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\begin{center}
\textbf{Mathematics 262} \\
\textbf{``Homework 4'' solutions}
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\vspace{-0.05in}

Assignment: Section 6.2, problems 1, 4, 6, 7

\S 6.2, \#1:
For the matrix 
\[
\begin{bmatrix}
1 & 0 \\ 0 & 0
\end{bmatrix},
\]
the characteristic polynomial is $(x-1)x$, so the characteristic
values are 1 and 0.  (This is true over both $\mathbf{R}$ and
$\mathbf{C}$.)  The vector $\begin{bmatrix} 1 \\ 0
\end{bmatrix}$ is an eigenvector for the eigenvalue 1, and the vector
$\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ is an eigenvector for 2.

\myline 

For the matrix 
\[
\begin{bmatrix}
2 & 3 \\ -1 & 1
\end{bmatrix},
\]
the characteristic polynomial is $x^{2}-3x+5$, which has no roots over
\textbf{R}, so $T$ has no characteristic values.  The operator $U$,
defined over \textbf{C}, has characteristic values $\frac{3\pm
i\sqrt{11}}{2}$.  The corresponding characteristic vectors are messy.
Whenever you have a $2 \times 2$ matrix $A$ with characteristic value
$c$, then the rows of $cI-A$ are linearly dependent; since there are
only two rows, this means that either one row is zero, or each row is
a scalar multiple of the other.  In general, this is a good way to
check your work; if the algebra is a bit complicated, as in this case,
you can also just look at, say, the second row.  For the
characteristic value $c=\frac{3+i\sqrt{11}}{2}$, $cI-A$ is 
\[
\begin{bmatrix}
\frac{3+i\sqrt{11}}{2}-2 & -3 \\
1 & \frac{3+i\sqrt{11}}{2}-1
\end{bmatrix}.
\]
From the second row, I can see that $(\frac{3+i\sqrt{11}}{2}-1, -1)$
is a characteristic vector.  I could instead have used the first row
to get the characteristic vector $(3, \frac{3+i\sqrt{11}}{2}-2)$,
which \emph{is} a scalar multiple of the first vector, even though it
may not look like it.

Similarly, the other characteristic value has associated
characteristic vector $(\frac{3-i\sqrt{11}}{2}-1, -1)$.

\myline 

The characteristic polynomial of the matrix 
\[
\begin{bmatrix}
1 & 1 \\ 1 & 1
\end{bmatrix}
\]
is $x(x-2)$, so the roots are 0 and 2.  0 has eigenvector
$\begin{bmatrix} 1 \\ -1 \end{bmatrix}$, and 2 has eigenvector
$\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.  (Same for $T$ and $U$.)

\newpage

\S 6.2, \#4:
First I'll find the characteristic values, then for each of those,
I'll find characteristic vectors.  The characteristic polynomial is
$f(x)=x^{3}-x^{2}-5x-3$.  By inspection (i.e., guessing), I find that
$-1$ is a root, so I factor out $x+1$ and see what's left; I get that
$f(x) = (x+1)(x^{2}-2x-3)=(x+1)^{2}(x-3)$.  For $-1$, we need to find
2 linearly independent characteristic vectors; we start by writing
down the matrix $-I-A$:
\[
\begin{bmatrix}
8 & -4 & -4 \\
8 & -4 & -4 \\
16 & -8 & -8
\end{bmatrix}.
\]
This obviously has rank 1, and hence nullity 2.  The nullspace is
spanned by the vectors $(1,2,0)$ and $(1,0,2)$ (for example).  A
characteristic vector for $3$ is $(1,1,2)$.  These three vectors
together form a basis for $\mathbf{R}^{3}$, with respect to which $T$
is diagonal.  In fact, the matrix for $T$ in this basis is 
\[
\begin{bmatrix}
-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3
\end{bmatrix}.
\]

\myline 

\S 6.2, \#6:
The given matrix has characteristic polynomial $x^{4}$, which has only
one root, $0$.  So for this to be diagonalizable, the eigenspace for
$0$ must have dimension 4.  Equivalently, the nullity of the matrix
$0I-A=-A$ must be 4.  Then $-A$ must be the zero matrix, so we must
have $a=b=c=0$.

\myline 

\S 6.2, \#7:
If $T$ has $n$ different eigenvalues $c_{1}$, \dots, $c_{n}$, then
since the characteristic polynomial for $T$ has degree $n$, it must be
$(x-c_{1}) \dots (x-c_{n})$.  To apply our theorem, we only have to
know that for each $j$, the eigenspace associated to $c_{j}$ has the
correct dimension, which in this case is $1$.  Eigenspaces always have
dimension at least 1, and their dimension is at most the multiplicity
of their eigenvalue as a root of the characteristic polynomial.  In
this case, that means that each eigenspace has dimension at most 1
(since there are no repeated roots of the characteristic polynomial),
and hence has dimension exactly 1.

By the way, this is probably the easiest way to check that a matrix is
diagonalizable, since in this case you just have to check that there
are $n$ distinct roots of the characteristic polynomial.  When there
are repeated roots, then you have to work harder to see whether it's
diagonalizable. 

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