From 6cf6c1977c1cda2a7c123368630cc7703ddc5172 Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Thu, 11 Nov 2021 18:13:00 -0500
Subject: more physics and linear homeworks

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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 5.1}
+
+\noindent{\bf 4.}
+
+$P$ has two eigenvalues $c_1=0$ and $c_2=1,$ giving us corresponding
+eigenvectors $v_1=(1,-1)$ and $v_2=(1,1).$
+This gives a general solution
+$$c_1v_1 + c_2v_2e^t,$$
+and our initial condition $u(0)$ can be written as the combination
+$v_1+4v_2,$
+giving our particular solution
+$$v_1 + 4v_2e^t.$$
+
+\noindent{\bf 14.}
+
+The matrix of ones has rank 1 because any pair of rows makes a linearly
+dependent set (by including a duplicate).
+$$\dmatrix{1-\lambda & 1 & 1 & 1\cr
+           1 & 1-\lambda & 1 & 1\cr
+           1 & 1 & 1-\lambda & 1\cr
+           1 & 1 & 1 & 1-\lambda\cr} = 0\Longrightarrow
+  \dmatrix{1-\lambda & 1 & 1 & 1\cr
+           \lambda & -\lambda & 0 & 0\cr
+           \lambda & 0 & -\lambda & 0\cr
+           \lambda & 0 & 0 & -\lambda\cr} = 0\Longrightarrow
+  \dmatrix{4-\lambda & 0 & 0 & 0\cr
+           1 & -1 & 0 & 0\cr
+           1 & 0 & -1 & 0\cr
+           1 & 0 & 0 & -1\cr} = 0,
+$$
+by row operations (including multiplications), which do not affect a
+zero determinant where $\lambda\neq0.$ Because this is triangular, the
+determinant is the product of the diagonal, so $\lambda = 4.$
+This eigenvalue corresponds to $(1,1,1,1)^T.$
+
+The checkerboard matrix has rank 2 because any triplet of rows makes a
+linearly dependent set (again by including a duplicate).
+
+$$\dmatrix{-\lambda&1&0&1\cr
+           1&-\lambda&1&0\cr
+           0&1&-\lambda&1\cr
+           1&0&1&-\lambda} = 0\Longrightarrow
+  \dmatrix{-\lambda&0&\lambda&0\cr
+           0&-\lambda&0&\lambda\cr
+           0&1&-\lambda&1\cr
+           1&0&1&-\lambda} = 0\Longrightarrow$$$$
+  \dmatrix{-1&0&1&0\cr
+           0&-1&0&1\cr
+           0&0&-\lambda&2\cr
+           0&0&2&-\lambda} = 0\Longrightarrow
+  \dmatrix{-1&0&1&0\cr
+           0&-1&0&1\cr
+           0&0&-\lambda&2\cr
+           0&0&0&-\lambda+4/\lambda} = 0
+$$
+This gives $\lambda^2 - 4 = 0 \to \lambda = \pm 2.$
+These eigenvalues correspond to $(1,1,1,1)$ and $(1,-1,1,-1).$
+
+\noindent{\bf 15.}
+
+The rank of an $n\times n$ matrix of ones is 1 (giving $\lambda=0$ with
+multiplicity $n-1$), and its remaining non-zero eigenvalue of $n$
+(because $A-nI$ has the sum of every row, a linear combination, equal to
+zero, therefore it is an eigenvalue).
+
+The rank of an $n\times n$ checkerboard matrix is 2 because it has only
+two unique columns (giving $\lambda=0$ with multiplicity $n-2$) and its
+non-zero eigenvalues being $\pm\sqrt n,$ determined from the cofactor
+expansion of $A-\lambda I$ where $A$ is the checkerboard matrix.
+
+    {\bf Section 5.2}
+
+\noindent{\bf 8.}
+
+{\it (a)}
+$Au = uv^Tu = u(v\cdot u) = (v\cdot u)u,$
+giving that $u$ is an eigenvector with corresponding eigenvalue $v^Tu.$
+
+{\it (b)}
+The other eigenvalues are zero because there is one non-zero eigenvalue
+already shown and the dimension of the null space of $A = A - 0I$ is
+$n-r = n-1,$ completing the eigenvalue set.
