AHL 1.15—Eigenvalues and eigenvectors

We need $A\mathbf{v}=\lambda \mathbf{v}$. Compute the left side: $\left(\begin{array}{cc}5& -2\\ -2& 5\end{array}\right)\left(\begin{array}{c}1\\ -1\end{array}\right)$.

Row 1: $5(1)+(-2)(-1)=5+2=7$. Row 2: $(-2)(1)+5(-1)=-2-5=-7$. So $A\mathbf{v}=\left(\begin{array}{c}7\\ -7\end{array}\right)$.

We have $\left(\begin{array}{c}7\\ -7\end{array}\right)=7\left(\begin{array}{c}1\\ -1\end{array}\right)=7\mathbf{v}$. Therefore $\lambda =7$.

The eigenvalue corresponding to eigenvector $\left(\begin{array}{c}1\\ -1\end{array}\right)$ is $\lambda =7$.

We need to check which value of $\lambda$ satisfies $A\mathbf{v}=\lambda \mathbf{v}$ for $\mathbf{v}=\left(\begin{array}{c}1\\ -1\end{array}\right)$.

$3\mathbf{v}=\left(\begin{array}{c}3\\ -3\end{array}\right)$, but $A\mathbf{v}=\left(\begin{array}{c}7\\ -7\end{array}\right)\ne \left(\begin{array}{c}3\\ -3\end{array}\right)$. Eliminated.

$5\mathbf{v}=\left(\begin{array}{c}5\\ -5\end{array}\right)\ne \left(\begin{array}{c}7\\ -7\end{array}\right)$. Note that $\lambda =5$ is just the diagonal entry and does not match. Eliminated.

$-7\mathbf{v}=\left(\begin{array}{c}-7\\ 7\end{array}\right)\ne \left(\begin{array}{c}7\\ -7\end{array}\right)$. The sign is wrong. Eliminated.

$7\mathbf{v}=\left(\begin{array}{c}7\\ -7\end{array}\right)=A\mathbf{v}$. The correct eigenvalue is $\lambda =7$.

The characteristic equation is $\det (B-\lambda I)=0$, where $B-\lambda I=\left(\begin{array}{cc}a-\lambda & b\\ c& d-\lambda \end{array}\right)$.

$\det (B-\lambda I)=(a-\lambda )(d-\lambda )-bc=ad-a\lambda -d\lambda +{\lambda }^2-bc$.

$={\lambda }^2-(a+d)\lambda +(ad-bc)=0$. Note: $a+d=\text{tr}(B)$ and $ad-bc=\det (B)$.

The characteristic equation is ${\lambda }^2-(a+d)\lambda +(ad-bc)=0$, which matches the first option.

The characteristic polynomial of a $2\times 2$ matrix is ${\lambda }^2-\text{tr}(B)\lambda +\det (B)=0$.

The option ${\lambda }^2+(a+d)\lambda +(ad-bc)=0$ has the wrong sign on the trace term. The coefficient of $\lambda$ should be $-(a+d)$, not $+(a+d)$. Eliminated.

The option ${\lambda }^2-(a+d)\lambda -(ad-bc)=0$ has the wrong sign on the determinant term. It should be $+(ad-bc)$. Eliminated.

The option ${\lambda }^2-(ad-bc)\lambda +(a+d)=0$ has the roles of trace and determinant swapped. Eliminated.

Only ${\lambda }^2-(a+d)\lambda +(ad-bc)=0$ correctly states the characteristic equation with $-\text{tr}(B)$ as the $\lambda$-coefficient and $\det (B)$ as the constant term.