Problem Page 59 # 9:
Prove that if \(T\) is a linear map from \(\textbf{F}^4\) to
\(\textbf{F}^2\) such that
\[
\text{null}(T) = \{(x_1,x_2,x_3,x_4)\in\textbf{F}^4 : x_1=5x_2
\text{ and } x_3=7x_4 \},
\]
then \(T\) is surjective.
Solution:
The null space of \(T\) consists of all vectors that look like
\[
(5x_2,x_2,7x_4,x_4) = x_2(5,1,0,0) + x_4(0,0,7,1),
\]
so
\[
\text{null}(T) = \text{Span} \{ (5,1,0,0),(0,0,7,1) \}.
\]
(In fact, it's easy to show that \((5,1,0,0)\) and \((0,0,7,1)\) are
linearly independent, so they are a basis for null\((T)\), but we don't
need to know that.) This shows that \(\dim\text{null}(T)\le 2\), since
any spanning set contains a basis. Hence the dimension theorem tells
us that
\[
\dim\text{range}(T)
= \dim\textbf{F}^4 - \dim\text{null}(T)
= 4 - \dim\text{null}(T)
\ge 4 - 2
= 2
= \dim\textbf{F}^2.
\]
So range\((T)\) is a subspace of \(\textbf{F}^2\) that has the same dimension
as \(\textbf{F}^2\), hence it is equal to all of \(\textbf{F}^2\). This proves that \(T\)
is surjective.
Problem Page 59 # 10:
Prove that there does not exist a linear map from \(\textbf{F}^5\) to
\(\textbf{F}^2\) whose null space equals
\[
\{(x_1,x_2,x_3,x_4,x_5)\in\textbf{F}^5
: x_1=3x_2 \text{ and } x_3=x_4=x_5\}.
\]
Solution:
Let
\[
U = \{(x_1,x_2,x_3,x_4,x_5)\in\textbf{F}^5
: x_1=3x_2 \text{ and } x_3=x_4=x_5\}.
\]
Then \(U\) consists of all vectors of the form
\[
(3x_2,x_2,x_3,x_3,x_3) = x_2(3,1,0,0,0)+x_3(0,0,1,1,1),
\]
so
\[
U = \text{Span}\{(3,1,0,0,0),(0,0,1,1,1)\}.
\]
But it is also clear that
\((3,1,0,0,0)\) and \((0,0,1,1,1)\) are linearly independent, since
if
\[
a(3,1,0,0,0)+ b(0,0,1,1,1) = (0,0,0,0,0),
\]
then
\[
(3a,a,b,b,b)=(0,0,0,0,0),
\]
so \(a=b=0\). This shows that \(\dim(U)=2\).
Suppose now that \(T:\textbf{F}^5\to\textbf{F}^2\) were a linear map
whose null space equaled \(U\). Then the dimension theorem would tell us
that
\[
2 = \dim(U)
= \dim \text{null}(T)
= \dim(\textbf{F}^5) - \dim\text{range}(T)
\ge \dim(\textbf{F}^5) - \dim(\textbf{F}^2)
= 5 - 2
= 3.
\]
This contradiction shows that no such \(T\) can exist.
Problem Page 59 # 16:
Suppose that \(U\) and \(V\) are finite-dimensional vector spaces,
and let \(S\in\mathcal{L}(V,W)\) and \(T\in\mathcal{L}(U,V)\) be
linear maps. Prove that
\[
\dim\text{null}(ST) \le \dim\text{null}(S) + \dim\text{null}(T).
\]
Solution:
We first observe that if \(\mathbf{u}\in\text{null}(ST)\), then
\(S(T(\mathbf{u}))=0\), so \(T(\mathbf{u})\in\text{null}(S)\). In other words,
if we view \(T\) as a map whose domain is \(\text{null}(ST)\), then \(T\)
is a linear map
\[
T : \text{null}(ST) \longrightarrow \text{null}(S).
\]
It is now easy to get confused, because the letter \(T\) could mean
the linear map whose domain is \(U\) and whose target space is \(V\), or it
could mean the linear map whose domain is null\((ST)\) and whose target
space is null\((S)\). In order to avoid confusion, we will use the
following alternative notation:
\[
U' = \text{null}(ST),\qquad
V' = \text{null}(S),\qquad
T' \in \mathcal{L}(U',V').
\]
We now want to compute null\((T')\). I claim that
\[
\text{null}(T') = \text{null}(T).
