We are asked to discuss the eigenvalues, eigenvectors and algebraic and geometric multiplicity of the matrix \[A = \left[ \begin{array}{rr} 1 & 1 \\ 1 & 1 \end{array} \right]\].
Solving the characteristic equation for \(A\) we get:
\[ det\left( \begin{array}{rr} 1 - \lambda & 1 \\ 1 & 1 - \lambda \end{array} \right) = (1-\lambda)^2 - 1 = \lambda^2 - 2\lambda + 1 - 1 = (\lambda-0)(\lambda-2)\]
Thus, the eigenvalues of \(A\) are \(\lambda = 0 \text{ and } 2\). Solving for the eigenvector of \(\lambda=0\) we get:
\[ \left[ \begin{array} {rr} 1 & 1 \\ 1 & 1 \\ \end{array} \right] \left( \begin{array} {r} x \\ y \end{array} \right) = \left( \begin{array}{r} 0 \\ 0 \end{array} \right)\] The solution for this is \(x + y = 0\) which implies \(x = -y\) and therefore the nullspace is spanned by the column vector \[ \left( \begin{array}{r} -1 \\ 1 \end{array} \right)\].
Similarly, the eigenvector for \(\lambda = 2\) is defined by the equations:
\[ \begin{align} x + y &= 2x \\ x + y &= 2y \end{align} \] This implies that $x = y $ is the defining relation. Thus the eigenspace of \(\lambda = 2\) is spanned by the vector \[ \left( \begin{array}{r} 1 \\ 1 \end{array} \right)\]
The algebraic and geometric multiplicity of \(\lambda = 0\) are both 1. The algebraic and geometric multiplicities of \(\lambda = 2\) are both 1. The algebraic multiplicities are defined by the factorization of the characteristic polynomial. The geometric multiplicities are defined by the size of the span of the eigenspaces.
This problem asks to prove the eigenvalue of \(A^-1\) is \(1/\lambda\) if \(A\) is nonsingular with eigenvalue \(\lambda\). The proof is asked to use the characteristic equation.
We see that
\[ \begin{align} det( A- \lambda I ) & = 0 & \text{ definition of characteristic equation} \\ det( A^{-1} ( A - \lambda I ) ) & = 0 & \text{ multiply by } A^{-1} \text{ on both sides} \\ det( \frac{1}{\lambda} A A^{-1} - \frac{1}{\lambda} ( \lambda I A )^{-1} ) & = 0 & \text{ then by } \frac{1}{\lambda} \\ det( \frac{1}{\lambda} I - A^{-1}) & = 0 \\ det( A^{-1} - \frac{1}{\lambda}I) & = 0 \\ \end{align} \] Clearly,the latter is the characteristic equation of \(A^{-1}\) which eigenvalue \(\lambda^{-1}\).