Question

Let \(A{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{{a_{{\bf{11}}}}}&{{a_{{\bf{12}}}}}\\{{a_{{\bf{21}}}}}&{{a_{{\bf{22}}}}}\end{aligned}} \right)\). Recall from Exercise \({\bf{25}}\) in Section \({\bf{5}}{\bf{.4}}\) that \({\rm{tr}}\;A\) (the trace of \(A\)) is the sum of the diagonal entries in \(A\). Show that the characteristic polynomial of \(A\) is \({\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). Then show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).

Short answer

It is proved that \(\det \left( {A - \lambda I} \right) = {\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). It is also proved that the eigenvalues of a \(2 \times 2\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{trA}}{2}} \right)^2}\).

Step-by-step solution

Step 1: Find the characteristic polynomial.

Determine\(\det \left( {A - \lambda I} \right)\).

\(\begin{aligned}{c}\det \left( {A - \lambda I} \right) &= \det \left( {\begin{aligned}{*{20}{c}}{{a_{11}} - \lambda }&{{a_{12}}}\\{{a_{21}}}&{{a_{22}} - \lambda }\end{aligned}} \right)\\ &= \left( {{a_{11}} - \lambda } \right)\left( {{a_{22}} - \lambda } \right) - {a_{12}}{a_{21}}\\ &= {\lambda ^2} - {a_{11}}\lambda - {a_{22}}\lambda + {a_{11}}{a_{22}} - {a_{12}}{a_{21}}\\ &= {\lambda ^2} - \left( {{a_{11}} + {a_{22}}} \right)\lambda + \left( {{a_{11}}{a_{22}} - {a_{12}}{a_{21}}} \right)\end{aligned}\)

Now use a trace of \(A\), that is, \({\rm{tr}}\;A = {a_{11}} + {a_{22}}\) then we have,

\(\det \left( {A - \lambda I} \right) = {\lambda ^2} - \left( {trA} \right)\lambda + \det A\)

Hence it is proved that\(\det \left( {A - \lambda I} \right) = {\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\).

Step 2: Show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\)

Determine the eigenvalues of \(A\).

\(\begin{aligned}{c}\det \left( {A - \lambda I} \right) &= 0\\{\lambda ^2} - \left( {{\rm{tr}}\;A} \right)\lambda + \det A &= 0\\\lambda &= \frac{{tr\;A \pm \sqrt {{{\left( {tr\;A} \right)}^2} - 4\det A} }}{2}\end{aligned}\)

As eigenvalues are positive if the discriminant is positive.

\(\begin{aligned}{c}{\left( {{\rm{tr}}A} \right)^2} - 4\det A \ge 0\\{\left( {{\rm{tr}}A} \right)^2} \ge 4\det A\\\frac{{{{\left( {{\rm{tr}}A} \right)}^2}}}{4} \ge \det A\\{\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2} \ge \det A\end{aligned}\)

Thus, \(\det A \le {\left( {\frac{{trA}}{2}} \right)^2}\).

Hence it is proved that the eigenvalues of a \(2 \times 2\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{trA}}{2}} \right)^2}\).