6.10.2E: Kernel and Image of a Linear Transformation Exercises

Exercises

Exercise $\PageIndex{1}$

For each matrix $A$, find a basis for the kernel and image of $T_{A}$, and find the $rank \;$ and $nullity \;$ of $T_{A}$.

1. $\left[ \begin{array}{rrrr} 1 & 2 & -1 & 1 \\ 3 & 1 & 0 & 2 \\ 1 & -3 & 2 & 0 \end{array} \right]$
2. $\left[ \begin{array}{rrrr} 2 & 1 & -1 & 3 \\ 1 & 0 & 3 & 1 \\ 1 & 1 & -4 & 2 \end{array} \right]$
3. $\left[ \begin{array}{rrr} 1 & 2 & -1 \\ 3 & 1 & 2 \\ 4 & -1 & 5 \\ 0 & 2 & -2 \end{array} \right]$
4. $\left[ \begin{array}{rrr} 2 & 1 & 0 \\ 1 & -1 & 3 \\ 1 & 2 & -3 \\ 0 & 3 & -6 \end{array} \right]$

Answer

2. $\left\{\left[\begin{array}{r}-3 \\ 7 \\ 1 \\ 0\end{array}\right],\left[\begin{array}{r}1 \\ 1 \\ 0 \\ -1\end{array}\right]\right\} ;\left\{\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right],\left[\begin{array}{r}0 \\ 1 \\ -1\end{array}\right]\right\} ; 2,2$
3. $\left\{\left[\begin{array}{r}-1 \\ 2 \\ 1\end{array}\right]\right\} ;\left\{\left[\begin{array}{l}1 \\ 0 \\ 1 \\ 1\end{array}\right],\left[\begin{array}{r}0 \\ 1 \\ -1 \\ -2\end{array}\right]\right\} ; 2,1$

Exercise $\PageIndex{2}$

In each case, (i) find a basis of $\text{ker }T$, and (ii) find a basis of $im \;T$. You may assume that $T$ is linear.

1. $T :{P}_{2} \to \mathbb{R}^2$; $T(a + bx + cx^{2}) = (a, b)$
2. $T :{P}_{2} \to \mathbb{R}^2$; $T(p(x)) = (p(0), p(1))$
3. $T : \mathbb{R}^3 \to \mathbb{R}^3$; $T(x, y, z) = (x + y, x + y, 0)$
4. $T : \mathbb{R}^3 \to \mathbb{R}^4$; $T(x, y, z) = (x, x, y, y)$
5. $T :{M}_{22} \to\|{M}_{22}$; $T\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} a + b & b + c \\ c + d & d + a \end{array} \right]$
6. $T :{M}_{22} \to \mathbb{R}$; $T\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = a + d$
7. $T :{P}_{n} \to \mathbb{R}$; $T(r_{0} + r_{1}x + \cdots + r_{n}x^{n}) = r_{n}$
8. $T : \mathbb{R}^n \to \mathbb{R}$; $T(r_{1}, r_{2}, \dots, r_{n}) = r_{1} + r_{2} + \cdots + r_{n}$
9. $T :{M}_{22} \to\|{M}_{22}$; $T(X) = XA - AX$, where
10. $A = \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right]$

Answer

2. $\left\{x^2-x\right\} ;\{(1,0),(0,1)\}$
3. $\{(0,0,1)\} ;\{(1,1,0,0),(0,0,1,1)\}$
4. $\left\{\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right]\right\} ;\{1\}$
5. $\{(1,0,0, \ldots, 0,-1),(0,1,0, \ldots, 0,-1)$, $\ldots,(0,0,0, \ldots, 1,-1)\} ;\{1\}$
6. $\begin{array}{l}\{ {\left.\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\right\} } \\ \left\{\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 1\end{array}\right]\right\}\end{array}$

Exercise $\PageIndex{1}$

Let $P : V \to \mathbb{R}$ and $Q : V \to \mathbb{R}$ be linear transformations, where $V$ is a vector space. Define $T : V \to \mathbb{R}^2$ by $T(\mathbf{v}) = (P(\mathbf{v}), Q(\mathbf{v}))$.

