From 6ce1db1f867c545ebdb1afb705580514b356f883 Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Wed, 5 Jan 2022 00:02:12 -0500 Subject: more math stuff --- li/hw9.tex | 185 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 185 insertions(+) create mode 100644 li/hw9.tex (limited to 'li/hw9.tex') diff --git a/li/hw9.tex b/li/hw9.tex new file mode 100644 index 0000000..ee99439 --- /dev/null +++ b/li/hw9.tex @@ -0,0 +1,185 @@ +\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 -- cgit