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\def\bv{{\bf v}}
\def\bw{{\bf w}}
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\centerline{Mathematics 244 -- Linear Algebra}
\centerline{Solutions -- Review True-False Questions for Exam 2}
\centerline{March 24, 2004}
\bigskip
\noindent
A)  If $T: V\to W$ is linear, $\dim(V) = 5$, $\dim(W) = 7$, 
then $T$ is surjective.  
\bigskip
\item{} This is {\it FALSE}.  For instance, letting $T$ be
the zero mapping gives a counterexample since ${\rm Im}(T) = \{ {\bf 0}_W\}
\ne W$.
\bigskip
\noindent
B)  Every linear $T: V\to W$, $\dim(V) = 5$, $\dim(W) = 7$, 
is $T$ is injective.  
\bigskip
\item{} This is {\it FALSE}.  For instance, letting $T$ be
the zero mapping gives a counterexample since ${\rm Ker}(T) = V
\ne \{{\bf 0}_V\}$.
\bigskip
\noindent
C)  Some linear $T: V\to W$ is linear, $\dim(V) = 5$, $\dim(W) = 7$, 
then $T$ is injective.  
\bigskip
\item{} This is {\it TRUE}.  For instance, let 
$\beta = \{{\bf v}_1,\ldots,{\bf v}_5\}$ be a basis for $V$, 
and let $\gamma = \{{\bf w}_1,\ldots,{\bf w}_5\}$ be any linearly independent
subset of $W$.  Then the unique linear
$T: V \to W$ with $T({\bf v}_i) = {\bf w}_i$,
$i = 1,\ldots, 5$ is injective.  {\it Proof}:  We show this
by showing $T$ satisfies ${\rm Ker}(T) = \{{\bf 0}_V\}$.
Suppose $T({\bf v}) = {\bf 0}_W$.  Then since $\beta$ is a
basis for $V$, 
$${\bf v} = a_1 {\bf v}_1+ \cdots + a_5{\bf v}_5$$
for some $a_i \in {\bf R}$.  But then, since $T$ is linear,
$${\bf 0}_W = T({\bf v}) = a_1 T({\bf v}_1) + \cdots + a_5 T({\bf v}_5)$$
Since $\gamma$ is linearly independent, this implies
$a_i = 0$ for all $i$, which says ${\bf v} = {\bf 0}_V$.  Hence
${\rm Ker}(T) = \{{\bf 0}_V\}$.
\bigskip
\noindent
D)  If $\{{\bf v}_1,\ldots,{\bf v}_n\}$ in V is linearly dependent,
and $T : V \to W$ is linear then $\{T({\bf v}_1),\ldots,T({\bf v}_n)\}$
is linearly dependent.  
\bigskip
\item{} This is {\it TRUE}.  If $\{{\bf v}_1,\ldots,{\bf v}_n\}$ 
is linearly dependent, then there exist scalars $a_i$, not all 
equal to zero, such that 
$$a_1{\bf v}_1 + \cdots + a_n {\bf v}_n = {\bf 0}_V$$ 
But then $T$ linear shows
$$T(a_1{\bf v}_1 + \cdots + a_n {\bf v}_n) = 
a_1T({\bf v}_1) + \cdots + a_n T({\bf v}_n) = T({\bf 0}_V) = {\bf 0}_W$$
This implies that $\{T({\bf v}_1),\ldots,T({\bf v}_n)\}$
is linearly dependent since some $a_i \ne 0$.
\bigskip
\noindent
E)  If $T: {\bf R}^n \to {\bf R}^n$ is linear then 
${\rm Ker}(T) \cap {\rm Im}(T) = \{ {\bf 0}\}$.  
\bigskip
\item{} This is {\it FALSE}.  A counterexample is given by the mapping
$T: {\bf R}^2 \to {\bf R}^2$ with standard matrix
$$[T] = \pmatrix{0&1\cr 0&0}$$
(as in a problem from the last problem set).  For this $T$,
${\rm Ker}(T) = {\rm Im}(T) = {\rm Span}((1,0))$, 
so ${\rm Ker}(T) \cap {\rm Im}(T) = {\rm Span}((1,0))$.
\bigskip
\noindent
F)   If $T : V\to V$ is linear and $\beta = \{{\bf v}_i\}$
and $\beta' = \{c{\bf v}_i\}$ for some scalar $c \ne 0$, 
then $[T]_\beta^\beta = [T]_{\beta'}^{\beta'}$.  
\bigskip
\item{} This is {\it TRUE}.  The scalars in the $j$th column of 
$[T]_\beta^\beta$ are the $a_{ij}$ in 
$$T({\bf v}_j) = a_{1j} {\bf v}_1 + \cdots + a_{nj}{\bf v}_n$$
By linearity and one of the ``distributive laws'' for scalar multiplication:
$$\eqalign{
T(c{\bf v}_j) &= cT({\bf v}_j)\cr
              &= c(a_{1j} {\bf v}_1 + \cdots + a_{nj}{\bf v}_n)\cr
              &= a_{1j} (c{\bf v}_1) + \cdots + a_{nj}(c{\bf v}_n)\cr}$$
Hence the $j$th column of $[T]_{\beta'}^{\beta'}$ is the same as
the $j$th column of $[T]_\beta^\beta$ for all $j$, so the matrices are
equal.
