\magnification=\magstep1
\def\Sol{\bigskip\item{}{\it Solution\/:}  }
\nopagenumbers
\centerline{Mathematics 244, section 1 -- Linear Algebra}
\centerline{Solutions for Exam 2 -- March 23, 2007}
\bigskip
\noindent
I. (15) Let $V$ be a vector space and let $S$ be a finite
subset of $V$ with ${\rm Span}(S) = V$.  If $T$ is a linearly
independent subset of $V$, show that $|T| \le |S|$.  (Here
$|S|$ denotes the number of elements in $S$, and similarly for
$|T|$.) 
\Sol  Say $S = \{v_1,\ldots,v_m\}$ and $T = \{w_1,\ldots,w_n\}$ in $V$, 
where ${\rm Span}(S) = V$.
We will show that if $n > m$, then $T$ must be linearly dependent.
Since ${\rm Span}(S) = V$, for each $j$, $1 \le j \le n$, there
exist scalars $a_{ij}$ such that 
$$w_j = a_{1j} v_1 + \cdots + a_{mj} v_m.\leqno(1)$$
Suppose we have scalars $c_1, \ldots, c_n$ such that 
$$0 = c_1 w_1 + c_2 w_2 + \cdots + c_n w_n.\leqno(2)$$
Then substituting from the equations (1), we have 
$$0 = c_1 (a_{11} v_1 + \cdots + a_{m1}v_m) + c_2(a_{12} v_1 + \cdots + a_{m2}v_m) 
+ \cdots + c_n(a_{1n} v_1 + \cdots + a_{mn}v_m).$$
Rearranging this equation gives:
$$0 = (a_{11}c_1 + \cdots + a_{1n}c_n)v_1 + \cdots + (a_{m1}c_1 + \cdots +  a_{mn}c_n)v_n.$$
This will be satisfied if
$$\eqalign{a_{11}c_1 + \cdots + a_{1n}c_n &= 0\cr
                      \vdots\qquad\qquad &\ \, \vdots \cr
           a_{m1}c_1 + \cdots + a_{mn}c_n &= 0\cr}$$
But this is a homogeneous 
system of $m$ linear equations in the variables $c_1, \ldots, c_n$.
Since we assume that $n > m$, there are free variables, hence nontrivial
solutions.  This implies that $T$ is {\it linearly dependent} if $n > m$
from (2).  Hence if $T$ is linearly independent, then $n = |T| \le |S| = m$.  
\bigskip
\noindent
II.  All parts of this question refer to the matrix
$$A = \pmatrix{1 & -2 & 0 &  s  & 2\cr
               -2 & -1 & -1 & 2s & -1\cr
                1 & 3 & 1 & s^2 & -1\cr}$$
\item{A)} (15) For which value(s) of the scalar $s$ does $A$ satisfy
$\dim {\rm Nul}(A) = 2$?  
\Sol
We begin by applying row operations $R_2 \mapsto R_2+2R_1$, 
$R_3 \mapsto R_3 - R_1$, and then $R_3 \mapsto R_3 + R_2$ 
the matrix $A$.
$$\eqalign{
 \pmatrix{1 & -2 & 0 &  s  & 2\cr
         -2 & -1 & -1 & 2s & -1\cr
          1 & 3 & 1 & s^2 & -1\cr} &\rightarrow 
 \pmatrix{1 & -2 & 0 &  s  & 2\cr
          0 & -5 & -1 & 4s & 3\cr
          0 &  5 & 1 & s^2 - s & -3\cr}\cr
&\rightarrow
\pmatrix{1 & -2 & 0 &  s  & 2\cr
          0 & -5 & -1 & 4s & 3\cr
          0 &  0 & 0 & s^2 + 3s & 0\cr}\cr}$$
From this, we can see that there are always pivots in columns 1 and 2.
