notes and homework

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+\def\bmatrix#1{\left[\matrix{#1}\right]}
+{\bf\noindent Section 1.6}
+{\bf\noindent 4.}
+{\it (a)}
+That means $A^{-1}$ exists s.t. $A^{-1}A = I.$
+$$AB = AC \to A^{-1}AB = A^{-1}AC \to B = C$$
+
+{\it (b)}
+$$B = \pmatrix{1 & 1\cr 1 & 0}$$
+$$C = \pmatrix{1 & 2\cr 1 & 1}$$
+
+$$AB = AC.$$ % double check
+
+{\bf\noindent 10.}
+
+$$A_1^{-1} = \bmatrix{0 & 0 & 0 & 1\cr 0 & 0 & 1/2 & 0\cr 0 & 1/3 & 0 &
+0\cr 1/4 & 0 & 0 & 0}$$
+
+$$A_2^{-1} = \bmatrix{1 & 0 & 0 & 0\cr 1/2 & 1 & 0 & 0\cr 0 & 2/3 & 1 &
+0\cr 0 & 0 & 3/4 & 1}$$
+
+\def\rowone{{d\over ad-bc}&{-b\over ad-bc}}
+\def\rowtwo{{-c\over ad-bc}&{a\over ad-bc}}
+$$A_3^{-1} = \bmatrix{\rowone & 0 & 0\cr \rowtwo & 0 & 0\cr 0 & 0 &
+\rowone \cr 0 & 0 & \rowtwo}$$
+
+{\bf\noindent 11.}
+
+If $A = I$ and $B = -I,$ both are invertible, but $A + B$ is not.
+
+If $A = \pmatrix{1&0\cr 0&0}$ and $B = \pmatrix{0&0\cr 0&1},$ $A + B =
+I$ is invertible, but neither $A$ nor $B$ are.
+
+$A = B = I$ gives $A + B,$ $A,$ and $B$ invertibility.
+$A^{-1} = B^{-1} = I \to A^{-1}(A+B)B^{-1} = B^{-1} + A^{-1} \to 2I =
+2I.$
+
+{\bf\noindent 12.}
+
+{\it (a)}
+
+This one remains.
+
+{\it (b)}
+
+This one remains.
+
+{\it (c)}
+
+This one remains.
+
+{\it (d)}
+
+This one doesn't remain.
+
+{\it (e)}
+
+This one remains.
+
+% I don't really have any coherent reasons for this.
+% But these all make a lot of sense to me.
+% This isn't graded, but I would like to understand it better. Office
+% hours?
+
+{\bf\noindent 23.}
+
+$$\bmatrix{10&20\cr 20&50}\bmatrix{x\cr y} = \bmatrix{1\cr 0} \to
+\bmatrix{10&20\cr 0&10}\bmatrix{x \cr y} = \bmatrix{1\cr -2}$$$$\to
+\bmatrix{1&2\cr 0&1}\bmatrix{x \cr y} = \bmatrix{1/10\cr -2/10} \to
+\bmatrix{1&0\cr 0&1}\bmatrix{x \cr y} = \bmatrix{5/10\cr -2/10}.
+$$
+
+$$\bmatrix{10&20\cr 20&50}\bmatrix{x\cr y} = \bmatrix{0\cr 1} \to
+\bmatrix{1&2\cr 0&1}\bmatrix{x\cr y} = \bmatrix{0\cr 1/10} \to
+\bmatrix{1&0\cr 0&1}\bmatrix{x\cr y} = \bmatrix{-2/10\cr 1/10}.
+$$
+
+$$A^{-1} = {1\over10}\bmatrix{5&-2\cr -2&1}$$
+
+{\bf\noindent 25.}
+
+{\it (a)}
+$$\bmatrix{\hbox{row 1}\cr\hbox{row 2}\cr\hbox{row 1 + row
+2}}\bmatrix{x\cr y\cr z} = \bmatrix{1\cr0\cr0}
+\to \bmatrix{\hbox{row 1}\cr\hbox{row 2}\cr0}\bmatrix{x\cr y\cr z} =
+\bmatrix{1\cr0\cr-1}.$$
+$0 \neq -1,$ so this is inconsistent.
+
+{\it (b)}
+
+If $b_3 = b_2 + b_1,$ this matrix admits a solution.
+
+{\it (c)}
+
+As already shown, row 3 becomes a zero row during elimination.
+
+{\bf\noindent 27.}
+
+$B = E_12A,$
+and the elementary matrix which switches the first two rows is by
+definition invertible ($E_12E_12 = I$) so if $A$ is invertible, $B$ is
+also invertible, and $B^{-1} = A^{-1}E_12$ (I think this is a switch of
+columns).