+
+{\it (c)}
+The trace is equal to the sum of eigenvalues, which is $v^Tu + 0 =
+v^Tu,$ and it's equal to the sum of the diagonal, of which each entry is
+an element-wise product of $v$ and $u.$
+
+\noindent{\bf 12.}
+
+{\it (a)}
+False.
+$$\bmatrix{1&1&0\cr0&1&1\cr0&0&1}$$
+is invertible but has only the eigenvector set mentioned.
+
+{\it (b)}
+True. Every eigenvalue corresponds to at least one eigenvector (counting
+complex eigenvectors), so there must be only one (multiple) eigenvalue
+of $A.$
+
+{\it (c)}
+
+A is not diagonalizable because its eigenvector set does not span $R^3.$
+
+\noindent{\bf 32.}
+$$A = \bmatrix{2&1\cr1&2}.$$
+This matrix has eigenvalues $1$ and $3,$ and they correspond to
+eigenvectors $(1,-1)$ and $(1,1)$ respectively, giving a diagonalization
+with
+$$S = \bmatrix{1&1\cr-1&1} \Longrightarrow S^{-1} = \fr12\bmatrix{1&-1\cr1&1}$$
+and
+$$\Lambda = \bmatrix{1&0\cr0&3}$$
+$$A^k = S\Lambda^kS^{-1} =
+\bmatrix{1&1\cr-1&1}\bmatrix{1&0\cr0&3^k}\fr12\bmatrix{1&-1\cr1&1} =
+\fr12\bmatrix{1&1\cr-1&1}\bmatrix{1&-1\cr3^k&3^k} =
+\fr12\bmatrix{1+3^k&-1+3^k\cr-1+3^k&1+3^k}.
+$$
+
+    {\bf Section 5.3}
+
+\noindent{\bf 4.}
+
+$$A = \bmatrix{1/2 & 1/2\cr 1 & 0}.$$
+
+{\it (a)}
+$A$ has characteristic polynomial $\lambda^2 - \lambda/2 - 1/2 =
+(\lambda-1)(\lambda+1/2) \to \lambda = 1, -1/2,$ with eigenvectors
+respectively $(1,1)$ and $(1/2, -1).$
+
+{\it (b)}
+
+$$A = -\fr23\bmatrix{1&1/2\cr 1&-1}\bmatrix{1&0\cr0&-1/2}
+\fr13\bmatrix{2&1\cr2&-2}.$$
+As $k\to\infty,$
+$$A^k = \fr13\bmatrix{1&1/2\cr 1&-1}\bmatrix{1&0\cr0&-1/2}^k
+\bmatrix{2&1\cr2&-2} =
+\bmatrix{1&1/2\cr 1&-1}\bmatrix{2&1\cr0&0} =
+\bmatrix{2&1\cr2&1}.$$
+
+{\it (c)}
+
+$$A\bmatrix{1\cr0} = -\fr13\bmatrix{2&1\cr2&1}\bmatrix{1\cr0} =
+\bmatrix{2/3\cr2/3},$$
+giving us the steady state limit of 2/3 from this initial condition.
+
+\noindent{\bf 18.}
+
+These matrices are stable/neutrally stable if both eigenvalues are less
+than or equal to 1. Therefore, the maximal value will be when the larger
+eigenvalue equals 1.
+
+For the first matrix with characteristic equation
+$\lambda^2 - (.2+a)\lambda + .2a + .64 = 0 \to
+(\lambda-(.1+a/2))^2 = .01 + .1a + a^2/4 -.2a + .64,$
+$\lambda = .1+a/2 \pm (.65-.1a+a^2/4) =
+.75 - .4a + .25a^2 = 1 \to a = 2.081$
+
+The eigenvalues of this matrix are $.2$ and $b,$ so the largest
+neutrally stable value of $b$ is 1.
+
+For the third matrix with characteristic equation $\lambda^2 -
+2c\lambda + c^2 - .16 = 0,$ $\lambda = c \pm .4,$ giving maximum $c$ of
+$0.6.$
+
+\bye
-- 
cgit 

