path: root/li/hw3.tex

\def\bmatrix#1{\left[\matrix{#1}\right]}

    {\noindent\bf Section 2.2}

{\noindent\bf 12.}

{\it (a)}

This is correct. It's equal to the number of linearly independent
rows/dimension of the row space/rank because if any of the non-zero rows
were linearly dependent, they would have been eliminated to a zero row
when forming $R.$

{\it (b)}

This is false. A zero matrix has rank zero but can have a nonzero number
in this property if it has more columns than rows.

{\it (c)}

This is true. All columns are either pivot columns or free columns, and
the rank is the number of pivot columns.

{\it (d)}

No. The following matrix has four ones but rank one:
$$\bmatrix{1&1&1&1}$$

{\noindent\bf 26.}

The maximum rank of a matrix is the smaller of its number of rows and
its number of columns because the pivot columns and rows are strictly
less than the total number of each. Therefore, $C$ and $A$ have at most
rank 2, and $CA$ also has at most rank 2 (column space of $A$ is a
superset of the column space of $CA,$ which becomes obvious if they're
treated like functions). $CA$ is a $3\times 3$ matrix, and $I_3$ has
rank 3, so $CA \neq I.$

$AC = I$ if
$$A = \bmatrix{1 & 0 & 0\cr
               0 & 1 & 0}$$
               and
$$C = \bmatrix{1 & 0\cr
               0 & 1\cr
               0 & 0}$$

{\noindent\bf 42.}

If $Ax = b$ has infinitely many solutions, then there exists infinitely
many solutions to $Ay = 0$ if $y = x - x_0$ where $x_0$ is a particular
solution to $Ax_0 = b.$ If there exists one particular solution $x_1$ to
$Ax_1 = B,$ then there must be an infinite number $A(x_1+y) = B$ where
$y$ is in the null space of $A$ as noted earlier.

However, $Ax = B$ could have zero solutions. The matrix
$$A = \bmatrix{1&0\cr 0&0}$$
does not include $b_0 = \bmatrix{0\cr 1}$ in its column space, so $Ax =
b_0$ would have zero solutions even though $Ax = \bmatrix{1\cr 0}$ has
an infinite number of solutions.

\iffalse % practice problems

{\noindent\bf 7.}

$$R_3 = R_2 + R_1 \to c = 5 + 2.$$

{\noindent\bf 9.}

{\it (a)}

$$\bmatrix{1&2&3&4\cr 0&0&1&2\cr 0&0&0&0}\bmatrix{x_1\cr x_2\cr x_3\cr
x_4} = \bmatrix{0\cr 0\cr 0} \to x = \bmatrix{-4\cr 0\cr -2\cr 1}x_4 +
\bmatrix{-2\cr 1\cr 0\cr 0}x_2$$
$$R = \bmatrix{1&2&0&-2\cr 0&0&0&1&2\cr 0&0&0&0}.$$
$$Rx = 0 \to x = \bmatrix{2&0&-2&1}x_4 + \bmatrix{-2&1&0&0}x_2.$$

{\it (b)}

If the right-hand side is $(a, b, 0),$ the solution set will be the null
space plus a particular solution. In the case of $U,$ a particular
solution would be $(a, 0, b, 0).$

{\noindent\bf 10.}

$$\bmatrix{0&1&-1\cr 1&0&-1}x = \bmatrix{1\cr -2\cr 0}.$$
$$\bmatrix{0&1&-1\cr 1&0&-1\cr 1&1&-2}x = \bmatrix{1\cr -2\cr 0}.$$

{\noindent\bf 14.}

$$R_A = \bmatrix{1&2&0\cr 0&0&1\cr 0&0&0}.$$
$$R_B = \bmatrix{1&2&0&1&2&0\cr 0&0&1&0&0&1\cr 0&0&0&0&0&0}.$$
$$R_C = \bmatrix{1&2&0&0&0&0\cr 0&0&1&0&0&0\cr 0&0&0&0&0&0\cr
0&0&0&1&2&0\cr 0&0&0&0&0&1\cr 0&0&0&0&0&0}.$$

{\noindent\bf 21.}

The rank $r$ is the number of pivot rows and the number of pivot
columns, so the subset of these rows and columns would be an $r\times r$
matrix. They are by definition linearly independent, so each spans/forms
a basis for $R^r,$ giving them invertibility.

