Exercise 3.1.1
(a) $T$ is not a linear transformation because $T(0,0)=(1,0)$ and according to the comments after Example 5 on page 68 we know that it must always be that $T(0,0)=(0,0)$.
(b) $T$ is a linear transformation. Let $\alpha=(x_1,x_2)$ and $\beta=(y_1,y_2)$. Then \begin{align*}T(c\alpha+\beta)&=T((cx_1+y_1,cx_2+y_2))\\&=(cx_2+y_2,cx_1+y_1)\\&=c(x_2,x_1)+(y_2,y_1)=cT(\alpha)+T(\beta).\end{align*}
(c) $T$ is not a linear transformation. If $T$ were a linear transformation then we’d have $$(1,0)=T((-1,0))=T(-1\cdot(1,0))=-1\cdot T(1,0)=-1\cdot(1,0)=(-1,0)$$ which is a contradiction, $(1,0)\not=(-1,0)$.
(d) $T$ is not a linear transformation. If $T$ were a linear transformation then $$(0,0)=T(\pi,0)=T(2(\pi/2,0))=2T((\pi/2,0))=2(\sin(\pi/2),0)=2(1,0)=(2,0)$$ which is a contradiction, $(0,0)\not=(2,0)$.
(e) $T$ is a linear transformation. Let $Q=\left[\begin{array}{cc}1&0\\-1&0\end{array}\right]$. Then (identifying $\mathbb R^2$ with $\mathbb R^{1\times 2}$) $T(x_1,x_2)=[x_1\ \ x_2]Q$ so from Example 4, page 68, (with $P$ being the identity matrix), it follows that $T$ is a linear transformation.
Exercise 3.1.2
Suppose $V$ has dimension $n$. The range of the zero transformation is the zero subspace $\{0\}$; the range of the identity transformation is the whole space $V$. The rank of the zero transformation is the dimension of the range which is zero; the rank of the identity transformation is the rank of the whole space $V$ which is $n$. The null space of the zero transformation is the whole space $V$; the null space of the identity transformation is the zero subspace $\{0\}$. The nullity of the zero transformation is the dimension of its null space, which is the whole space, so is $n$; the nullity of the identity transformation is the dimension of its null space, which is the zero space, so is $0$.
Exercise 3.1.3
$V$ is the space of polynomals. The range of the differentiiation transformation is all of $V$ since if $$f(x)=c_0+c_1x+\cdots+c_nx^n$$ then $f(x)=(Dg)(x)$ where $$g(x)=c_0x+\frac{c_1}{2}x^2+\frac{c_2}{3}x^3+\cdots+\frac{c_n}{n+1}x^{n+1}.$$ The null space of the differentiation transformation is the set of constant polynomials since $(Dc)(x)=0$ for constants $c\in F$.
The range of the integration transformation is all polynomials with constant term equal to zero. Let $$f(x)=c_1x+c_2x^2+\cdots+x_nc^{n}.$$ Then $f(x)=(Tg)(x)$ where $$g(x)=c_1+2c_2x+3c_3x^2+\cdots+nc_nx^{n-1}.$$ Clearly the integral transoformation of a polynomial has constant term equal to zero, so this is the entire range of the integration transformation. The null space of the integration transformation is the zero space $\{0\}$ since the (indefinite) integral of any other polynomial is non-zero.
Exercise 3.1.4
Yes, there is such a linear transformation. Clearly $\alpha_1=(1,-1,1)$ and $\alpha_2=(1,1,1)$ are linearly independent. By Corollary 2, page 46, $\exists$ a third vector $\alpha_3$ such that $\{\alpha_1,\alpha_2,\alpha_3)$ is a basis for $\mathbb R^3$. By Theorem 1, page 69, there is a linear transformation that takes $\alpha_1,\alpha_2,\alpha_3$ to any three vectors we want. Therefore we can find a linear transformation that takes $\alpha_1\mapsto(1,0)$, $\alpha_2\mapsto(0,1)$ and $\alpha_3\mapsto(0,0)$. (We could have used any vector instead of $(0,0)$.)
