Back to the class
Section 6.4 #20: Show that there is no linear transformation $T \colon \mathbb{R}^{3 \times 1} \rightarrow \mathscr{P}_2$ such that
$$T \left( \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} \right) = 1 +x,$$
$$T \left( \begin{bmatrix} 3\\ 0 \\ 2 \end{bmatrix} \right)= 2-x+x^2,$$
and
$$T \left( \begin{bmatrix} 0 \\ 6 \\ -8 \end{bmatrix} \right) = -2+2x^2.$$
Solution: First let us write the vector $\begin{bmatrix} 0 \\ 6 \\ -8 \end{bmatrix}$ as a linear combination of $\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix}$, i.e. solve the following vector equation:
$$c_1 \begin{bmatrix} 2\\ 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 6 \\ -8 \end{bmatrix}.$$
To solve it, write the natural augmented matrix and put it into reduced echelon form: compute
$$\begin{bmatrix} 2&3&0 \\ 1&0&6 \\ 0&2&-8 \end{bmatrix} \sim \begin{bmatrix} 1&0&6 \\ 0&1&-4 \\ 0&0&0 \end{bmatrix}.$$
This implies that $c_1=6$ and $c_2=-4$ is a solution to the vector equation. Therefore if $T$ were a linear transformation, we would be able to compute
$$\begin{array}{ll}
T \left( 6 \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} + (-4) \begin{bmatrix}3 \\ 0 \\ 2 \end{bmatrix} \right) &\stackrel{T \mathrm{\hspace{2pt} linear}}{=} 6 T \left( \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} \right) - 4 T \left( \begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix} \right) \\
&\stackrel{\mathrm{given}}{=} 6 (1+x) -4 (2-x+x^2) \\
&= 6+6x -8 +4x -4x^2 \\
&= -2 + 10x -4x^2,
\end{array}$$
but this contradicts the given value of $T \left( \begin{bmatrix} 0 \\ 6 \\ -8 \end{bmatrix} \right)$, and so we must conclude that $T$ cannot be a linear transformation.
Section 6.5 #2: Let $T \colon \mathbb{R}^{2 \times 2} \rightarrow \mathbb{R}$ be the linear transformation defind by $T(A)=\mathrm{tr}(A)$.
a.) Which, if any, of the following matrices are in $\mathrm{ker}(T)$?:
i.) $\begin{bmatrix} 1 &2 \\ -1&3 \end{bmatrix}$
ii.) $\begin{bmatrix} 0 & 4 \\ 2 & 0 \end{bmatrix}$
iii.) $\begin{bmatrix} 1 & 3 \\ 0 & -1 \end{bmatrix}$
b.) Which, if any, of the following scalars are in $\mathrm{range}(T)$?:
i.) $0$
ii.) $2$
iii.) $\dfrac{\sqrt{2}}{2}$
c.) Describe $\mathrm{ker}(T)$ and $\mathrm{range}(T)$.
Solution: For part a.), we just compute where the given matrices go to under the function $T$:
(a) i.) Compute
$$T \left( \begin{bmatrix} 1 &2 \\ -1&3 \end{bmatrix} \right) = \mathrm{tr} \left( \begin{bmatrix} 1 &2 \\ -1&3 \end{bmatrix} \right) = 1+3 = 4.$$
Since this is not zero, we conclude that $\begin{bmatrix} 1 &2 \\ -1&3 \end{bmatrix}$ is not in the kernel of $T$.
(a) ii.) Compute
$$T \left( \begin{bmatrix} 0 & 4 \\ 2 & 0 \end{bmatrix} \right) = \mathrm{tr} \left( \begin{bmatrix} 0 & 4 \\ 2 & 0 \end{bmatrix} \right) = 0+0 = 0.$$
Since this is zero, we conclude that $\begin{bmatrix} 0 & 4 \\ 2 & 0 \end{bmatrix}$ is in the kernel of $T$.
(a) iii.) Compute
$$T \left( \begin{bmatrix} 1 & 3 \\ 0 & -1 \end{bmatrix} \right) = \mathrm{tr} \left( \begin{bmatrix} 1 & 3 \\ 0 & -1 \end{bmatrix} \right) = 1 + (-1) = 0.$$
Since this is zero, we conclude that $\begin{bmatrix} 1 & 3 \\ 0 & -1 \end{bmatrix}$ is in the kernel of $T$.
b.) All are in the range of $T$. For i.) just that $T \left( \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} \right)=0$. For ii.), just note that $T \left( \begin{bmatrix} 1&0 \\ 0 &2 \end{bmatrix} \right)=2$, and for iii.) just note that $T \left( \begin{bmatrix} 1 & 0 \\ 0 & \frac{\sqrt{2}}{2} \end{bmatrix} \right) = \dfrac{\sqrt{2}}{2}$.