Moreover, if $s^2 + 3s \ne 0$, then there is a pivot in column 4 as
well.  To get $\dim {\rm Nul}(A) = 2$ (not 3), we need three pivot 
columns (2 free variables), so the condition is $s^2 + 3s \ne 0$.
$\dim {\rm Nul}(A) = 2$ for all real $s$ other than $s = 0, s =-3$.  
\bigskip
\item{B)} (10) Say $s = 1$.  Give bases of ${\rm Col}(A)$ and 
${\rm Nul}(A)$.
\Sol
We substitute $s = 1$ in the last matrix above and continue
the reduction to row-reduced echelon form.  The result is:
$$\pmatrix{1 &  0 & 2/5 &  0  & 4/5\cr
           0 &  1 & 1/5 & 0 & -3/5\cr
           0 &  0 & 0 & 1 & 0\cr}$$
Thus, a basis for ${\rm Col}(A)$ is given by columns 1,2,4 of 
the original matrix $A$ (taking $s = 1$ of course):
$$\left\{\pmatrix{1 \cr -2\cr 1}, \pmatrix{-2\cr -1\cr 3\cr}, \pmatrix{1\cr 2\cr 1\cr}\right\}.$$
A basis for ${\rm Nul}(A)$ comes from parametrizing the solutions of $Ax = 0$ 
with the free variables $x_3,x_5$ as usual:
$$\left\{\pmatrix{-2/5 \cr -1/5 \cr 1 \cr 0 \cr 0\cr}, \pmatrix{-4/5\cr 3/5 \cr 0 \cr 0 \cr 1}\right\}$$ 
\bigskip
\noindent
III.  (10)  Use Cramer's Rule to solve the system
$$\eqalign{3x_1 + 2 x_2 &= 9\cr
          -8x_1 + 3 x_2 &= -12\cr}$$
\Sol
We have 
$$x_1 = { \det\pmatrix{9 & 2\cr -12 & 3\cr} \over \det \pmatrix{3 & 2\cr -8 & 3\cr}} = 51/25$$
and 
$$x_2 = { \det\pmatrix{3 & 9\cr -8 & -12\cr} \over \det \pmatrix{3 & 2\cr -8 & 3\cr}} = 
36/25$$
\bigskip\noindent
IV.  Let $V = M_{2\times 2}({\bf R})$, the vector space of all 
$2\times 2$ matrices with real entries.  Let $W = {\bf R}^2$.
\bigskip
\item{A)} (10) Is $T: V \to W$ defined by $T(A) = \pmatrix{\det(A) \cr \det(A^2)\cr}$
a linear mapping?  Why or why not?
\Sol
No, $T$ is not a linear mapping.  The easiest way to see this is to note
that if we multiply the matrix $A$ by a scalar $c$, then
$$T(cA) = \pmatrix{\det(cA) \cr \det((cA)^2)\cr} = \pmatrix{c^2 \det(A) \cr c^4 \det(A^2)}.$$
If $c \ne 1$, then this is not the same as $cT(A)$.  For instance if $A = I$
and $c = 2$, we get 
$$T(2I) = \pmatrix{4\cr 16\cr} \ne 2 T(I) = \pmatrix{2\cr 2\cr}.$$
\item{B)} (10) Show that
$$H = \left\{A = \pmatrix{a & b\cr c & d\cr} \in V : a + b = c + d\right\}$$
is a vector subspace of $V$.  
\Sol
The zero matrix $\pmatrix{0 & 0 \cr 0 & 0\cr}$ is in $H$ since $0 + 0 = 0 + 0$.
If $A = \pmatrix{a & b\cr c & d\cr}$ and $B = \pmatrix{e & f\cr g & h\cr}$
are in $H$, then $a + b = c + d= 0$ and $e + f = g + h = 0$.  Then 
$A + B = \pmatrix{a + e & b + f\cr c + g & d + h\cr}$ and 
$$(a+e) + (b + f) = (a+b) + (e+f) = (c+d) + (g+h) = (c+g) + (d+h)$$
(using commutativity and associativity of addition in ${\bf R}$).