Exercise 3.1.5
No there is no such transformation. If there was then since $\{\alpha_1,\alpha_2\}$ is a basis for $\mathbb R^2$ their images determine $T$ completely. Now $\alpha_3=-\alpha_1-\alpha_2$, thus it must be that \begin{align*}T(\alpha_3)&=T(-\alpha_1-\alpha_2)=-T(\alpha_1)-T(\alpha_2)\\&=-(1,0)-(0,1)=(-1,-1)\not=(1,1).\end{align*} Thus no such $T$ can exist.
Exercise 3.1.6
I’m not 100% sure I understand what they want here. Let $A$ be the matrix $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$. Then the range of $T$ is the row-space of $A$ which can have dimension $0,1,$ or $2$ depending on the row-rank. Explicitly it is all vectors of the form $$x(a,b)+y(c,d)=(ax+cy,bx+dy)$$ where $x,y$ are arbitrary elements of $F$. The rank is the dimension of this row-space, which is $0$ if $a=b=c=d=0$ and if not all $a,b,c,d$ are zero then by Exercise 1.6.8, page 27, the rank is $2$ if $ad-bc\not=0$ and equals $1$ if $ad-bc=0$.
Now let $A$ be the matrix $\left[\begin{array}{cc}a&c\\b&d\end{array}\right]$. Then the null space is the solution space of $AX=0$. Thus the nullity is $2$ if $a=b=c=d=0$, and if not all $a,b,c,d$ are zero then by Exercise 1.6.8, page 27 and Theorem 13, page 23, is $0$ if $ad-bc\not=0$ and is $1$ if $ad-bc=0$.
Exercise 3.1.7
(a) Let
$$P=\left[\begin{array}{ccc}1&-1&2\\2&1&0\\-1&-2&2\end{array}\right].$$Then $T$ can be represented by
$$\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]\mapsto P\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right].$$By Example 4, page 68, this is a linear transformation, where we’ve identified $F^3$ with $F^{3\times 1}$ and taken $Q$ in Example 4 to be the identity matrix.
(b) The range of $T$ is the column space of $P$, or equivalently the row space of
$$P^{\text{T}}=\left[\begin{array}{ccc}1&2&-1\\-1&1&-2\\2&0&2\end{array}\right].$$We row reduce the matrix as follows
$$\rightarrow\left[\begin{array}{ccc}1&2&-1\\0&3&-3\\0&-4&4\end{array}\right].$$$$\rightarrow\left[\begin{array}{ccc}1&2&-1\\0&1&-1\\0&0&0\end{array}\right].$$$$\rightarrow\left[\begin{array}{ccc}1&0&1\\0&1&-1\\0&0&0\end{array}\right].$$Let $\rho_1=(1,0,1)$ and $\rho_2=(0,1,-1)$. Then elements of the row space are elements of the form $b_1\rho_1+b_2\rho_2=(b_1,b_2,b_1-b_2)$. Thus the rank of $T$ is two and $(a,b,c)$ is in the range of $T$ as long as $c=a-b$.
Alternatively, we can row reduce the augmented matrix
$$\left[\begin{array}{ccc|c}1&-1&2&a\\2&1&0&b\\-1&-2&2&c\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|c}1&-1&2&a\\0&3&-4&b-2a\\0&-3&4&a+c\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|c}1&-1&2&a\\0&3&-4&b-2a\\0&0&0&-a+b+c\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|c}1&-1&2&a\\0&1&-4/3&(b-2a)/4\\0&0&0&-a+b+c\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|c}1&0&2/3&(b+2a)/4\\0&1&-4/3&(b-2a)/4\\0&0&0&-a+b+c\end{array}\right]$$from which we arrive at the condition $-a+b+c=0$ or equivalently $c=a-b$.