c.) We may describe $\mathrm{ker}(T)$ by considering the formula
$$T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = a+d = 0.$$
This forces $a=-d$. Therefore we may say that
$$\mathrm{ker}(T) = \left\{ \begin{bmatrix} -d & b \\ c&d \end{bmatrix} \colon b,c,d \in \mathbb{R} \right\}=\mathrm{span} \left\{ \begin{bmatrix} -1 & 0 \\ 0&1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0& 0 \\ 1 &0 \end{bmatrix} \right\} .$$
The range will be the whole of $\mathbb{R}$ because if $\alpha \in \mathbb{R}$ is arbitrary, consider
$$T \left( \begin{bmatrix} a & b \\ c&d \end{bmatrix} \right) = a+d = \alpha.$$
This equation just requires $a = \alpha - d$. This can always be done, and so
$$\mathrm{range}(T) = \mathbb{R}.$$
(note: notice that $\mathbb{R}^{2 \times 2}$ is dimension $4$, the kernel we found is dimension $3$ ("nullity"), and the range we found was dimension $1$ ("rank") -- compare these values to the rank-nullity theorem!)
Section 6.5 #4: Let $T \colon \mathscr{P}_2 \rightarrow \mathscr{P}_2$ be the linear transformation defind by $T(p(x))=xp'(x)$.
a.) Which, if any, of the following polynomials are in $\mathrm{ker}(T)$?:
i.) $1$
ii.) $x$
iii.) $x^2$
b.) Which, if any, of the polynomials in part (a) are in $\mathrm{range}(T)$?
c.) Describe $\mathrm{ker}(T)$ and $\mathrm{range}(T)$.
Solution: For part a.), we just compute where the given polynomials go under the function $T$:
(a) i.) Compute
$$T(1) = x \dfrac{\mathrm{d}}{\mathrm{d}x} 1 = x (0) = 0.$$
Therefore $1 \in \mathrm{ker}(T)$.
(a) ii.) Compute
$$T(x) = x \dfrac{\mathrm{d}}{\mathrm{d}x} x = x(1) = x.$$
Therefore $x \not\in \mathrm{ker}(T)$.
(a) iii.) Compute
$$T(x^2) = x \dfrac{\mathrm{d}}{\mathrm{d}x} x^2 = x(2x) = 2x^2.$$
Therefore $x^2 \not\in \mathrm{ker}(T)$.
(b) i.) We ask "is $1 \in \mathrm{range}(T)$?". For this to happen, we must have a polynomial $p=ax^2+bx+c \in \mathscr{P}_2$ such that $T(p)=1$. So calculate
$$T(p)=x \dfrac{\mathrm{d}}{\mathrm{d}x} (ax^2+bx+c) = x(2ax+b)=2ax^2+bx.$$
Is it possible to pick $a$ and $b$ so that $2ax^2+bx=1$? No! Therefore $1 \not\in \mathrm{range}(T)$.
(b) ii.) We ask "is $x \in \mathrm{range}(T)$?". For this to happen, we must have a polynomial $p=ax^2+bx+c \in \mathscr{P}_2$ such that $T(p)=x$. So calculate
$$T(p) = 2ax^2+bx.$$
Is it possible ot pick $a$ and $b$ so that $2ax^2+bx = x$? Yes -- take $a=0$ and $b=1$. Therefore $x \in \mathrm{range}(T)$.
(b) iii.) We ask "is $x^2 \in \mathrm{range}(T)$?". For this to happen we must pick $a$ and $b$ so that $2ax^2+bx=x^2$. This happens when $a=1$ and $b=0$. Therefore $x^2 \in \mathrm{range}(T)$.
(c) To describe $\mathrm{ker}(T)$ we want to characterize the polynomials $p=ax^2+bx+c$ such that $T(p)=0$. So calculate
$$T(p) = 2ax^2+bx =0.$$
This yields the requirements $2a=0$ (hence $a=0$) and $b=0$. Therefore we observe that
$$\mathrm{ker}(T) = \{ ax^2+bx+c \colon a=0,b=0, c \in \mathbb{R} \} = \{c \colon c \in \mathbb{R} \} = \mathrm{span}\{1\}.$$
For the range, we want to characterize the polynomials $q=\alpha x^2+\beta bx+\gamma$ such that there is a polynomial $p=ax^2+bx+c$ with the property that $T(p)=q$. This means
$$2ax^2 + bx = \alpha x^2 + \beta x + \gamma.$$
Equating coefficients shows that $2a=\alpha$ (hence $a=\dfrac{\alpha}{2}$), $b=\beta$, and $\gamma=0$. Therefore we may write
$$\mathrm{range}(T) = \left\{ \dfrac{\alpha}{2}x^2 + \beta x + \gamma \colon \alpha, \beta \in \mathbb{R}, \gamma=0 \right\} = \left\{ \dfrac{\alpha}{2} x^2 + \beta x \colon \alpha, \beta \in \mathbb{R} \right\} = \mathrm{span}\{x^2,x\}.$$
(note: notice that the dimension of $\mathscr{P}_2$ is $3$ and the dimension of $\mathrm{ker}(T)$ is $1$ while the dimension of $\mathrm{range}(T)=2$ -- rank-nullity is apparent again!)
