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+\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