wrote linear homeworks

2 files changed, 167 insertions, 0 deletions

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