Section 6.5 #9: Find either the nullity or rank of $T$ and then use the Rank-Nullity Theorem to find the other: $T \colon \mathbb{R}^{2 \times 2} \rightarrow \mathbb{R}^{2 \times 1}$ defined by
$$T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} a-b \\ c-d \end{bmatrix}.$$
Solution: We will find the nullity of $T$ -- to do this we must find the dimension of the kernel of $T$. A matrix $\begin{bmatrix} a&b \\ c&d \end{bmatrix}$ is in the kernel of $T$ provided that
$$T \left( \begin{bmatrix} a&b \\ c&d \end{bmatrix} \right) = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$
This means
$$\begin{bmatrix} a-b \\ c-d \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$
This means $a=b$ and $c=d$. Therefore the kernel is of the following form:
$$\mathrm{ker}(T) = \left\{ \begin{bmatrix} a & a \\ c&c \end{bmatrix} \colon a,c \in \mathbb{R} \right\} = \mathrm{span} \left\{ \begin{bmatrix} 1&1 \\ 0&0 \end{bmatrix}, \begin{bmatrix} 0&0 \\ 1&1 \end{bmatrix} \right\}.$$
The set $\left\{ \begin{bmatrix} 1&1 \\ 0&0 \end{bmatrix}, \begin{bmatrix} 0&0 \\ 1&1 \end{bmatrix} \right\}$ is a basis for $\mathrm{ker}(T)$ and so $\mathrm{nullity}(T)=\mathrm{dim}(\mathrm{ker}(T))=2$. The vector space $\mathbb{R}^{2 \times 2}$ has dimension $4$. The rank-nullity theorem says that
$$\mathrm{dim}(\mathbb{R}^{2 \times 2}) = \mathrm{rank}(T) + \mathrm{nullity}(T),$$
and plugging in our known values gives
$$4 = \mathrm{rank}(T) + 2.$$
Hence $\mathrm{rank}(T)=4-2=2$.
Section 6.5 #11: Find either the nullity or rank of $T$ and then use the Rank-Nullity Theorem to find the other: $T \colon \mathbb{R}^{2 \times 2} \rightarrow \mathbb{R}^{2 \times 2}$ defined by $T(A)=AB-BA$, where $B = \begin{bmatrix} 1 & 1 \\ 0&1 \end{bmatrix}$.
Solution: We will find the rank of $T$ -- to do this we must find the dimension of the range of $T$. A matrix $\begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix} \in \mathrm{range}(T)$ provided there is a matrix $\begin{bmatrix} a&b \\ c&d \end{bmatrix}$ such that $T \left( \begin{bmatrix} a&b \\ c&d \end{bmatrix} \right) = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}$. Applying the definition of $T$ to this formula leads to the following equation:
$$\begin{bmatrix} a&b \\ c&d \end{bmatrix} B - B \begin{bmatrix} a&b \\ c&d \end{bmatrix} = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}.$$
Plugging in the value of $B$ yields
$$(*) \hspace{35pt} \begin{bmatrix} a&b \\ c&d \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0&1 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 0&1 \end{bmatrix} \begin{bmatrix} a&b \\ c&d \end{bmatrix} = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}.$$
Now calculate
$$\begin{bmatrix} a&b \\ c&d \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0&1 \end{bmatrix}=\begin{bmatrix} a & a+b \\ c & c+d \end{bmatrix}$$
and calculate
$$\begin{bmatrix} 1 & 1 \\ 0&1 \end{bmatrix} \begin{bmatrix} a&b \\ c&d \end{bmatrix} = \begin{bmatrix} a+c & b+d \\ c & d \end{bmatrix}.$$
Therefore $(*)$ can be written as
$$\begin{bmatrix} a & a+b \\ c& c+d \end{bmatrix} - \begin{bmatrix} a+c & b+d \\ c& d \end{bmatrix} = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}.$$
Performing the matrix addition on the left yields
$$\begin{bmatrix} -c & a-d \\ 0 & c \end{bmatrix} = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}.$$
This is equivalent to the following system of four equations:
$$\left\{ \begin{array}{rrrrl}
&&-c& &= \alpha \\
a&&&-d&=\beta \\
&&0&&=\gamma \\
&&c&&=\delta.
\end{array} \right.$$
From the first and last line we get that $c=\delta=-\alpha$ and from the third line we get $\gamma=0$. The second line $a-d=\beta$ tells us how to pick the value of $a$ (so it doesn't restrict the range). Therefore, the range of $T$ is comprised of matrices of the form
$$\begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix} = \begin{bmatrix} \alpha & \beta \\ 0 & -\alpha \end{bmatrix} = \mathrm{span} \left\{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0&0 \end{bmatrix} \right\},$$
which is a basis. Therefore
$$\mathrm{rank}(T)=\mathrm{dim}(\mathrm{range}(T)) = 2.$$
Hence by the rank-nullity theorem, we have $4=2+\mathrm{nullity}(T)$, and we must conclude that $\mathrm{nullity}(T)=2$.
