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\textbf{Exercises on eigenvalues and eigenvectors}
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\begin{problem} (6.1 \#19.  {\em Introduction to Linear Algebra:} Strang) A three by three matrix $B$ is known to have eigenvalues 0, 1 and 2. This information is enough to find three of these (give the answers where possible):

\begin{enumerate}[a)]

\item The rank of $B$

\item The determinant of $B^T B$

\item The eigenvalues of $B^T B$

\item The eigenvalues of $(B^2 + I)^{-1}$
\end{enumerate}
\end{problem}

\solution{

\begin{enumerate}[a)]

\item $B$ has 0 as an eigenvalue and is therefore singular (not invertible). Since $B$ is a three by three matrix, this means that its rank can be at most 2. Since $B$ has two distinct nonzero eigenvalues, its rank is exactly 2. 

\item Since $B$ is singular,  $\det(B) =0$ . Thus $ \det (B^T B) = \det (B^T) \det(B) = 0$. 

\item There is not enough information to find the eigenvalues of $B^T B.$ For example:
\begin{small}
\begin{align*}
\text{If } B &= \threebythree 0 {} {} {} 1 {} {} {} 2 \text{ then } B^T B = \threebythree 0 {} {} {} 1 {} {} {} 4 \text{ and } \mb B \text{ has eigenvalues 0, 1, 4.}\\
\text{If } B &= \threebythree 0 1 {} {} 1 {} {} {} 2 \text{ then } B^T B = \threebythree 0 {} {} {} 2 {} {} {} 4 \text{ and } \mb B \text{ has eigenvalues 0, 2, 4.} 
\end{align*}
\end{small}

\item If $p(t)$ is a polynomial and if $\vb x$ is an
eigenvector of $A$ with eigenvalue $\lambda$, then $$p(A) \vb x =
p(\lambda) \vb x.$$ We also know that if $\lambda$ is an eigenvalue of $\mb A$
then $1/\lambda$ is an eigenvalue of $\mb A^{-1}$.  Hence the
eigenvalues of $(B^2 +I)^{-1}$ are $\frac{1}{0^2 + 1}$, $\frac{1}{1^2 + 1}$ and $\frac{1}{2^2 + 1}$, or \textbf{1, 1/2} and \textbf{1/5}.
\end{enumerate}
}

\begin{problem} (6.1 \#29.) Find the eigenvalues of $A, B,$ and $C$ when
$$\mb A = \threebythree 1 2 3 0 4 5 0 0 6, \; B = \threebythree 0 0 1 0 2 0 3 0 0 \text{ and } C = \threebythree 2 2 2 2 2 2 2 2 2. $$
\end{problem}

\solution{ Since the eigenvalues of a triangular matrix are its diagonal entries, the eigenvalues of $A$ are 1,4, and 6. For $B$ we have:
\begin{align*}
\det (B - \lambda I) &= (-\lambda)(2-\lambda)(-\lambda) -3(2-\lambda)\\
&= (\lambda^2 - 3)(2-\lambda).
\end{align*}
Hence the eigenvalues of $B$ are $\pm \sqrt{3}$ and 2. Finally, for $C$ we have:
\begin{align*}
\det(C-\lambda I) &= (2-\lambda)[(2-\lambda)^2 -4] - 2[2(2-\lambda)-4] + 2 [4-2(2-\lambda)] \\
&= \lambda^3 - 6\lambda^2 = \lambda^2 (\lambda -6).
\end{align*}
The eigenvalues of $C$ are 6, 0, and 0.

We can quickly check our answers by computing the determinants of $\mb A$ and $\mb B$ and by noting that $\mb C$ is singular.
}



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