Let \[ A = \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{3} \end{bmatrix}. \]
Show that \(A\) is positive.
Let \(V\) be the space of \(2 \times 1\) real matrices, with the inner product \((X \mid Y) = Y^tAX\). Find an orthonormal basis for \(V\), by applying the Gram-Schmidt process to the basis \(\{X_1, X_2\}\) defined by \[ X_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad X_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}. \]
Find an invertible \(2 \times 2\) real matrix \(P\) such that \(A = P^tP\).
Solution
1. The matrix \(A\) is positive.
The matrix is symmetric. Its leading principal minors are \[ \Delta_1=1>0 \] and \[ \Delta_2 =\det A =\frac13-\frac14 =\frac1{12}>0. \] By Sylvester’s criterion, \(A\) is positive definite.
2. Apply Gram–Schmidt to \(X_1,X_2\).
Since \[ \|X_1\|^2=X_1^tAX_1=1, \] the first orthonormal vector is \[ E_1=X_1=\binom10. \] Also \[ (X_2\mid E_1)=E_1^tAX_2=\frac12. \] Thus \[ U_2=X_2-\frac12E_1 =\binom{-1/2}{1}. \] A direct calculation gives \[ \|U_2\|^2=U_2^tAU_2=\frac1{12}. \] Hence \[ E_2=\frac{U_2}{\|U_2\|} =\binom{-\sqrt3}{2\sqrt3}. \] Therefore an orthonormal basis is \[ \boxed{ \left\{ \binom10, \binom{-\sqrt3}{2\sqrt3} \right\}.} \]
3. Construct \(P\) with \(A=P^tP\).
Let \(Q\) be the matrix whose columns are the orthonormal basis from step 2: \[ Q= \begin{pmatrix} 1&-\sqrt3\\ 0&2\sqrt3 \end{pmatrix}. \] The orthonormality relation for the \(A\)-inner product is \[ Q^tAQ=I. \] Thus \[ A=(Q^{-1})^tQ^{-1}. \] Now \[ Q^{-1} =\begin{pmatrix} 1&1/2\\ 0&1/(2\sqrt3) \end{pmatrix}. \] Therefore one valid choice is \[ \boxed{ P= \begin{pmatrix} 1&1/2\\ 0&1/(2\sqrt3) \end{pmatrix},} \] and direct multiplication gives \[ P^tP= \begin{pmatrix} 1&1/2\\ 1/2&1/3 \end{pmatrix}=A. \]