+
+{\bf\noindent 36.}
+
+$$\bmatrix{2 & 1 & 0 & 1 & 0 & 0\cr
+           1 & 2 & 1 & 0 & 1 & 0\cr
+           0 & 1 & 2 & 0 & 0 & 1}
+\to
+\bmatrix{2 & 1 & 0 & 1 & 0 & 0\cr
+           0 & 1.5 & 1 & .5 & 1 & 0\cr
+           0 & 1 & 2 & 0 & 0 & 1}
+\to
+\bmatrix{2 & 1 & 0 & 1 & 0 & 0\cr
+           0 & 1 & 0 & .5 & 1 & .5\cr
+           0 & 1 & 2 & 0 & 0 & 1}
+$$$$\to
+\bmatrix{2 & 0 & 0 & .5 & -1 & -.5\cr
+           0 & 1 & 0 & .5 & 1 & .5\cr
+           0 & 1 & 2 & 0 & 0 & 1}
+\to
+\bmatrix{2 & 0 & 0 & .5 & -1 & -.5\cr
+           0 & 1 & 0 & .5 & 1 & .5\cr
+           0 & 0 & 2 & -.5 & -1 & .5}
+$$
+
+Inverse is
+$$
+\bmatrix{.25 & -.5 & -.25\cr
+         .5 & 1 & .5\cr
+         -.25 & -.5 & .25}
+$$
+
+% {\bf\noindent 40.}
+
+{\bf\noindent Section 2.1}
+{\bf\noindent 2.}
+
+{\it (a)}
+
+This is. $k(0, b_2, b_3) = (0, kb_2, kb_3),$ and similar for addition.
+
+{\it (b)}
+
+This isn't. $2(0, 1, 0) = (0, 2, 0),$ which isn't in the subset.
+
+{\it (c)}
+
+This isn't. $(1, 1, 0) + (1, 0, 1) = (2, 1, 1),$ which isn't in the
+subset, even though its two starting components are.
+
+{\it (d)}
+
+This is, by definition.
+
+{\it (e)}
+
+Yes, this is the null space of a homogenous matrix with one row. It is a
+subspace.
+
+{\bf\noindent 7.}
+
+{\it (a)}
+
+This is not a subspace. The given sequence plus the given sequence with
+a zero prepended would sum to a sequence without a zero, meaning it is
+not closed under addition.
+
+{\it (b)}
+
+This is a subspace because the sums and multiples are closed.
+
+{\it (c)}
+
+$x_{j+1} \leq x_j \land y_{j+1} \leq y_j \to x_{j+1} + y_{j+1} \leq x_j
++ y_j,$
+and similar for multiplication.
+
+{\it (d)}
+
+Yes, the sum of two convergent sequences will tend to the sum of their
+limits, and each can be multiplied by a real.
+
+{\it (e)}
+
+Yes, the sum of two arithmetic series will be an arithmetic series, and
+they work with multiplications by a real.
+
+{\it (f)}
+
+This isn't because $x_1 = 1,$ $k = 2,$ and $x_1 = 1,$ $k = 3$ sum to
+$$2 + 5 + 14 + \cdots,$$ which is not a geometric sequence.
+
+{\bf\noindent 8.} % required
+
+The two rows are linearly independent, so with 3 unknows, this forms a
+line (intersection of two equations/planes).
+
+{\bf\noindent 18.}
+
+{\it (a)}
+
+It's probably a line, but it could be a plane.
+
+{\it (b)}
+
+It's probably $(0, 0, 0)$, but it could be a line.
+
+% {\it (c)}
+
+% I don't know.
+
+{\bf\noindent 22.} % required
+
+{\it (a)}
+
+This matrix has a column space with basis $(1, 2, -1)^T,$ so it is only
+solvable if $b = (b_1, b_2, b_3)^T$ is a multiple of that vector.
+
+{\it (b)}
+
+These two columns are linearly independent, so any b within the span of
+them has a solution.
+
+{\bf\noindent 25.}
+
+Unless $b$ is in the span of $A.$ If A is the zero $3\times 1$ matrix,
+and $b$ is the first column of $I_3,$ then the column space increases.
+If $A = b,$ then it would not extend the column space. Iff $b$ is
+already in the column space, then $Ax = b$ has a solution.
+
+{\bf\noindent 26.}
+
+They are not equal only if $B$ is not invertible (it is a necessary but
+not sufficient condition). If $A$ is invertible, like $A = I,$ then
+non-invertible $B$ gives $AB = B$ with smaller column space.
+
+{\bf\noindent 28.} % required
+
+{\it (a)}
+
+False. Counterexample: The complement of $(0, 0, 0)$ does not pass the
+test $0v \in V.$
+
+{\it (b)}
+
+True. The space with only the zero vector has a basis of only zero
+vectors.
+
+{\it (c)}
+
+True. Each of $2A$'s columns can be transformed into the columns of $A$
+by division by 2.
+
+{\it (d)}
+
+False. The column space of $-I$ is the full space, unlike the zero
+matrix it's ``based on.''
+
+{\bf\noindent 30.} % required
+
+$${\bf C}(A) = {\bf R}^9.$$
+
+\bye