path: root/li/hw9.tex

\def\bmatrix#1{\left[\matrix{#1}\right]}
\def\dmatrix#1{\left|\matrix{#1}\right|}
\def\fr#1#2{{#1\over #2}}

    {\bf Section 5.4}

\noindent{\bf 8.}

$${dr\over dt} = 4r-2w.$$
$${dw\over dt} = r+w.$$
$${du\over dt} = \bmatrix{4&-2\cr1&1}u.$$

The system has eigenvalues $\lambda = 2, 3,$ with eigenvectors
$(1, 1)^T$ and $(2, 1)^T,$

{\it (a)} This system is unstable because all eigenvalues are greater
than zero.

{\it (b)} $$\bmatrix{300\cr 200} = 100\bmatrix{1\cr 1} +
100\bmatrix{2\cr 1},$$
giving
$$u = \bmatrix{1&2\cr 1&1}\bmatrix{e^{2t}&0\cr 0&e^{3t}}\bmatrix{100\cr100} =
\bmatrix{100e^{2t} + 200e^{3t}\cr 100e^{2t} + 100e^{3t}}.$$

{\it (c)}

The dominant growth is $e^{3t},$ so the ratio will eventually be
2 to 1.

\noindent{\bf 24.}

$v+w$ is constant if $${d\over dt}(v+w) = {dv\over dt} + {dw\over dt} =
w-v + v-w = 0.$$

$${du\over dt} = \bmatrix{-1&1\cr1&-1}u,$$
with eigenvalues $-2$ and $0$ and corresponding eigenvectors $(1, -1)^T$
and $(1, 1)^T$ respectively. This gives a particular solution (with
$v(0) = 30$ and $w(0) = 10$)
$$u = \bmatrix{-1&1\cr1&-1}\bmatrix{20\cr10e^{-2t}} = \bmatrix{-1\cr
1}20 + \bmatrix{1\cr -1}10e^{-2t}$$

\noindent{\bf 42.}

The reason this (falsely) appears to work is because the right side of
the matrix is still multiplying from
$$u = \bmatrix{x\cr y},$$ while the left side is supposed to be
$${du\over dt} = \bmatrix{dy/dt\cr dx/dt}.$$
In order to make this correct, there should also be a column exchange
(changing $u$ to represent $(y, x)^T,$)
giving
$$\bmatrix{2&-2\cr-4&0},$$
which is unstable.

    {\bf Section 5.5}

\noindent{\bf 11.}

$$P = 0\bmatrix{1/2\cr -1/2}\bmatrix{1/2&-1/2}
    + 1\bmatrix{1/2\cr 1/2}\bmatrix{1/2&1/2}.$$

$$Q = 1\bmatrix{1/2\cr -1/2}\bmatrix{1/2&-1/2}
     -1\bmatrix{1/2\cr 1/2}\bmatrix{1/2&1/2}.$$

$$R = 5\bmatrix{2/\sqrt5\cr 1/\sqrt5}\bmatrix{2/\sqrt5&1/\sqrt5}
     -5\bmatrix{1/\sqrt5\cr -2/\sqrt5}\bmatrix{1/\sqrt5&-2/\sqrt5}.$$

\noindent{\bf 13.}

{\it (a)}

The eigenvectors $u,$ $v,$ and $w$ are pairwise orthogonal because $A$
is symmetric.

{\it (b)}
The null space is $\mathop{\rm sp}(u)$ because it's, by definition, in
$\mathop{\rm null}(A-0I),$ and the column space is $\mathop{\rm
sp}(v,w),$ because an eigenvalue's multiple must be a valid output of
the matrix.
The left null space is also $\mathop{\rm sp}(u)$ by
orthogonality to the column space, and the row space is $\mathop{\rm
sp}(v,w)$ by orthogonality to the null space.

{\it (c)}

No. $x+u$ also satisfies. $A(x+u) = Ax + Au = v + w + 0.$

{\it (d)}

If $b\in\mathop{\rm sp}(v,w),$ this equation has a solution, by
definition of the column space.

