From f1733a2433a4780322a7d74ce9cbe36deb9375c7 Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Wed, 6 Oct 2021 15:18:51 -0400
Subject: notes and homeworks for math

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 li/hw4.tex | 140 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
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+\def\bmatrix#1{\left[\matrix{#1}\right]}
+
+    {\noindent\bf Section 2.4}
+
+{\noindent\bf 6.}
+
+{\it (a)}
+
+It has a two-sided inverse if $r = m = n.$
+
+{\it (b)}
+
+It has infinitely many solutions if $r = m < n.$
+
+{\noindent\bf 12.}
+
+{\it (a)}
+
+Matrix $A$ is of rank 1 and equal to
+$$\bmatrix{1\cr 0\cr 2}\bmatrix{1&0&0&3}$$
+
+{\it (b)}
+
+Matrix $A$ is of rank 1 and equal to
+$$\bmatrix{2\cr 6}\bmatrix{1&-1}$$
+
+{\noindent\bf 18.}
+
+The row space has basis:
+$$\{\bmatrix{0\cr1\cr2\cr3\cr4}, \bmatrix{0\cr0\cr0\cr1\cr2}\},$$
+the null space has basis:
+$$\{\bmatrix{1\cr0\cr0\cr0\cr0}, \bmatrix{0\cr-2\cr1\cr0\cr0},
+\bmatrix{0\cr0\cr0\cr-2\cr0}\},$$
+the column space has basis:
+$$\{\bmatrix{1\cr1\cr0},\bmatrix{3\cr4\cr1}\},$$
+and the left null space has basis:
+$$\{\bmatrix{1\cr-1\cr1}\}.$$
+
+{\noindent\bf 32.}
+
+$A$ has column space of the xy-plane, and left null space of the z-axis.
+It also has row space of the yz-plane, and null space of the x-axis.
+
+$I+A$ has full rank, so its column space and row space are ${\bf R}^3,$
+and its null space and left null space are the zero vector.
+
+\iffalse % practice problems
+
+{\noindent\bf 2.}
+
+{\noindent\bf 3.}
+
+{\noindent\bf 8.}
+
+{\noindent\bf 9.}
+
+{\noindent\bf 10.}
+
+{\noindent\bf 16.}
+
+{\noindent\bf 17.}
+
+{\noindent\bf 21.}
+
+{\noindent\bf 25.}
+
+{\noindent\bf 27.}
+
+{\noindent\bf 35.}
+
+{\noindent\bf 37.}
+
+\fi
+
+    {\noindent\bf Section 2.6}
+
+{\noindent\bf 16.}
+
+$$\bmatrix{0&1&0&0\cr 0&0&1&0\cr 0&0&0&1\cr 1&0&0&0}$$
+If $A$ maps $(x_1, x_2, x_3, x_4)$ to $(x_2, x_3, x_4, x_1),$ $A^2$ maps
+$x$ to $(x_3, x_4, x_1, x_2),$ and $A^3$ takes $x$ to $(x_4, x_1, x_2,
+x_3),$ and $AA^3 = I = A^4,$ so $A^3 = A^{-1}$ by definition of the
+identity.
+
+{\noindent\bf 28.}
+
+{\it (a)}
+
+Range is $V^2,$ and kernel is $0.$
+
+{\it (b)}
+
+Range is $V^2,$ and kernel has basis $(0, 0, 1).$
+
+{\it (c)}
+
+Range is $0,$ and kernel is $V^2$
+
+{\it (d)}
+
+Range is the subspace with basis $(1, 1)$ and kernel has basis $(0, 1).$
+
+{\noindent\bf 36.}
+
+{\it (a)}
+
+$$\bmatrix{2&5\cr 1&3}$$
+
+{\it (b)}
+
+$$\bmatrix{3&-5\cr -1&2}$$
+
+{\it (c)}
+
+Because, by linearity, if $(2, 6) \mapsto (1, 0),$ $.5(2,6) = (1, 3)
+\mapsto (.5, 0).$
+
+{\noindent\bf 44.}
+
+This is equivalent to a 180$^\circ$ rotation.
+
+\iffalse % practice problems
+
+{\noindent\bf 6.}
+
+{\noindent\bf 7.}
+
+{\noindent\bf 8.}
+
+{\noindent\bf 9.}
+
+{\noindent\bf 17.}
+
+{\noindent\bf 40.}
+
+{\noindent\bf 45.}
+
+\fi
+
+\bye
-- 
cgit