From 6cf6c1977c1cda2a7c123368630cc7703ddc5172 Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Thu, 11 Nov 2021 18:13:00 -0500 Subject: more physics and linear homeworks --- li/hw8.tex | 169 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 169 insertions(+) create mode 100644 li/hw8.tex (limited to 'li/hw8.tex') diff --git a/li/hw8.tex b/li/hw8.tex new file mode 100644 index 0000000..9b2938b --- /dev/null +++ b/li/hw8.tex @@ -0,0 +1,169 @@ +\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