Hence $A + B \in H$.  Finally, if $A\in H$ and $r \in {\bf R}$, then
$rA = \pmatrix{ra & rb\cr rc & rd\cr}$ and $ra + rb = r(a+b) = r(c+d) = rc + rd$.
Therefore $rA \in H$.
\bigskip
\item{C)} (5 {\bf Extra Credit})  Find a basis for $H$ from part B and 
determine its dimension.
\Sol
In the equation $a + b - c - d = 0$, $b,c,d$ are free variables.  Hence
$\dim H = 3$, and a basis consists of 
$$\pmatrix{-1 & 1 \cr 0 & 0\cr}, \pmatrix{1 & 0\cr 1 & 0\cr}, \pmatrix{1 & 0\cr 0 & 1\cr}.$$
\bigskip
\noindent
V.  (True-False)  For each true statement, give a short proof.
For each false statement, give a counterexample.  (Note:
As always, ``true'' means ``true in every case.'' Partial credit {\it will}
be given for definitions or theorems that apply even if you don't find a 
complete solution.)
\bigskip
\item{A)} (10) Let $A$ be a $4\times 4$ matrix in which 
$R_1 + 2 R_2 = R_3 + R_4$  ($R_i$ is row $i$ of the matrix). 
Then $\det(A) = 0$.  
\Sol This is {\it True}.  For a proof, we will use the fact that
``replacement operations'' $R_i \mapsto R_i + cR_j$ do not
change the value of the determinant.  Hence if we apply the row
operations $R_1 \mapsto R_1 + 2R_2$, and $R_4 \mapsto R_4 + R_3$,
we will obtain matrix with two equal rows (rows 1 and 4).  Hence
if we apply another operation such as $R_4 \mapsto R_4 - R_1$,
we obtain a row of zeroes.  Expanding the determinant along
row 4, we obtain $\det(A) = 0$.
\bigskip
\item{B)} (10) There exist linear mappings
$T : {\bf R}^2 \to {\bf R}^2$ with standard matrices $A$ satisfying
${\rm Nul}(A) = {\rm Col}(A)$.
\Sol  This is {\it True}.  Since the statement says ``there exists
such a matrix,'' it suffices to produce one.  From the 
equation $\dim {\rm Nul}(A) + \dim {\rm Col}(A) = 2$, we can see
that to get equality, we want both terms $ = 1$.  This means that
we want a $2\times 2$ matrix $A$ whose columns are scalar multiples,
and such that $Ax = 0$ for $x = $ both of its columns.
The condition $\dim {\rm Col}(A) = 1$ means that 
$$A = \pmatrix{a & ca\cr
               b & cb\cr}$$
for some $a,b,c$. Then 
$$  \pmatrix{a & ca\cr
               b & cb\cr} \pmatrix{a \cr b\cr} = \pmatrix{0 \cr 0 \cr}$$
implies that $a^2 + cab = 0$ and $ab + cb^2 = 0$. 
There are infinitely many different solutions of these equations.
If we take $c = -1$, for instance, then $a = 1, b = 1$ gives one:
$$A = \pmatrix{1 & -1\cr 1 & -1\cr}$$ 
is such a matrix, with
$${\rm Col}(A) = {\rm Span}\left\{\pmatrix{1 \cr 1\cr}\right\} = {\rm Nul}(A).$$
(There are infinitely many others too.) 
\bigskip
\item{C)} (10) Let $A,B,C$ be $n \times n$ matrices.  If $\det(A^t B C^5) = 23$,
then $A,B,C$ are {\it all} invertible matrices.  
\Sol
This is {\it True}.  From properties of determinants we know
$$23 = \det(A^t B C^5) = \det(A)\det(B)(\det(C))^5.$$
Since the product is $23 \ne 0$, $\det(A), \det(B), \det(C)$
must all be $\ne 0$.  This means that $A,B,C$ are all invertible. 
\bye