\]
Proof of Claim:
We first take a vector \(\mathbf{u}\in\text{null}(T)\). Then
\(T(\mathbf{u})=0\), so \(ST(\mathbf{u})=S(0)=0\), so \(\mathbf{u}\in
U'\). Since also \(T'(\mathbf{u})=T(\mathbf{u})=0\), this shows that
\(\mathbf{u}\in\text{null}(T')\), which proves the inclusion
\(\text{null}(T)\subset\text{null}(T')\). Next we take a
vector \(\mathbf{u}'\in\text{null}(T')\). Then
\(T(\mathbf{u}')=T'(\mathbf{u}')=0\), so
\(\mathbf{u}'\in\text{null}(T)\). This proves the other inclusion
\(\text{null}(T')\subset\text{null}(T)\), which
completes the proof of the claim.
We now apply the dimension theorem to the linear map \(T':U'\to V'\).
It tells us that
\[
\begin{aligned}
\dim\text{null}(ST)
&= \dim U'
&&\text{since \(U'=\text{null}(ST)\),} \\
&= \dim \text{null}(T') + \dim \text{range}(T')
&&\text{by the dimension theorem applied to \(T':U'\to V'\),} \\
&\le \dim \text{null}(T') + \dim V'
&&\text{since \(\text{range}(T')\subset V'\),} \\
&= \dim \text{null}(T) + \dim V'
&&\text{since the claim says that \(\text{null}(T') = \text{null}(T)\)}\\
&= \dim \text{null}(T) + \dim \text{null}(S)
&&\text{since \(V'=\text{null}(S)\).}
\end{aligned}
\]
This proves the desired inequality.
Problem # A.4:
Find the matrix associated to each of the following
linear transformations relative to the given bases.
(a)
\(T:\mathbb{F}^3\to\mathbb{F}^3\) defined by
\(T(x,y,z)=(2x-3y,3x+5z,7x+2y-z)\),
using the standard basis for \(\mathbb{F}^3\).
(b)
\(T:\mathbb{F}^2\to\mathbb{F}^3\) defined by
\(T(x,y,z)=(2x-3y,3x+5y,-x)\),
using the standard bases for \(\mathbb{F}^2\) and \(\mathbb{F}^3\).
(c)
Let \(\mathcal{P}_n(\mathbb{F})\) be the \(\mathbb{F}\)-vector space
of polynomials of degree at most \(n\) with coefficients in \(\mathbb{F}\).
Use the basis \(\{1,t,t^2,\ldots,t^n\}\). What is the matrix associated
to the linear transformation
\(T:\mathcal{P}_n(\mathbb{F})\to\mathcal{P}_n(\mathbb{F})\) defined
by differentiation \(T(f(t)) = f'(t)\)?
[Hint. If you don't see immediately how to do it, try
doing \(n=2\) and \(n=3\) to get an idea what's going on.]
(d)
Continuing with the notation from (c), what is the matrix
for the linear transformation
\(T:\mathcal{P}_n(\mathbb{F})\to\mathcal{P}_{n+2}(\mathbb{F})\) defined
\(T(f(t))=(t^2+1)f(t)\)?
(e)
Let \(V\) be the real vector space consisting of all functions of
the form \(f(t)=ae^{2t}+bte^{2t}\) with \(a\) and \(b\) in \(\mathbb{R}\).
Using the basis \(\{e^{2t},te^{2t}\}\), what is the matrix associated
to the linear transformation \(T:V\to V\) defined
by \(T(f(t))=f'(t)\)?
Solution:
(a) and (b) are easy.
(a)
\begin{align*}
T(\mathbf{e}_1)
&=T(1,0,0)=(2,3,7)=2\mathbf{e}_1+3\mathbf{e}_2+7\mathbf{e}_3, \\
T(\mathbf{e}_2)
&=T(0,1,0)=(-3,0,2)=-3\mathbf{e}_1+0\mathbf{e}_2+2\mathbf{e}_3,\\
T(\mathbf{e}_3)
&=T(0,0,1)=(0,5,-1)=0\mathbf{e}_1+5\mathbf{e}_2-\mathbf{e}_3.
\end{align*}
So
\(\mathcal{M}(T)=\begin{pmatrix} 2&-3&0\\ 3&0&5\\ 7&2&-1\\ \end{pmatrix}\).
(b)
\(\mathcal{M}(T)=\begin{pmatrix} 2&-3\\ 3&5\\ -1&0\\ \end{pmatrix}\).