1. Show that $T$ is a linear transformation.
2. Show that $\text{ker }T = \text{ker }P \cap \text{ker }Q$, the set of vectors in both $\text{ker }P$ and $\text{ker }Q$.

Answer

b. $T(\mathbf{v}) = \mathbf{0} = (0, 0)$ if and only if $P(\mathbf{v}) = 0$ and $Q(\mathbf{v}) = 0$; that is, if and only if $\mathbf{v}$ is in $\text{ker }P \cap \text{ker }Q$.

Exercise $\PageIndex{4}$

In each case, find a basis
$B = \{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}, \mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}$ of $V$ such that $\{\mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}$ is a basis of $\text{ker }T$, and verify Theorem [thm:021572].

1. $T : \mathbb{R}^3 \to \mathbb{R}^4$; $T(x, y, z) = (x - y + 2z, x + y - z, 2x + z, 2y - 3z)$
2. $T : \mathbb{R}^3 \to \mathbb{R}^4$; $T(x, y, z) = (x + y + z, 2x - y + 3z, z - 3y, 3x + 4z)$

Answer

b. $\text{ker }T = span \;\{(-4, 1, 3)\}$; $B = \{(1, 0, 0), (0, 1, 0), (-4, 1, 3)\}$, $im \;T = span \;\{(1, 2, 0, 3), (1, -1, -3, 0)\}$

Exercise $\PageIndex{5}$

Show that every matrix $X$ in $\mathbf{M}_{nn}$ has the form $X = A^{T} - 2A$ for some matrix $A$ in $\mathbf{M}_{nn}$. [Hint: The dimension theorem.]

Exercise $\PageIndex{6}$

In each case either prove the statement or give an example in which it is false. Throughout, let $T : V \to W$ be a linear transformation where $V$ and $W$ are finite dimensional.

1. If $V = W$, then $\text{ker }T \subseteq im \;T$.
2. If $dim \;V = 5$, $dim \;W = 3$, and $dim \;(\text{ker }T) = 2$, then $T$ is onto.
3. If $dim \;V = 5$ and $dim \;W = 4$, then $\text{ker }T \neq \{\mathbf{0}\}$.
4. If $\text{ker }T = V$, then $W = \{\mathbf{0}\}$.
5. If $W = \{\mathbf{0}\}$, then $\text{ker }T = V$.
6. If $W = V$, and $im \;T \subseteq \text{ker }T$, then $T = 0$.
7. If $\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\}$ is a basis of $V$ and $T(\mathbf{e}_{1}) = \mathbf{0} = T(\mathbf{e}_{2})$, then $dim \;(im \;T) \leq 1$.
8. If $dim \;(\text{ker }T) \leq dim \;W$, then $dim \;W \geq \frac{1}{2} dim \;V$.
9. If $T$ is one-to-one, then $dim \;V \leq dim \;W$.
10. If $dim \;V \leq dim \;W$, then $T$ is one-to-one.
11. If $T$ is onto, then $dim \;V \geq dim \;W$.
12. If $dim \;V \geq dim \;W$, then $T$ is onto.
13. If $\{T(\mathbf{v}_{1}), \dots, T(\mathbf{v}_{k})\}$ is independent, then $\{\mathbf{v}_{1}, \dots, \mathbf{v}_{k}\}$ is independent.
14. If $\{\mathbf{v}_{1}, \dots, \mathbf{v}_{k}\}$ spans $V$, then $\{T(\mathbf{v}_{1}), \dots, T(\mathbf{v}_{k})\}$ spans $W$.