{\noindent\bf 24.}

The rank of $A$ is the same as the rank of $A^T,$ so
$${\rm rank}(AB) \leq {\rm rank}(A) \to {\rm rank}((AB)^T) \leq
{\rm rank}(A^T) \to {\rm rank}(B^TA^T) \leq {\rm rank}(A^T) \to
{\rm rank}(AB) \leq {\rm rank}(B).$$

{\noindent\bf 25.}



{\noindent\bf 36.}

{\it (a)}

All vectors in $R^3$ are in the column space, so only the trivial
combination of the rows of $A$ gives zero.

{\it (b)}

Only vectors where $b_3 = 2b_2$ are within the column space. This means
that $2x_2 = -x_3$ gives a zero combination. % double check this.

{\noindent\bf 40.}

$x_5$ is a free variable, the zero vector isn't the only solution to
$Ax=0,$ and if $Ax=b$ has a solution, then it has infinite solutions.

{\noindent\bf 43.}

{\it (a)}

$q=6$ gives a rank of 1 for $B,$ and $q=3$ gives a rank of 1 for the
frist matrix.

{\it (b)}

$q = 7$ gives a rank of 2 for both matrices.

{\it (c)}

A rank of 3 is impossible for both matrices.

{\noindent\bf 45.}
% idk come back to this.

{\it (a)}

$r < n.$

{\it (b)}

$r > m.$ $r\geq n.$ % ???

{\it (c)}

$r < n.$

{\it (d)}

{\noindent\bf 53.}

{\it (a)}

False. The zero matrix has $n$ free variables.

{\it (b)}

True. If the linear function corresponding to the matrix can be
inverted, it must not have a non-zero null-space (i.e. a non-injective
relation).

{\noindent\bf 60.}

$$\bmatrix{1&0&-2&-3\cr0&1&-2&-1}$$ has this nullspace.

{\noindent\bf 61.}

% simple enough to construct

{\noindent\bf 62.}



{\noindent\bf 63.}
{\noindent\bf 64.}

{\noindent\bf 65.}

$$\bmatrix{0&0\cr 1&0}$$

{\noindent\bf 66.}

Dimension of null space is $n-r = 3-r,$ and dimension of column space is
$r,$ so they cannot have the same dimension and therefore cannot be
equal.

\fi

    {\noindent\bf Section 2.3}

{\noindent\bf 22.}

{\it (a)}

They might not span ${\bf R}^4$ if, for example, they are all the zero
vector, but they could span it, like if the first four were elementary
vectors $e_1$ to $e_4.$

{\it (b)}

They are not linearly independent because 4 is the maximal independent
set.

{\it (c)}

Any four might be a basis for ${\bf R}^4,$ because they could be
linearly independent and four vectors in ${\bf R}^4$ could span it.

{\it (d)}

$Ax = b$ might not have a solution. It could have a solution depending
on the $b,$ but $0x = e_1,$ where $0$ refers to the zero vector for $A$
has zero solutions.

{\noindent\bf 27.}

The column space of $A$ has basis in $\{(1, 0, 1)^T, (3, 1, 3)^T\}$ and
the column space of $U$ has basis in $\{(1, 0, 0)^T, (3, 1, 0)^T\}.$
The two matrices have the same row space, based in $\{(1, 3, 2), (0, 1,
1)\}.$
They also have the same null space, based in $\{(-1, 1, -1)\}.$

{\noindent\bf 32.}

{\it (a)}

The dimension is 3 because this is the set of vectors on ${\bf R}^4$
under one linear constraint: $v_4 = -(v_3 + v_2 + v_1).$

{\it (b)}

The dimension is 0 because the identity matrix, by definition only
returns 0 if given 0.

