From 6d18ebaa31ed960b051826a2c44c1b770324da2a Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Tue, 19 Oct 2021 21:42:14 -0400
Subject: wrote linear homeworks

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+\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
-- 
cgit