Answer

2. Yes. $dim \;(im \;T) = 5 - dim \;(\text{ker }T) = 3$, so $im \;T = W$ as $dim \;W = 3$.
3. No. $T = 0: \mathbb{R}^2 \to \mathbb{R}^2$
4. No. $T : \mathbb{R}^2 \to \mathbb{R}^2$, $T(x, y) = (y, 0)$. Then $\text{ker }T = im \;T$
5. Yes. $dim \;V = dim \;(\text{ker }T) + dim \;(im \;T) \leq dim \;W + dim \;W = 2 dim \;W$
6. No. Consider $T : \mathbb{R}^2 \to \mathbb{R}^2$ with $T(x, y) = (y, 0)$.
7. No. Same example as (j).
8. No. Define $T : \mathbb{R}^2 \to \mathbb{R}^2$ by $T(x, y) = (x, 0)$. If $\mathbf{v}_{1} = (1, 0)$ and $\mathbf{v}_{2} = (0, 1)$, then $\mathbb{R}^2 = span \;\{\mathbf{v}_{1}, \mathbf{v}_{2}\}$ but $\mathbb{R}^2 \neq span \;\{T(\mathbf{v}_{1}), T(\mathbf{v}_{2})\}$.

Exercise $\PageIndex{7}$

Show that linear independence is preserved by one-to-one transformations and that spanning sets are preserved by onto transformations. More precisely, if $T : V \to W$ is a linear transformation, show that:

1. If $T$ is one-to-one and $\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\}$ is independent in $V$, then $\{T(\mathbf{v}_{1}), \dots, T(\mathbf{v}_{n})\}$ is independent in $W$.
2. If $T$ is onto and $V = span \;\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\}$, then $W = span \;\{T(\mathbf{v}_{1}), \dots, T(\mathbf{v}_{n})\}$.

Answer

b. Given $\mathbf{w}$ in $W$, let $\mathbf{w} = T(\mathbf{v})$, $\mathbf{v}$ in $V$, and write $\mathbf{v} = r_{1}\mathbf{v}_{1} + \cdots + r_{n}\mathbf{v}_{n}$. Then $\mathbf{w} = T(\mathbf{v}) = r_{1}T(\mathbf{v}_{1}) + \cdots + r_{n}T(\mathbf{v}_{n})$.

Exercise $\PageIndex{8}$

Given $\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\}$ in a vector space $V$, define $T : \mathbb{R}^n \to V$ by $T(r_{1}, \dots, r_{n}) = r_{1}\mathbf{v}_{1} + \cdots + r_{n}\mathbf{v}_{n}$. Show that $T$ is linear, and that:

1. $T$ is one-to-one if and only if $\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\}$ is independent.
2. $T$ is onto if and only if $V = span \;\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\}$.

Answer

b. $im \;T = \left\lbrace \sum_{i} r_i\mathbf{v}_i \mid r_i \mbox{ in } \mathbb{R} \right\rbrace = span \;\{\mathbf{v}_i\}$.

Exercise $\PageIndex{9}$

Let $T : V \to V$ be a linear transformation where $V$ is finite dimensional. Show that exactly one of (i) and (ii) holds:

(i) $T(\mathbf{v}) = \mathbf{0}$ for some $\mathbf{v} \neq \mathbf{0}$ in $V$;

(ii) $T(\mathbf{x}) = \mathbf{v}$ has a solution $\mathbf{x}$ in $V$ for every $\mathbf{v}$ in $V$.

Exercise $\PageIndex{10}$

Let $T :\|{M}_{nn} \to \mathbb{R}$ denote the trace map: $T(A) = tr \;A$ for all $A$ in $\mathbf{M}_{nn}$. Show that
$dim \;(\text{ker }T) = n^{2} - 1$.

Answer

$T$ is linear and onto. Hence $1 = dim \;\mathbb{R} = dim \;(im \;T) = dim \;(\mathbf{M}_{nn}) - dim \;(\text{ker }T) = n^{2} - dim \;(\text{ker }T)$

Exercise $\PageIndex{11}$

Show that the following are equivalent for a linear transformation $T : V \to W$.