(c)
Our basis is
\[
\mathbf{v}_0=1, \; \mathbf{v}_1=t,
\; \mathbf{v}_2=t^2, \;\dots\; \mathbf{v}_n=t^n.
\]
The linear map \(T\) is differentiation, so
\[
\begin{aligned}
T(\mathbf{v}_0) &= d1/dt = 0,\\
T(\mathbf{v}_1) &= dt/dt = 1 = \mathbf{v}_0,\\
T(\mathbf{v}_2) &= dt^2/dt = 2t = 2\mathbf{v}_1,\\
T(\mathbf{v}_3) &= dt^3/dt = 3t^2 = 3\mathbf{v}_2,\\
\vdots & \qquad \vdots \\
T(\mathbf{v}_k) &= dt^k/dt = kt^{k-1} = k\mathbf{v}_{k-1},\\
\vdots & \qquad \vdots \\
T(\mathbf{v}_n) &= dt^n/dt = nt^{n-1} = n\mathbf{v}_{n-1}.\\
\end{aligned}
\]
So the matrix of \(T\) is the following \((n+1)\)-by-\((n+1)\) matrix
\[
\mathcal{M}(T) =
\begin{pmatrix}
0 & 1 & 0 & 0 & \cdots & 0 \\
0 & 0 & 2 & 0 & \cdots & 0 \\
0 & 0 & 0 & 3 & \cdots & 0 \\
& \vdots & & & \ddots & \vdots \\
0 & 0 & 0 & 0 & \cdots & n \\
0 & 0 & 0 & 0 & \cdots & 0 \\
\end{pmatrix}
\]
(c)
We compute
\[
\begin{aligned}
T(\mathbf{v}_0) &= (t^2+1)\cdot 1 = 1+t^2 = \mathbf{v}_0+\mathbf{v}_2 \\
T(\mathbf{v}_1) &= (t^2+1)\cdot t = t+t^3 = \mathbf{v}_1+\mathbf{v}_3 \\
T(\mathbf{v}_2) &= (t^2+1)\cdot t^2 = t^2+t^4
= \mathbf{v}_2+\mathbf{v}_4 \\
T(\mathbf{v}_3) &= (t^2+1)\cdot t^3 = t^3+t^5
= \mathbf{v}_3+\mathbf{v}_5 \\
\vdots & \quad \vdots \\
T(\mathbf{v}_k) &= (t^2+1)\cdot t^k = t^k+t^{k+2}
= \mathbf{v}_k+\mathbf{v}_{k+2} \\
\vdots & \qquad \vdots \\
T(\mathbf{v}_n) &= (t^2+1)\cdot t^n = t^n+t^{n+2}
= \mathbf{v}_n+\mathbf{v}_{n+2} \\
\end{aligned}
\]
So the matrix of \(T\) is
\[
\mathcal{M}(T) =
\begin{pmatrix}
1 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & \cdots & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & \cdots & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & \cdots & 0 & 0 & 0 \\
& \vdots & & & \ddots & & &\vdots \\
0 & 0 & 0 & 0 & \cdots & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & \cdots & 1 & 0 & 1 \\
\end{pmatrix}
\]
The matrix of \(T\) has \(n+3\) rows and \(n+1\) columns.
(e) First it's worth checking that \(T\) actually send vectors in \(V\) to vectors in \(V\). So let \(f(t)=ae^{2t}+bte^{2t}\) be a vector in \(V\). Then \[ T(f(t)) = f'(t) = 2ae^{2t} + be^{2t} + 2bte^{2t} = (2a+b)e^{2t}+2bte^{2t}, \] so \(T(f(t))\) is in \(V\), since it is a constant multiple of \(e^{2t}\) added to a constant multiple of \(te^{2t}\). We now compute the matrix of \(T\) with respect to the basis is \(\mathbf{v}_1=e^{2t}\) and \(\mathbf{v}_2=te^{2t}\). Then \[ \begin{aligned} T(\mathbf{v}_1)&=\frac{de^{2t}}{dt} = 2e^{2t} =2\mathbf{v}_1 \\ T(\mathbf{v}_2)&=\frac{d (te^{2t})}{dt} = e^{2t}+2te^{2t} =\mathbf{v}_1+2\mathbf{v}_2 \\ \end{aligned} \] So the matrix of \(T\) is \[ \mathcal{T}=\begin{pmatrix} 2 & 1 \\ 0 & 2 \\ \end{pmatrix}. \]
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