{\it (e)}

$S^TS = I,$ because all the vectors are orthogonal, and normal/unit, so
the dot product with eachother is zero and the dot product with
themselves is one. Thus, $S^{-1} = S^T.$

$$S^{-1}AS = \Lambda,$$
where $\Lambda$ is the diagonal matrix of eigenvalues
$$\bmatrix{0&0&0\cr 0&1&0\cr 0&0&2}.$$

\noindent{\bf 14.}

$A$ is orthogonal (all columns have norm one and are orthogonal to each
other), permutation (all rows and columns have exactly one non-zero
entry which is a one) and therefore invertible, diagonalizable (on the
complex numbers), and Markov.
By row reducing $A-\lambda I,$ we get
$$\bmatrix{-\lambda&1&0&0\cr
           0&-\lambda&1&0\cr
           0&0&-\lambda&1\cr
           0&0&0&-\lambda+\lambda^{-3}},$$
which has determinant equal to product of the diagonal, or
characteristic polynomial
$$\lambda^4 - 1 = 0,$$
giving $A$ eigenvalues of the fourth roots of unity, $\{1,-1,i,-i\}.$

$B$ is a projection (because $B^2 = B,$ and symmetric), Hermitian (by
symmetric), rank-1 (because all columns are identical), diagonalizable
(by symmetric), and Markov (all columns sum to one).
Rank-1/singular excludes orthogonal and invertible.
$B$ has eigenvalues $0$ and $1.$

    {\bf Section 5.6}

\noindent{\bf 30.}

$M$ is the matrix which transforms the eigenvectors of $A$ to the
corresponding eigenvectors (based on eigenvalues) of $B.$

{\it (a)}
$$M = M^{-1} = \bmatrix{0&1\cr1&0}.$$
$$B = M^{-1}AM =
\bmatrix{0&1\cr1&0}\bmatrix{0&1\cr0&1}\bmatrix{0&1\cr1&0} =
\bmatrix{1&0\cr1&0}\bmatrix{0&1\cr1&0} = \bmatrix{0&1\cr0&1}.$$

{\it (b)}

$$M = M^{-1} = \bmatrix{1&0\cr0&-1}.$$
$$B = M^{-1}AM =
\bmatrix{1&0\cr0&-1}\bmatrix{1&1\cr1&1}\bmatrix{1&0\cr0&-1} =
\bmatrix{1&0\cr0&-1}\bmatrix{1&-1\cr1&-1} =
\bmatrix{1&-1\cr-1&1}.$$

{\it (c)}

$$A = \bmatrix{1&2\cr3&4}$$
has eigenvalues $\lambda^2 - 5l - 2 = 0 \to \lambda = 5/2\pm
\sqrt{33}/2,$
corresponding to eigenvectors $(\fr16(-3+\sqrt33), 1)^T$ and
$(\fr16(-3-\sqrt33),1)^T$ in decreasing order.
$$B = \bmatrix{4&3\cr2&1},$$
with the same eigenvalues but corresponding eigenvectors (also in
decreasing order of eigenvalue)
$(\fr14(3+\sqrt33),1)$ and
$(\fr14(3-\sqrt33),1).$

This corresponds to
$$M = \bmatrix{3/2&3/2\cr0&1}.$$
For this $M,$ $B = M^{-1}AM.$

\noindent{\bf 32.}

There is the zero family (with one matrix), the family with all ones
(with one matrix), the permutation matrix $\bmatrix{0&1\cr1&0}$ (with
a family of one), and the identity matrix (with one in its family).

Then, there are the families with two members: those with one nonzero
entry on the antidiagonal, like
$$\bmatrix{0&1\cr0&0},$$
those with a full diagonal and one entry on the antidiagonal, like
$$\bmatrix{1&1\cr0&1},$$
and those with a full antidiagonal and one entry on the diagonal, like
$$\bmatrix{1&1\cr1&0}.$$

Then, there is the largest family, with six members, consisting of one
entry on the diagonal and zero or one entry on the antidiagonal, like
$$\bmatrix{1&1\cr0&0}.$$

I found these distinct families with the number of possible
characteristic polynomials, based on trace having a value $0,1,2$ and
determinant having a value $-1,0,1,$ and then splitting up repeated
eigenvalue matrices based on the number of distinct eigenvalues.

\bye