$\text{ker }T = V$

$im \;T = \{\mathbf{0}\}$

$T = 0$

Exercise $\PageIndex{12}$

Let $A$ and $B$ be $m \times n$ and $k \times n$ matrices, respectively. Assume that $A\mathbf{x} = \mathbf{0}$ implies $B\mathbf{x} = \mathbf{0}$ for every $n$-column $\mathbf{x}$. Show that $rank \;A \geq rank \;B$.

Answer

The condition means $\text{ker }(T_{A}) \subseteq \text{ker}(T_{B})$, so $dim \;\left[\text{ker}(T_{A})\right] \leq dim \;\left[\text{ker}(T_{B})\right]$. Then Theorem [thm:021499] gives $dim \;\left[im \;(T_{A})\right] \geq dim \;\left[im \;(T_{B})\right]$; that is, $rank \;A \geq rank \;B$.

Exercise $\PageIndex{13}$

Let $A$ be an $m \times n$ matrix of $rank \;r$. Thinking of $\mathbb{R}^n$ as rows, define $V = \{\mathbf{x}$ in $\mathbb{R}^m \mid \mathbf{x}A = \mathbf{0}\}$. Show that $dim \;V = m - r$.

Exercise $\PageIndex{14}$

Consider

$$V = \left\lbrace \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] \, \middle| \, a + c = b + d \right\rbrace \nonumber$$

1. Consider $S :\|{M}_{22} \to \mathbb{R}$ with $S\left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] = a + c - b - d$. Show that $S$ is linear and onto and that $V$ is a subspace of $\mathbf{M}_{22}$. Compute $dim \;V$.
2. Consider $T : V \to \mathbb{R}$ with $T\left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] = a + c$. Show that $T$ is linear and onto, and use this information to compute $dim \;(\text{ker }T)$.

Exercise $\PageIndex{15}$

Define $T :\|{P}_{n} \to \mathbb{R}$ by $T\left[p(x)\right] =$ the sum of all the coefficients of $p(x)$.

1. Use the dimension theorem to show that $dim \;(\text{ker }T) = n$.
2. Conclude that $\{x - 1, x^{2} - 1, \dots, x^{n} - 1\}$ is a basis of $\text{ker }T$.

Answer

b. $B = \{x - 1, \dots, x^{n} - 1\}$ is independent (distinct degrees) and contained in $\text{ker }T$. Hence $B$ is a basis of $\text{ker }T$ by (a).

Exercise $\PageIndex{16}$

Use the dimension theorem to prove Theorem 1.3.1: If $A$ is an $m \times n$ matrix with $m < n$, the system $A\mathbf{x} = \mathbf{0}$ of $m$ homogeneous equations in $n$ variables always has a nontrivial solution.

Exercise $\PageIndex{17}$

Let $B$ be an $n \times n$ matrix, and consider the subspaces $U = \{A \mid A \mbox{ in }\|{M}_{mn}, AB = 0\}$ and $V = \{AB \mid A \mbox{ in }\|{M}_{mn}\}$. Show that $dim \;U + dim \;V = mn$.

Exercise $\PageIndex{18}$

Let $U$ and $V$ denote, respectively, the spaces of even and odd polynomials in $\mathbf{P}_{n}$. Show that $dim \;U + dim \;V = n + 1$. [Hint: Consider $T :\|{P}_{n} \to\|{P}_{n}$ where $T\left[p(x)\right] = p(x) - p(-x)$.]

Exercise $\PageIndex{19}$

Show that every polynomial $f(x)$ in $\mathbf{P}_{n-1}$ can be written as $f(x) = p(x + 1) - p(x)$ for some polynomial $p(x)$ in $\mathbf{P}_{n}$. [Hint: Define $T :\|{P}_{n} \to\|{P}_{n-1}$ by $T\left[p(x)\right] = p(x + 1) - p(x)$.]

Exercise $\PageIndex{20}$

Let $U$ and $V$ denote the spaces of symmetric and skew-symmetric $n \times n$ matrices. Show that $dim \;U + dim \;V = n^{2}$.

