From 6d18ebaa31ed960b051826a2c44c1b770324da2a Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Tue, 19 Oct 2021 21:42:14 -0400 Subject: wrote linear homeworks --- li/hw5.tex | 85 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ li/hw6.tex | 82 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 2 files changed, 167 insertions(+) create mode 100644 li/hw5.tex create mode 100644 li/hw6.tex (limited to 'li') diff --git a/li/hw5.tex b/li/hw5.tex new file mode 100644 index 0000000..3236e49 --- /dev/null +++ b/li/hw5.tex @@ -0,0 +1,85 @@ +\def\bmatrix#1{\left[\matrix{#1}\right]} +\def\fr#1#2{{#1\over #2}} + + {\bf Section 3.1} + +\noindent{\bf 6.} + +An orthonormal basis would be + +$$\{\fr1{\sqrt6}\bmatrix{1\cr1\cr-2}\}.$$ + +This was found by inspection starting with the second vector, which is +orthogonal to $(1,-1,k),$ where $k$ is any number and then choosing $k$ +such that the dot product with the first vector was zero. Then, it was +normalized. It is the complete orthogonal space because a space +orthogonal to a 2-dimensional space in $R^3$ has dimension 1. + +\noindent{\bf 12.} + +A basis would be +$$\{\bmatrix{-2\cr-2\cr 1}\}.$$ + +This was found by a similar method to 6, and because it is also in +$R^3,$ is proved complete by the same means. + +\noindent{\bf 19.} + +{\it (a)} + +The trivial space $\{0\}$ is orthogonal to itself, but $\{0\}^\perp = +R^n$ is not orthogonal to itself. + +{\it (b)} + +Orthogonality is reflexive, so if $V=Z,$ this statement is false if $V$ +is nontrivial. + +\noindent{\bf 32.} + +$A$ and $B$ have the same column spaces and left null spaces, the column +space having basis +$$\{\bmatrix{1\cr 3}\}$$ +and the left null space having orthogonal basis +$$\{\bmatrix{-3\cr 1}\}.$$ + +However, they have different row and null spaces, $A$ having row space +with basis +$$\{\bmatrix{1\cr 2}\},$$ +and null space with basis +$$\{\bmatrix{-2\cr 1}\}.$$ +and $B$ having row space with basis +$$\{\bmatrix{1\cr 0}\},$$ +and null space with basis +$$\{\bmatrix{0\cr 1}\}.$$ + + {\bf Section 3.2} + +\noindent{\bf 12.} + +Orthonormal basis is +$$Q = \bmatrix{-2/\sqrt{5}\cr 1/\sqrt{5}},$$ +which is on the line. +$$P = QQ^T = \fr1{5}\bmatrix{4&-2\cr-2&1}.$$ + +\noindent{\bf 14.} + +The point $(x,y,t) = (1,-2,1)$ is a basis for this subspace. +By the same method as in 12, + +$$P = \fr16\bmatrix{1\cr -2\cr 1}\bmatrix{1&-2&1} = +\fr16\bmatrix{1&-2&1\cr -2&4&-2\cr 1&-2&1}.$$ + +\noindent{\bf 17(a)} % only (a) + +The projection of $b$ onto the subspace through $a$ is $a^Tb/||b|| * a,$ +giving +$$(5/9)\bmatrix{1\cr1\cr1}$$ + +\noindent{\bf 24.