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\title{MATH 54\\Linear Algebra and Diff. Eqns.\\
Section 1 (Lenstra), assignment \# 5}
\author{William Stein}

\begin{document}
\maketitle

\begin{center}
{\bf {\em Anton}: V.1, \# 6,12; V.2, \# 2bc, 4ace, 7abcd}
\end{center}

\prob{V.1.6}
{\em Is the set $W=\{(x,y):x\geq 0, x,y\in\R\}$ with the standard
operations a vector space?}

No, axiom 5 fails since $-(1,0)=(-1,0)\not\in W$. Axiom 6
fails since $-2\cdot(2,3)=(-4,-6)\not\in W$.

\prob{V.1.12}
{\em Is the set of all $2\times 2$ matrices of the form
$\bigl(\begin{smallmatrix}a&a+b\\a+b&b\end{smallmatrix}\bigr)$ with
matrix addition and scalar multiplication a vector space?}

Yes, just check all of the axioms.

\prob{V.2.2}

{\em (b) Is the set of matrices
$\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)$ with
$a+b+c+d=0$ a subspace of $M_{22}$?}

Yes. Apply theorem 5.2.1. First check that the sum of
two such matrices is again such a matrix. If the sum of
the entries of $A$ is zero and the sum of the entries
of $B$ is zero then the sum of the entries of $A+B$ is zero.
Next check that if the sum of the entries of $A$ is zero
then the sum of the entries of $kA$ is zero. Do this by
assuming $A=\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)$
and $a+b+c+d=0$. Then $kA=\bigl(\begin{smallmatrix}ka&kb\\kc&kd\end{smallmatrix}\bigr)$
so the sum of the entries of $kA$ is
$$ka+kb+kc+kd=k(a+b+c+d)=k0=0.$$

{\em (c) Is the set of all $2\times 2$ matrix $A$ such that
$\det(A)=0$ a subspace?}

No. Although property (b) of theorem 5.2.1 is satisfied, property
(a) is not. For example, if
$A=\bigl(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\bigr)$
and
$B=\bigl(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\bigr)$
then $\det(A)=\det(B)=0$ but $\det(A+B)=1$.

\prob{V.2.4}
{\em (a) Is the set of $f$ such that $f(x)\leq 0$ for all $x$
a subspace of $F(-\infty,\infty)$?}

No, since property (b) of theorem 5.2.1 fails. For example,
if $f(x)$ is the function $f(x)=-3$ and $k=-7$ then $k\cdot{}f(x)=21$
so $kf(x)$ does not lie in the set under consideration.

{\em (c) Is the set of $f$ such that $f(0)=2$ a subspace of
$F(-\infty,\infty)$?}

No, since property (a) of theorem 5.2.1 fails (in fact, property
(b) fails as well). We can show that property (a) fails by
exhibiting one example. Let $f$ be the function $f(x)=2$ and
let $g$ be the function $g(x)=x+2$. Then $f$ and $g$ lie
in the subset under consideration, but $(f+g)(0)=f(0)+g(0)=2+2=4$
so $f+g$ does not lie in the subset.

{\em (e) Is the set
$$W=\{k_1+k_2\sin x:k_1,k_2\in\R\}$$
a subspace of $F(-\infty,\infty)$?}

Yes. To see this check properties (a) and (b) in theorem 5.2.1.
Let $k_1+k_2\sin x$ and $k_3+k_4\sin x$ be elements of $W$.
Then $$(k_1+k_2\sin x) +(k_3+k_4\sin x)=(k_1+k_3)+(k_2+k_4)\sin x\in W.$$
If $a\in\R$ and $k_1+k_2\sin x\in W$ then
$$a(k_1+k_2\sin x)=ak_1+ak_2\sin x\in W.$$

\prob{V.2.7}
{\em (a) Write $x=(2,2,2)$ as a linear combination of $u=(0,-2,2)$ and
$v=(1,3,-1)$. }

We just need to find $k_1$ and $k_2$ so that
$$x=k_1u+k_2v.$$
This amounts to solving a system of 3 linear equations in
the two unknown $k_1$ and $k_2$. To see this expand both
sides and simplify to get
$$(2,2,2)=(k_2,-2k_1+3k_2,2k_1-k_2).$$
Then equating corresponding entries gives the system
\begin{align*}
2&=k_2\\
2&=-2k_1+3k_2\\
2&=2k_1-k_2
\end{align*}
Solving gives $k_1=2$ and $k_2=2$. Check that this works.
Note that if you are working a similiar problem and the system
turns out to have no solution, then this means that there is
no way to write $x$ as a linear combination of $u$ and $v$ (or

{\em (b) Write $x=(3,1,5)$ as a linear combination of $u=(0,-2,2)$ and
$v=(1,3,-1)$.}

Do exactly the same thing as you did for (a) above. You will
get $k_1=4$ and $k_2=3$.

{\em (d) Show that $x=(0,4,5)$ can not be written
as a linear combination of $u=(0,-2,2)$ and $v=(1,3,-1)$.}

As in part (a) we obtain a system of equations by equating corresponding
entries in the equation
$$(0,4,5)=(k_2,-2k_1+3k_2,2k_1-k_2).$$
The system is
\begin{align*}
0&=k_2\\
4&=-2k_1+3k_2\\
5&=2k_1-k_2
\end{align*}
We must show that this system has no solutions. If it did, then
we must have $k_2=0$ from the first equation. Then the third equation implies
$k_1=5/2$. But this contradicts the second equation which implies
$k_1=-2$.

\end{document}