Answer

Define $T :{M}_{nn} \to {M}_{nn}$ by $T(A) = A - A^{T}$ for all $A$ in $\mathbf{M}_{nn}$. Then $\text{ker }T = U$ and $im \;T = V$ by Example [exa:021376], so the dimension theorem gives $n^{2} = dim \;\mathbf{M}_{nn} = dim \;(U) + dim \;(V)$.

Exercise $\PageIndex{21}$

Assume that $B$ in $\mathbf{M}_{nn}$ satisfies $B^{k} = 0$ for some $k \geq 1$. Show that every matrix in $\mathbf{M}_{nn}$ has the form $BA - A$ for some $A$ in $\mathbf{M}_{nn}$. [Hint: Show that $T :{M}_{nn} \to {M}_{nn}$ is linear and one-to-one where
$T(A) = BA - A$ for each $A$.]

Exercise $\PageIndex{22}$

Fix a column $\mathbf{y} \neq \mathbf{0}$ in $\mathbb{R}^n$ and let
$U = \{A$ in $\mathbf{M}_{nn} \mid A\mathbf{y} = \mathbf{0}\}$. Show that $dim \;U = n(n - 1)$.

Answer

Define $T :\|{M}_{nn} \to \mathbb{R}^n$ by $T(A) = A\mathbf{y}$ for all $A$ in $\mathbf{M}_{nn}$. Then $T$ is linear with $\text{ker }T = U$, so it is enough to show that $T$ is onto (then $dim \;U = n^{2} - dim \;(im \;T) = n^{2} - n$). We have $T(0) = \mathbf{0}$. Let $\mathbf{y} = \left[ \begin{array}{cccc} y_{1} & y_{2} & \cdots & y_{n} \end{array} \right]^{T} \neq \mathbf{0}$ in $\mathbb{R}^n$. If $y_{k} \neq \mathbf{0}$ let $\mathbf{c}_{k} = y_{k}^{-1}\mathbf{y}$, and let $\mathbf{c}_{j} = \mathbf{0}$ if $j \neq k$. If $A = \left[ \begin{array}{cccc} \mathbf{c}_{1} & \mathbf{c}_{2} & \cdots & \mathbf{c}_{n} \end{array} \right]$, then $T(A) = A\mathbf{y} = y_{1}\mathbf{c}_{1} + \cdots + y_{k}\mathbf{c}_{k} + \cdots + y_{n}\mathbf{c}_{n} = \mathbf{y}$. This shows that $T$ is onto, as required.

Exercise $\PageIndex{23}$

If $B$ in $\mathbf{M}_{mn}$ has $rank \;r$, let $U = \{A$ in $\mathbf{M}_{nn} \mid BA = 0\}$ and $W = \{BA \mid A$ in $\mathbf{M}_{nn}\}$. Show that $dim \;U = n(n - r)$ and $dim \;W = nr$. [Hint: Show that $U$ consists of all matrices $A$ whose columns are in the $null$ space of $B$. Use Example $\PageIndex{7}$.]

Exercise $\PageIndex{24}$

Let $T : V \to V$ be a linear transformation where $dim \;V = n$. If $\text{ker }T \cap im \;T = \{\mathbf{0}\}$, show that every vector $\mathbf{v}$ in $V$ can be written $\mathbf{v} = \mathbf{u} + \mathbf{w}$ for some $\mathbf{u}$ in $\text{ker }T$ and $\mathbf{w}$ in $im \;T$. [Hint: Choose bases $B \subseteq \text{ker }T$ and $D \subseteq im \;T$, and use Exercise 6.3.33.]