} + +$a_1$ is a normal, so $b$ projected onto $a_1$ is $(1,0).$ +$b$ projected onto $a_2$ is $3a_2/\sqrt5.$ +The sum is $(1+3/\sqrt5, 6/\sqrt5).$ + +\bye diff --git a/li/hw6.tex b/li/hw6.tex new file mode 100644 index 0000000..e0ff219 --- /dev/null +++ b/li/hw6.tex @@ -0,0 +1,82 @@ +\def\bmatrix#1{\left[\matrix{#1}\right]} +\def\fr#1#2{{#1\over #2}} +\def\proj{\mathop{\rm proj}} + + {\bf Section 3.3} + +\noindent{\bf 6.} + +$$A = \bmatrix{1&1\cr 1&-1\cr -2&4}.$$ +$$A^TA = \bmatrix{6&-8\cr -8&18}.$$ +$$P = A(A^TA)^{-1}A^T = \bmatrix{1&1\cr 1&-1\cr -2&4} +\fr1{22}\bmatrix{9&4\cr4&3}\bmatrix{1&1&-2\cr1&-1&4} += \fr1{22}\bmatrix{20&6&2\cr 6&4&-6\cr2&-6&20}.$$ + +$$Pb = \fr1{22}\bmatrix{20&6&2\cr 6&4&-6\cr2&-6&20}\bmatrix{1\cr2\cr7} += \fr1{22}\bmatrix{46\cr-28\cr130}$$ + +\noindent{\bf 7.} + +$$A = \bmatrix{1&1\cr0&1\cr1&-1}$$ +$$A^TA = \bmatrix{2&0\cr0&3}$$ +$$A(A^TA)^{-1}A^T = +\bmatrix{1&1\cr0&1\cr1&-1}\bmatrix{1/2&0\cr0&1/3}\bmatrix{1&0&1\cr1&1&-1}.$$ + +\noindent{\bf 12.} + +{\it (a)} + +$$\left\{\pmatrix{1\cr-1\cr0\cr0}, \pmatrix{0\cr1\cr0\cr-1}\right\}$$ +is a basis for the orthogonal complement. + +{\it (b)} + +If $v_1$ and $v_2$ are the two given basis vectors for ${\bf V},$ they +are orthogonal and $v_1/\sqrt3$ and $v_2$ form an orthonormal basis, so +$$Q = \fr1{\sqrt3}\bmatrix{1&0\cr1&0\cr0&\sqrt3\cr1&0}$$ +$$P = QQ^T = +\bmatrix{1/3&1/3&0&1/3\cr1/3&1/3&0&1/3\cr0&0&1&0\cr1/3&1/3&0&1/3}$$ + +{\it (c)} + +This is the zero vector because an orthogonal vector $b$ has no parallel +component to the plane, and any parallel part in ${\bf V}$ would give a +longer distance. + + {\bf Section 3.4} + +\noindent{\bf 16.} + +A is a $3\times2$ matrix, $Q$ is also a $3\times2$ matrix, and $R$ is a +$2\times2$ matrix. + +$$A = \bmatrix{1&1\cr2&3\cr2&1} = \bmatrix{1/3&0\cr +2/3&1/\sqrt2\cr2/3&-1/\sqrt2}\bmatrix{3&3\cr0&\sqrt2} = QR.$$ + +\noindent{\bf 17.} + +$$Pb = QQ^Tb = \bmatrix{1/9&2/9&2/9\cr 2/9&17/18&-1/18\cr +2/9&-1/18&17/18}\bmatrix{1\cr1\cr1} = \bmatrix{5/9\cr10/9\cr10/9}$$ + + {\bf Section 3.5} + +\noindent{\bf 12.} + +$$y = F_8c.$$ +$$y' = F_4c'.$$ +$$(y')' = F_2(c')'.$$ +$$((y')')' = F_1((c')')' = 1.$$ +$$((y')')'' = F_1((c')')'' = 1.$$ +$$(y')' = \bmatrix{((y')')' + w_2^0((y')')''\cr ((y')')' - w_2^0((y')')''} = \bmatrix{2\cr0}.$$ +$$((y')'')' = F_1((c')'')' = 1.$$ +$$((y')'')'' = F_1((c')'')'' = 1.$$ +$$(y')'' = \bmatrix{2\cr 0}.$$ +$$y' = \bmatrix{4\cr0\cr0\cr0}$$ +$$y'' = F_4c'' = \underline 0.$$ +$$y = \bmatrix{4\cr0\cr0\cr0\cr4\cr0\cr0\cr0}.$$ + +The same computation with $(0,1,0,1,0,1,0,1)$ has $y'$ and $y''$ switch +values, giving +$$y = \bmatrix{4\cr0\cr0\cr0\cr-4\cr0\cr0\cr0}.$$ + +\bye -- cgit