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\textbf{Mathematics 262} \\
\textbf{Homework 6 solutions}
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Assignment: Section 6.2, problem 12 (cont'd); Section 6.4, problems 1,
4, 5, 9

\S 6.2, \#12:
When last we left our heroes, they had solved this problem except for
the case when the matrix $A$ has one eigenvalue $a$, with a
$1$-dimensional eigenspace.  The characteristic polynomial for $A$ is
$(x-a)^{2}$; by the Cayley-Hamilton theorem, we have $(A-aI)^{2}=0$.
Now we can apply problem 11 to the matrix $A-aI$, to conclude that
$A-aI$ is similar to either the zero matrix or to $\begin{bmatrix}
0 & 0 \\ 1 & 0 \end{bmatrix}$.  In other words, for some invertible
matrix $P$, we have
\[
P(A-aI)P^{-1} = PAP^{-1} - P(aI)P^{-1} = PAP^{-1} - aI = 
\left\{ \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \ \text{or} \ 
\begin{bmatrix}
0 & 0 \\ 1 & 0 
\end{bmatrix}
\right\}.
\]
Hence $PAP^{-1}$ is equal to either $aI$ or to $\begin{bmatrix} a & 0
\\ 1 & a \end{bmatrix}$.  The first case can't actually happen,
because then the eigenspace for $a$ would be 2-dimensional; therefore
$A$ must be similar to $\begin{bmatrix} a & 0 \\ 1 & a \end{bmatrix}$.

\S 6.4, \#1:
(a) If $T$ has an invariant subspace other than $0$ and
$\mathbf{R}^{2}$, then it must be one-dimensional, which means that
there must be a nonzero vector $\alpha$ so that $T\alpha = c\alpha$
for some $c$.  In other words, $T$ must have an eigenvalue.  But the
characteristic polynomial for $T$ is $x^{2}-3x+4$, which has no real
roots.

(b) The characteristic polynomial for $A$ has complex roots $\frac{3
\pm \sqrt{9-16}}{2}$.  These two distinct roots are the eigenvalues,
and their one-dimensional eigenspaces are both invariant subspaces.
(In fact, they are the only invariant one-dimensional subspaces, but
you weren't asked to show that.)

\S 6.4, \#4:
By the theorem from class, if we want to know whether $A$ is similar
to a triangular matrix, we should try to find the minimal polynomial
for $A$.  To do this, first we find the characteristic polynomial; it
ends up being $f(x) = x^{3}$.  Since the minimal polynomial divides
evenly into this, it must be $p(x)=x^{r}$, where $r$ is 1, 2, or 3.
In any case, it factors into linear terms, and so $A$ is similar to a
triangular matrix (with 0's on the diagonal---you always get the
eigenvalues on the diagonal).

To find that triangular matrix, I guess we should look for a basis, as
in the proof of the theorem.  $A$ has only one eigenvalue, 0, so the
first basis element should be an eigenvector for 0.  The eigenspace is
1-dimensional, spanned by $(1, 0, -1)$, so we'll let $\alpha_{1} = (1,
0, -1)$.  Now we look for a vector $\alpha_{2}$ so that
$(A-cI)\alpha_{2}$ is in the vector space spanned by $\alpha_{1}$,
where $c$ is ``some eigenvector of $A$.''  $A$ only has one
eigenvector, $0$, so we can rewrite this equation as 
\[
A \alpha_{2} = b\alpha_{1}.
\]
If you solve this equation for $\alpha_{2}$ (and in fact you might as
well let $b=1$, right?), you find that there are several possibilities
for $\alpha_{2}$, so you pick one: let $\alpha_{2} = (1, 1, 0)$.
(Another possibility is $(0,1,1)$, which is
$\alpha_{2}-\alpha_{1}$---any linear combination of $\alpha_{1}$ and
$\alpha_{2}$ would have worked.)  Now we want a vector $\alpha_{3}$,
not in the span of $(\alpha_{1}, \alpha_{2})$, which satisfies
\[
A \alpha_{3} = a \alpha_{1} + b \alpha_{2},
\]
for some scalars $a$ and $b$.  Actually, at this point you know you're
supposed to have a basis, so you could try anything for $\alpha_{3}$
which is linearly independent from the other two, and see what
happens.  For instance, the vector $\alpha_{3} = (1, 0, 0)$ is a
solution to 
\[
A \alpha_{3} = -2\alpha_{1} + 2\alpha_{2} = \begin{bmatrix}
0 \\ 2 \\ 2 \end{bmatrix}.
\]
With respect to the basis $(\alpha_{1}, \alpha_{2}, \alpha_{3})$, the
matrix $A$ is
\[
\begin{bmatrix}
0 & 1 & -2 \\ 0 & 0 & 2 \\ 0 & 0 & 0
\end{bmatrix}.
\]
(Note: the matrix $A$ has rank 2, and so does this matrix.  So this
answer may in fact be correct.  There are also lots of possible
correct answers, unfortunately for the grader.)

\S 6.4, \#5:
If the matrix $A$ satisfies $A^{2}=A$, then it satisfies the
polynomial $x^{2}-x = x(x-1)$.  Therefore the minimal polynomial for
$A$ divides evenly into $x(x-1)$, and so factors into distinct linear
terms.  Therefore $A$ is similar to a diagonal matrix.

\S 6.4, \#9:
Yes, the space of polynomials is invariant under $T$: the integral of
any polynomial is again a polynomial.

Yes, the space of differentiable functions is invariant under $T$: by
the fundamental theorem of calculus, $Tf$ is differentiable for every
$f$.

No, the space of functions which vanish at $x=\frac{1}{2}$ is not
invariant; for instance, if I let $f(x) =2x-1$, then $f(x)$ is in this
subspace.  On the other hand, $(Tf)(x)=x^{2}-x$, which is not.

(On the other hand, the space of functions which vanish at $0$ is
invariant.)

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