## Matrix with a parameter

### Task number: 2730

For which \(g\in \mathbb R\) is the matrix \( \mathbf G=\begin{pmatrix} g& 1& 0 \\ 1& g& 1 \\ 0& 1& g \\ \end{pmatrix} \) positive definite?

#### Solution 1

A recurrent method, with Gaussian elimination.

Clearly \(g\gt 0\).

Also the matrix \(\mathbf G_1=\mathbf G'-\frac{1}{g}(1{,}0)^T(1{,}0)= \begin{pmatrix} g- \frac{1}{g} & 1\\ 1& g\\ \end{pmatrix} \) must be also positive definite, i.e. \(g\gt 1\).

In the last step we get \(\mathbf G_2=\mathbf G'_1-\frac{1}{g}(1)^T(1)=\left(g-\frac{g}{g^2-1}\right)=\left(\frac{g^3-2g}{g^2-1}\right)\) that is a positive definite matrix of order one only for \(g\gt \sqrt{2}\).

The same calculation conducted by elementary transformations:

\( \begin{pmatrix} g& 1& 0 \\ 1& g& 1 \\ 0& 1& g \\ \end{pmatrix} \sim \begin{pmatrix} g& 1& 0 \\ 0& \frac{g^2-1}{g}& 1 \\ 0& 1& g \\ \end{pmatrix} \sim \begin{pmatrix} g& 1& 0 \\ 0& \frac{g^2-1}{g}& 1 \\ 0& 0& \frac{g(g^2-2)}{g^2-1}\\ \end{pmatrix} \)

#### Solution 2

Another method by the use of determinants:

\( \det(\mathbf G)= \begin{vmatrix} g& 1& 0 \\ 1& g& 1 \\ 0& 1& g \\ \end{vmatrix} =g(g^2-2)\gt 0\) for \(g\in (-\sqrt{2},0) \cup (\sqrt{2},\infty)\).

\( \begin{vmatrix} g& 1\\ 1& g\\ \end{vmatrix} =g^2-1\gt 0\) for \(g\in (-\infty,-1) \cup (1,\infty)\).

\(|g|=g\gt 0\) for \(g\in (0,\infty)\).

Th matrix is positive definite, when \(g\) is in the intersection of all three sets.

#### Answer

The matrix \(\mathbf G\) is positive definite only for \(g>\sqrt{2}\).