Question-88

Consider a square matrix \(\displaystyle A\) with real entries of order \(\displaystyle 2\) which has two equal eigenvalues. If \(\displaystyle t\) is its trace and \(\displaystyle d\) its determinant, which of the following are true?

• \(\displaystyle d\geqslant 0\)
• \(\displaystyle d< 0\)
• \(\displaystyle A=0\)
• \(\displaystyle A\) is a scalar matrix
• The diagonal entries of \(\displaystyle A\) are equal

Answer

• \(\displaystyle d\geqslant 0\)
• \(\displaystyle d< 0\)
• \(\displaystyle A=0\)
• \(\displaystyle A\) is a scalar matrix
• The diagonal entries of \(\displaystyle A\) are equal

Solution

Let:

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

The characteristic equation is given by:

\[ \begin{equation*} ( a-\lambda )( d-\lambda ) -bc=0 \end{equation*} \]

Simplifying:

\[ \begin{equation*} \lambda ^{2} -( a+d) \lambda +ad-bc=0 \end{equation*} \]

For two equal eigenvalues, the discriminant of the quadratic equation has to be zero:

\[ \begin{equation*} ( a+d)^{2} -4( ad-bc) =0 \end{equation*} \]

This simplifies to:

\[ \begin{equation*} ( a-d)^{2} +4bc=0 \end{equation*} \]

Alternatively, we can express the characteristic equation as:

\[ \begin{equation*} \lambda ^{2} -t\lambda +d=0 \end{equation*} \]

For equal real roots:

\[ \begin{equation*} t^{2} =4d \end{equation*} \]

We now see that \(\displaystyle d\geqslant 0\). \(\displaystyle A\) needn’t be a scalar matrix, nor is it necessary for the diagonal entries to be equal. A quick counter-example that addresses both cases: \(\displaystyle \begin{bmatrix} 2 & -1\\ 1 & 0 \end{bmatrix}\).