Activity 4.2.2

The eigenvalues of a square matrix are defined by the condition that there be a nonzero solution to the homogeneous equation $(A-\lambda I)\vvec=\zerovec$.

1. If there is a nonzero solution to the homogeneous equation $(A-\lambda I)\vvec = \zerovec$, what can we conclude about the invertibility of the matrix $A-\lambda I$?

2. If there is a nonzero solution to the homogeneous equation $(A-\lambda I)\vvec = \zerovec$, what can we conclude about the determinant $\det(A-\lambda I)$?

3. Let's consider the matrix $$A = \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ \end{array}\right]$$ from which we construct $$A-\lambda I = \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ \end{array}\right] - \lambda \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ \end{array}\right] = \left[\begin{array}{rr} 1-\lambda & 2 \\ 2 & 1-\lambda \\ \end{array}\right]$$. Find the determinant $\det(A-\lambda I)$. What kind of equation do you obtain when we set this determinant to zero to obtain $\det(A-\lambda I) = 0$?

4. Use the determinant you found in the previous part to find the eigenvalues $\lambda$ by solving the equation $\det(A-\lambda I) = 0$. We considered this matrix in this activity so we should find the same eigenvalues for $A$ that we found by reasoning geometrically there.

5. Consider the matrix $A = \left[\begin{array}{rr} 2 & 1 \\ 0 & 2 \\ \end{array}\right]$ and find its eigenvalues by solving the equation $\det(A-\lambda I) = 0$.

6. Consider the matrix $A = \left[\begin{array}{rr} 0 & -1 \\ 1 & 0 \\ \end{array}\right]$ and find its eigenvalues by solving the equation $\det(A-\lambda I) = 0$.

7. Find the eigenvalues of the triangular matrix $\left[\begin{array}{rrr} 3 & -1 & 4 \\ 0 & -2 & 3 \\ 0 & 0 & 1 \\ \end{array}\right]$. What is generally true about the eigenvalues of a triangular matrix?