(c) We must find all $X=\left[\begin{array}{c}a\\b\\c\end{array}\right]$ such that $PX=0$ where $P$ is the matrix from part (a). We row reduce the matrix
$$\left[\begin{array}{ccc}1&-1&2\\2&1&0\\-1&-2&2\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc}1&-1&2\\0&3&-4\\0&-3&4\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc}1&-1&2\\0&3&-4\\0&0&0\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc}1&-1&2\\0&1&-4/3\\0&0&0\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc}1&0&2/3\\0&1&-4/3\\0&0&0\end{array}\right]$$Therefore
$$\left\{\begin{array}{c}a+\frac23c=0\\b-\frac43c=0\end{array}\right.$$So elements of the null space of $T$ are of the form $(-\frac23c, \frac43c,c)$ for arbitrary $c\in F$ and the dimension of the null space (the nullity) equals one.
Exercise 3.1.8
By Theorem 1, page 69, (and its proof) there is a linear transformation $T$ from $\mathbb R^3$ to $\mathbb R^3$ such that $$T(1,0,0)=(1,0,-1), \quad T(0,1,0)=(1,0,-1)$$ and $T(0,0,1)=(1,2,2)$ and the range of $T$ is exactly the subspace generated by $$\{T(1,0,0),T(0,1,0),T(0,0,1)\}=\{(1,0,-1),(1,2,2)\}.$$
Exercise 3.1.9
We have $$T(cA_1+A_2)=(cA_1+A_2)B-B(cA_1+A_2)=cA_1B+A_2B-cBA_1-BA_2,$$$$=c(A_1B-BA_1)+(A_2B-BA_2)=cT(A_1)+T(A_2).$$
Exercise 3.1.10
Let $T:V\rightarrow V$ be given by $a+bi\mapsto a$. Let $z=a+bi$ and $w=a’+b’i$ and $c\in\Bbb R$. Then \begin{align*}T(cz+w)&=T((ca+a’)+(cb+b’)i)\\&=ca+a’=cT(a+bi)+T(a’+b’i)\\&=aT(z)+T(w).\end{align*} Thus $T$ is real linear. However, if $T$ were complex linear then we must have $$0=T(i)=T(i\cdot 1)=i\cdot T(1)=i\cdot 1=i.$$ But $0\not=i$ so this is a contradiction. Thus $T$ is not complex linear.
Exercise 3.1.11
If $A$ is the zero matrix then clearly $T$ is the zero transformation. Conversely, suppose $A$ is not the zero matrix, suppose the $k$-th column $A_k$ has a non-zero entry. Then $T(\epsilon_k)=A_k\not=0$.
Exercise 3.1.12
From Theorem 2, page 71, we know $$\text{rank}(T)+\text{nullity}(T)=\dim V.$$ In this case we are assuming both terms on the left hand side are equal, say equal to $m$. Thus $m+m=n$ or equivalently $n=2m$ which implies $n$ is even.
The simplest example is $V=\{0\}$ the zero space. Then trivially the range and null space are equal. To give a less trivial example assume $V=\mathbb R^2$ and define $T$ by $T(1,0)=(0,0)$ and $T(0,1)=(1,0)$. We can do this by Theorem 1, page 69 because $\{(1,0),(0,1)\}$ is a basis for $\mathbb R^2$. Then clearly the range and null space are both equal to the subspace of $\mathbb R^2$ generated by $(1,0)$.
Exercise 3.1.13
(a) $\Rightarrow$ (b): Statement (a) says that nothing in the range gets mapped to zero except for $0$. In other words if $x$ is in the range of $T$ then $Tx=0$ $\Rightarrow$ $x=0$. Now $T\alpha$ is in the range of $T$, thus $T(T\alpha)=0$ $\Rightarrow$ $T\alpha=0$.
(b) $\Rightarrow$ (a): Suppose $x$ is in both the range and null space of $T$. Since $x$ is in the range, $x=T\alpha$ for some $\alpha$. But then $x$ in the null space of $T$ implies $T(x)=0$ which implies $T(T\alpha)=0$. Thus statement (b) implies $T\alpha=0$ or equivalently $x=0$. Thus the only thing in both the range and null space of $T$ is the zero vector $0$.
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