{\it (c)}

The dimension is 16 because there are 16 unconstrained components.

{\noindent\bf 36.}

6 independent vectors satisfy $Ax=0$ by the rank theorem. $A^T$ has the
same rank, so 53 independent vectors satisfy $A^Ty = 0.$

{\noindent\bf 42.}

$\{x^3, x^2, x, 1\}$ form a basis of the polynomials of degree up to 3,
and this set restricted to those where $p(1) = 0$ has basis $\{x^3-1,
x^2-1, x-1\}.$

\iffalse % practice problems

{\noindent\bf 7.}

$$v_1 - v_2 + v_3 = w_2 - w_3 - w_1 + w_3 + w_1 - w_2 = 0,$$
proving dependence of these vectors.

{\noindent\bf 8.} % not an actual problem

$$c_1v_1 + c_2v_2 + c_3v_3 = c_1(w_2 + w_3) + c_2(w_1 + w_3) + c_3(w_1 +
w_2) = (c_2+c_3)w_1 + (c_1+c_3)w_2 + (c_1+c_2)w_3 = 0.$$
Because the set of $w$ vectors are independent, this sum is only equal
to zero if $c_2 + c_3 = 0 \to c_3 = -c_2,$ $c_1 + c_3 = 0 \to c_1 = -c_3
= +c_2,$ and $c_1+c_2 = 0 \to c_2 = c_1 = 0 \to c_3 = 0.$

{\noindent\bf 9.}

{\it (a)}

If $v_1$ to $v_3$ are linearly independent, the dimension of their
spanning set must be 3 (and the set equal to $R^3$), so $v_4 \in R^3$
can be written as a combination of the other three.

{\it (b)}

$v_2 = kv_1$ where $k\in\bf R$

{\it (c)}

$0v_1 + k(0,0,0) = 0,$ giving a non-trivial combination with the value
0.

{\noindent\bf 12.}

The vector $b$ is in the subspace spanned by the columns of $A$ when
there is a solution to $Ax = b.$ The vector $c$ is in the row space of
$A$ when there is a solution to $A^Tx = c$ or $x^TA = c.$

The zero vector is in every space, so the rows may still be independent.
(False)

{\noindent\bf 13.}

The dimensions of the column spaces and of the row spaces of $A$ and $U$
are the same (2), and the row spaces are the same between the two (and
conversely, the null space)

{\noindent\bf 21.}

% easy

{\noindent\bf 23.}

If they are linearly independent, the rank of $A$ is $n.$ If they span
$R^m,$ the rank is $m.$ If they are a basis for $R^m,$ then both are
true and $n = m.$

{\noindent\bf 25.}

{\it (a)}

The columns are linearly independent, so there is no nontrivial linear
combination equal to 0.

{\it (b)}

The columns of $A$ span $R^5,$ so there must be a linear combination
(value of $x$) equal to $b.$

{\noindent\bf 26.}

{\it (a)}

True. Thm in the book.

{\it (b)}

False. See 31.

{\noindent\bf 31.}

If we let $v_k = e_k,$ the subspace with basis $(0, 0, 1, 1)$ does not
have a basis in the elementary vectors.

{\noindent\bf 34.}

% seems simple enough, don't know if I can do it.

{\noindent\bf 35.}

{\it (a)}

False. The unit vector $e_1$'s single column is linearly independent,
but except in $R,$ it doesn't span $R^k,$ and $e_1x = e_2$ has no
solution.

{\it (b)}

True. The rank is at most $5,$ meaning there must be two free variables.

{\noindent\bf 41.}

{\it (a)}

For dimension 1, $y_k = kx.$

{\it (b)}

For dimension 2, $y_1 = x^2,$ $y_2 = 2x,$ and $y_3 = 3x.$

{\it (c)}

For dimension 3, $y_k = x^k.$

\fi

\bye