## Matrix with a parameter

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.

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