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Problems #11,pg.272 and #7,pg.279 are graded.
#11,pg.272: Find a basis for the eigenspace corresponding to each listed eigenvalue: $A=\begin{bmatrix} 1 & -3 \\ -4 & 5 \end{bmatrix}; \lambda=-1,7$.
Solution: Recall that given any eigenvalue $\lambda$, the eigenspace is $\mathrm{Nul}(A-\lambda I)$. First let us find the eigenspace associated with the eigenvalue $\lambda=-1$. This means we need to find a basis for
$$\mathrm{Nul}(A-(-1)I) = \mathrm{Nul} \left( \begin{bmatrix} 1-(-1) & -3 \\
-4 & 5-(-1) \end{bmatrix} \right) = \mathrm{Nul} \left( \begin{bmatrix} 2 & -3 \\ -4 & 6 \end{bmatrix} \right).$$
Recall that the basis of a nullspace can be easy found from a reduced echelon form. So we row reduce $\begin{bmatrix} 2 & -3 \\ -4 & 6 \end{bmatrix}$ and find
$$\begin{bmatrix} 2 & -3 \\ -4 & 6 \end{bmatrix} \sim \begin{bmatrix} 1 & -\dfrac{3}{2} \\ 0 & 0 \end{bmatrix}.$$
Hence we conclude that the solution vector $\vec{x}$ to the homogeneous equation $\begin{bmatrix} 2& -3 \\ -4 & 6 \end{bmatrix} \vec{x} = \vec{0}$ is
$$\vec{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \dfrac{3}{2}x_2 \\ x_2 \end{bmatrix} = x_2 \begin{bmatrix} \dfrac{3}{2} \\ 1\end{bmatrix}.$$
Hence we conclude that
$$\mathrm{Nul} \left( \begin{bmatrix} 2 & -3 \\ -4 & 6 \end{bmatrix} \right) = \mathrm{span} \left\{ \begin{bmatrix} \dfrac{3}{2} \\ 1 \end{bmatrix} \right\}.$$
Since the set $\left\{ \begin{bmatrix} \dfrac{3}{2} \\ 1 \end{bmatrix} \right\}$ is linearly independent, we conclude that
$$\mathscr{B} = \left\{ \begin{bmatrix} \dfrac{3}{2} \\ 1 \end{bmatrix} \right\}$$
is a basis for the eigenspace associated with $\lambda=-1$.
#7,pg.279: Find the characteristic polynomial and the real eigenvalues of the matrix $\begin{bmatrix} 5 & 3 \\ -4 & 4 \end{bmatrix}$.
Solution: Recall that the characteristic polynomial is the polynomial in variable $\lambda$ given by $\mathrm{det}(A-\lambda I)$. The roots of this polynomial are the eigenvalues, i.e. we must find $\lambda$ that solves $\mathrm{det}(A-\lambda I)=0.$ So compute
$$A - \lambda I = \begin{bmatrix} 5-\lambda & 3 \\ -4 & 4-\lambda \end{bmatrix},$$
so
$$\begin{array}{ll}
\mathrm{det} (A-\lambda I) &= (5-\lambda)(4-\lambda) - (3)(-4) \\
&=x^2-9x+32.
\end{array} $$
We can compute the roots of this polynomial via the quadratic formula:
$$\begin{array}{ll}
x &= \dfrac{9 \pm \sqrt{(-9)^2-4(1)(32)}}{2} \\
&= \dfrac{9}{2} \pm \dfrac{\sqrt{81-128}}{2} \\
&=\dfrac{9}{2} \pm \dfrac{\sqrt{47}}{2}i.
\end{array}$$
Hence there are no real eigenvalues of this matrix.