\bigskip
\noindent
G)  If $A$ is an $n\times n$ matrix and $c$ is a fixed real number
then the set $W = \{ {\bf v} \in {\bf R}^n : A{\bf v} = c{\bf v}\}$
is a vector subspace of ${\bf R}^n$.
\bigskip
\item{} This is {\it TRUE}.  Let ${\bf v}, {\bf w} \in W$ and let
$d$ be an arbitrary scalar.  Then by distributivity of matrix
multiplication over matrix sums,
$$\eqalign{
A(d{\bf v} +  {\bf w}) &= d A{\bf v} +  A{\bf w}\cr
                       &= d c{\bf v} +  c{\bf w}\cr
                       &= c(d{\bf v} +  {\bf w})\cr}$$
Hence $d{\bf v} +  {\bf w}\in W$, and $W$ is a vector subspace.
\bigskip
\noindent
H) If $T = I_{{\bf R}^2}$ is the identity mapping, and $\beta,\beta'$
are two bases for ${\bf R}^2$, then 
$$[T]_\beta^{\beta'} = \pmatrix{1&0\cr 0&1\cr}.$$
\bigskip
\item{} This is {\it FALSE}.  A counterexample is:  Let 
$\beta = \{(3,4),(1,-1)\}$, and let $\beta'$ be the standard basis.
Then 
$$[T]_\beta^{\beta'} = \pmatrix{3&1\cr 4&-1\cr}$$
(look at the columns).
\bigskip
\noindent
I) If $T : V \to W$ is an isomorphism, and $U$ is a vector subspace 
of $V$, then $\dim(U) = \dim(T(U))$.
\bigskip
\item{}  This is {\it TRUE}.
 If $\{\bv_1, ... , \bv_n\}$ is a basis for $U$, then we know 
 $\{T(\bv_1), ... , T(\bv_n)\}$ spans $T(U)$. Furthermore, since $T$ is 
 an isomorphism, it is injective.
 Hence, if $c_1 T(\bv_1) + \cdots + c_nT(\bv_n) = {\bf 0}$, 
 then $T(c_1\bv_1 + \cdots + c_n\bv_n) = {\bf 0}$ 
 which implies $c_1 \bv_1 + ... + c_n \bv_n = {\bf 0}$, 
 which implies $c_i = 0$ for $i = 1, ... n$. Thus
$\{T(\bv_1), ... , T(\bv_n)\}$ is linearly independent, hence a basis for 
${\rm Im}(U)$.  Hence the dimensions of $U$ and $T(U)$ are equal. 
\bigskip
\noindent
J) If $\bv_1, \ldots , \bv_n$ are vectors in a vector space $V$, 
and $\bw_1,\ldots,\bw_n$ are vectors in a vector space $W$, 
then there always exists some linear $T : V \to W$ satisfying 
$T(\bv_i) =\bw_i$ for all $i = 1,..., n$.
\bigskip
\item{}  This is {\it FALSE} because we could consider a case where 
the $\bv_i$ are linearly dependent, but the $\bw_i$ are linearly independent. 
There is no linear mapping that takes a linearly dependent set to a linearly 
independent set (see question D above).
\bigskip
\noindent
K) If $T: {\bf R}^n\to {\bf R}^n$ is linear, and 
${\rm Ker}(T) + {\rm Im}(T) = {\bf R}^n$, then 
${\rm Ker}(T) \cap {\rm Im}(T)=\{{\bf 0}\}$.
\bigskip
\item{}  This is {\it TRUE}. By the Dimension Theorem, 
$\dim {\rm Ker}(T) + \dim {\rm Im}(T) = n$. The equality 
${\rm Ker}(T) + {\it Im}(T) = {\bf R}^n$ is given. 
Hence if $\beta$ is a basis for ${\rm Ker}(T)$ and $\gamma$
is a basis for ${\rm Im}(T)$, then $\beta\cup \gamma$ is a 
basis for ${\bf R}^{n}$, because it spans ${\bf R}^n$ and contains
exactly $n$ vectors.  
If ${\rm Ker}(T) \cap {\rm Im}(T)$ contained some nonzero vector,
$\bv$, then there would be an equation $\bv = \bw$ for 
some $\bv\in {\rm Ker}(T)$ and some $\bw \in {\rm Im}(T)$
but then from $\bv - \bw = {\bf 0}$, expanding $\bv$ in 
terms of $\beta$ and $\bw$ in terms of $\gamma$, there would 
be a linear dependence on the vectors in $\beta \cup \gamma$.  
This is impossible, so ${\rm Ker}(T) \cap {\rm Im}(T) = \{{\bf 0}\}$.
\bye
