C23) Find the Eigenvalues, eigenspaces, algebraic and geometric multiplicities for \(A = \begin{bmatrix}1 & 1\\ 1 & 1\\ \end{bmatrix}\)

We’ll begin with getting \(A - \lambda I\), which will just be \(\begin{bmatrix}1-\lambda & 1 \\ 1 & 1-\lambda \end{bmatrix}\). Then we need the determinant \(det(A - \lambda I)\), which is \((1 - \lambda)(1 - \lambda) - 1 = 1 - \lambda - \lambda + \lambda^2 - 1 = \lambda^2 - 2\lambda + 0 = (\lambda + 0)(\lambda - 2)\)

The eigenvalues are then \(\lambda = 0\) and \(\lambda = 2\)

Now that we have our eigenvalues, we can find the eigenspaces. Let’s start with \(\lambda = 0\). Substituting it into \(A - \lambda I\) we have \(\begin{bmatrix}1 - 0 & 1 \\ 1 & 1 - 0 \end{bmatrix} = \begin{bmatrix}1 & 1 \\ 1 & 1 \end{bmatrix}\), which happens to be just \(A\). Now we must solve for \(x_1\) and \(x_2\) set to 0. We can simply do row reduction on the matrix \(\begin{bmatrix} 1 & 1 & 0\\1 & 1 & 0\end{bmatrix}\). Well, row 1 - row 2 = row 2 is the only move we need to make, leaving us with \(\begin{bmatrix}1 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\) or \(x_1 + x_2 = 0\). Very simply we can see that \(x_1 = -1\) and \(x_2 = 1\) will solve this, meaning that the eigenspace for \(\lambda = 0\) is the vector \(\begin{bmatrix}-1 \\ 1\end{bmatrix}\)

We’ll repeat the same process for \(\lambda = 2\). \(A - \lambda I\) becomes \(\begin{bmatrix} 1-2 & 1 \\1 & 1-2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\ 1 & -1 \end{bmatrix}\). To solve for \(x_1\) and \(x_2\) we set to \(0\) and row reduce. \(\begin{bmatrix} -1 & 1 & 0 \\ 1 & -1 & 0\end{bmatrix}\). Row 1 + Row 2 = Row 2 leaves us with \(\begin{bmatrix}-1 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\) or \(-x_1 + x_2 = 0\). Having both \(x_1\) and \(x_2\) be equal to \(1\) solves this equation, which means the eigenspace for \(\lambda = 2\) is the vector \(\begin{bmatrix}1 \\ 1 \end{bmatrix}\)

The algebraic multiplicities, \(\alpha A\), for both eigenvalues is 1 because the highest power for each \(\lambda + 0\) and \(\lambda - 2\) is 1.

The geometric multiplicities, \(\gamma A\), for both eigenvalues is also 1 because the eigenspaces for both are 1.