aligned}\)

For,\(B = PC{P^T}\), we have:

\(\begin{aligned}{}{B^T}B &= {\left( {PC{P^T}} \right)^T}\left( {PC{P^T}} \right)\\ &= \left( {{P^T}{C^T}{P^T}^T} \right)\left( {PC{P^T}} \right)\\ &= P{C^T}C{P^T}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{P^T}P = 1} \right\}\\ &= PD{P^T}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{C^T}C = D} \right\}\\ &= A\end{aligned}\)

Henceit is proved thatif a \(n \times n\) matrix \(A\) is positive definite, then there exists a positive definite matrix \(B\) such that \({B^T}B = A\).