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diff --git a/li/hw2.tex b/li/hw2.tex new file mode 100644 index 0000000..769cad3 --- /dev/null +++ b/li/hw2.tex @@ -0,0 +1,265 @@ +\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 |