Question-43

Consider the following matrix:

\[\begin{equation*} \begin{bmatrix} 1 & b\\ b & 9 \end{bmatrix} \end{equation*}\]

Under what conditions is this matrix positive semi-definite?

• \(b \in (\infty, -3] \cup [3, \infty)\)
• \(b \in \mathbb{R}\)
• \(b \geq 0\)
• \(b \in [-3, 3]\)

Hint

Characteristic polynomial

Answer

• \(b \in (\infty, -3] \cup [3, \infty)\)
• \(b \in \mathbb{R}\)
• \(b \geq 0\)
• \(b \in [-3, 3]\)

Solution

For the matrix to be positive semi-definite, all its eigenvalues should be non-negative. This happens when \(\displaystyle 9-b^{2} \geqslant 0\Longrightarrow b^{2} \leqslant 9\Longrightarrow b\in [ -3,3]\). In general:

Note

Consider:

\[\begin{equation*} A=\begin{bmatrix} a & b\\ b & c \end{bmatrix} \end{equation*}\]

\(\displaystyle A\) is positive semi-definite if and only if \(\displaystyle a\geqslant 0\) and \(\displaystyle ac-b^{2} \geqslant 0\). This can be derived as follows:

\[\begin{equation*} \begin{aligned} |A-\lambda I| & =( a-\lambda )( c-\lambda ) -b^{2}\\ & =\lambda ^{2} -( a+c) \lambda +ac-b^{2} \end{aligned} \end{equation*}\]

For the roots to be non-negative, their sum and product have to be non-negative:

\[\begin{equation*} a+c\geqslant 0\ \ \ \ \ \text{and} \ \ \ \ \ ac-b^{2} \geqslant 0 \end{equation*}\]

This can happen only if \(\displaystyle a\geqslant 0\) and \(\displaystyle c\geqslant 0\). When clubbed with the second condition, it is enough to have just \(\displaystyle a\geqslant 0\) and \(\displaystyle ac-b^{2} \geqslant 0\).