Exercise $\PageIndex{25}$

Let $T : \mathbb{R}^n \to \mathbb{R}^n$ be a linear operator of $rank \;1$, where $\mathbb{R}^n$ is written as rows. Show that there exist numbers $a_{1}, a_{2}, \dots, a_{n}$ and $b_{1}, b_{2}, \dots, b_{n}$ such that $T(X) = XA$ for all rows $X$ in $\mathbb{R}^n$, where

$$A = \left[ \begin{array}{cccc} a_{1}b_{1} & a_{1}b_{2} & \cdots & a_{1}b_{n} \\ a_{2}b_{1} & a_{2}b_{2} & \cdots & a_{2}b_{n} \\ \vdots & \vdots & & \vdots \\ a_{n}b_{1} & a_{n}b_{2} & \cdots & a_{n}b_{n} \end{array} \right] \nonumber$$

[Hint: $im \;T = \mathbb{R}\mathbf{w}$ for $\mathbf{w} = (b_{1}, \dots, b_{n})$ in $\mathbb{R}^n$.]

Exercise $\PageIndex{26}$

Prove Theorem $\PageIndex{5}$.

Exercise $\PageIndex{27}$

Let $T : V \to \mathbb{R}$ be a nonzero linear transformation, where $dim \;V = n$. Show that there is a basis $\{\mathbf{e}_{1}, \dots, \mathbf{e}_{n}\}$ of $V$ so that $T(r_{1}\mathbf{e}_{1} + r_{2}\mathbf{e}_{2} + \cdots + r_{n}\mathbf{e}_{n}) = r_{1}$.

Exercise $\PageIndex{28}$

Let $f \neq 0$ be a fixed polynomial of degree $m \geq 1$. If $p$ is any polynomial, recall that
$(p \circ f)(x) = p\left[f(x)\right]$. Define $T_{f} : P_{n} \to P_{n+m}$ by
$T_{f}(p) = p \circ f$.

1. Show that $T_{f}$ is linear.
2. Show that $T_{f}$ is one-to-one.

Exercise $\PageIndex{29}$

Let $U$ be a subspace of a finite dimensional vector space $V$.

1. Show that $U = \text{ker }T$ for some linear operator $T : V \to V$.

2. Show that $U = im \;S$ for some linear operator
   $S : V \to V$. [Hint: Theorem [thm:019430] and Theorem [thm:020916].]

Answer

b. By Lemma [lem:019415], let $\{\mathbf{u}_{1}, \dots, \mathbf{u}_{m}, \dots, \mathbf{u}_{n}\}$ be a basis of $V$ where $\{\mathbf{u}_{1}, \dots, \mathbf{u}_{m}\}$ is a basis of $U$. By Theorem [thm:020916] there is a linear transformation $S : V \to V$ such that $S(\mathbf{u}_{i}) = \mathbf{u}_{i}$ for $1 \leq i \leq m$, and $S(\mathbf{u}_{i}) = \mathbf{0}$ if $i > m$. Because each $\mathbf{u}_{i}$ is in $im \;S$, $U \subseteq im \;S$. But if $S(\mathbf{v})$ is in $im \;S$, write $\mathbf{v} = r_{1}\mathbf{u}_{1} + \cdots + r_{m}\mathbf{u}_{m} + \cdots + r_{n}\mathbf{u}_{n}$. Then $S(\mathbf{v}) = r_{1}S(\mathbf{u}_{1}) + \cdots + r_{m}S(\mathbf{u}_{m}) = r_{1}\mathbf{u}_{1} + \cdots + r_{m}\mathbf{u}_{m}$ is in $U$. So $im \;S \subseteq U$.

Exercise $\PageIndex{30}$

Let $V$ and $W$ be finite dimensional vector spaces.

1. Show that $dim \;W \leq dim \;V$ if and only if there exists an onto linear transformation $T : V \to W$. [Hint: Theorem [thm:019430] and Theorem [thm:020916].]
2. Show that $dim \;W \geq dim \;V$ if and only if there exists a one-to-one linear transformation $T : V \to W$. [Hint: Theorem [thm:019430] and Theorem [thm:020916].]

Answer

Add texts here. Do not delete this text first.

Let $A$ and $B$ be $n \times n$ matrices, and assume that $AXB=0$, $X \in\|{M}_{nn}$, implies $X=0$. Show that $A$ and $B$ are both invertible. [Hint: Dimension Theorem.]