6.10.2E: Kernel and Image of a Linear Transformation Exercises
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( \newcommand{\kernel}{\mathrm{null}\,}\)
Exercises
Exercise 6.10.2E.1
For each matrix A, find a basis for the kernel and image of TA, and find the rank and nullity of TA.
- [12−1131021−320]
- [21−13103111−42]
- [12−13124−1502−2]
- [2101−1312−303−6]
- Answer
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- {[−3710],[110−1]};{[101],[01−1]};2,2
- {[−121]};{[1011],[01−1−2]};2,1
Exercise 6.10.2E.2
In each case, (i) find a basis of ker T, and (ii) find a basis of imT. You may assume that T is linear.
- T:P2→R2; T(a+bx+cx2)=(a,b)
- T:P2→R2; T(p(x))=(p(0),p(1))
- T:R3→R3; T(x,y,z)=(x+y,x+y,0)
- T:R3→R4; T(x,y,z)=(x,x,y,y)
- T:M22→‖M22; T[abcd]=[a+bb+cc+dd+a]
- T:M22→R; T[abcd]=a+d
- T:Pn→R; T(r0+r1x+⋯+rnxn)=rn
- T:Rn→R; T(r1,r2,…,rn)=r1+r2+⋯+rn
- T:M22→‖M22; T(X)=XA−AX, where
- A=[0110]
- Answer
-
- {x2−x};{(1,0),(0,1)}
- {(0,0,1)};{(1,1,0,0),(0,0,1,1)}
- {[100−1],[0100],[0010]};{1}
- {(1,0,0,…,0,−1),(0,1,0,…,0,−1), …,(0,0,0,…,1,−1)};{1}
- {[0100],[0001]}{[1100],[0011]}
Exercise 6.10.2E.1
Let P:V→R and Q:V→R be linear transformations, where V is a vector space. Define T:V→R2 by T(v)=(P(v),Q(v)).
- Show that T is a linear transformation.
- Show that ker T=ker P∩ker Q, the set of vectors in both ker P and ker Q.
- Answer
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b. T(v)=0=(0,0) if and only if P(v)=0 and Q(v)=0; that is, if and only if v is in ker P∩ker Q.
Exercise 6.10.2E.4
In each case, find a basis
B={e1,…,er,er+1,…,en} of V such that {er+1,…,en} is a basis of ker T, and verify Theorem [thm:021572].
- T:R3→R4; T(x,y,z)=(x−y+2z,x+y−z,2x+z,2y−3z)
- T:R3→R4; T(x,y,z)=(x+y+z,2x−y+3z,z−3y,3x+4z)
- Answer
-
b. ker T=span{(−4,1,3)}; B={(1,0,0),(0,1,0),(−4,1,3)}, imT=span{(1,2,0,3),(1,−1,−3,0)}
Exercise 6.10.2E.5
Show that every matrix X in Mnn has the form X=AT−2A for some matrix A in Mnn. [Hint: The dimension theorem.]
Exercise 6.10.2E.6
In each case either prove the statement or give an example in which it is false. Throughout, let T:V→W be a linear transformation where V and W are finite dimensional.
- If V=W, then ker T⊆imT.
- If dimV=5, dimW=3, and dim(ker T)=2, then T is onto.
- If dimV=5 and dimW=4, then ker T≠{0}.
- If ker T=V, then W={0}.
- If W={0}, then ker T=V.
- If W=V, and imT⊆ker T, then T=0.
- If {e1,e2,e3} is a basis of V and T(e1)=0=T(e2), then dim(imT)≤1.
- If dim(ker T)≤dimW, then dimW≥12dimV.
- If T is one-to-one, then dimV≤dimW.
- If dimV≤dimW, then T is one-to-one.
- If T is onto, then dimV≥dimW.
- If dimV≥dimW, then T is onto.
- If {T(v1),…,T(vk)} is independent, then {v1,…,vk} is independent.
- If {v1,…,vk} spans V, then {T(v1),…,T(vk)} spans W.
- Answer
-
- Yes. dim(imT)=5−dim(ker T)=3, so imT=W as dimW=3.
- No. T=0:R2→R2
- No. T:R2→R2, T(x,y)=(y,0). Then ker T=imT
- Yes. dimV=dim(ker T)+dim(imT)≤dimW+dimW=2dimW
- No. Consider T:R2→R2 with T(x,y)=(y,0).
- No. Same example as (j).
- No. Define T:R2→R2 by T(x,y)=(x,0). If v1=(1,0) and v2=(0,1), then R2=span{v1,v2} but R2≠span{T(v1),T(v2)}.
Exercise 6.10.2E.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→W is a linear transformation, show that:
- If T is one-to-one and {v1,…,vn} is independent in V, then {T(v1),…,T(vn)} is independent in W.
- If T is onto and V=span{v1,…,vn}, then W=span{T(v1),…,T(vn)}.
- Answer
-
b. Given w in W, let w=T(v), v in V, and write v=r1v1+⋯+rnvn. Then w=T(v)=r1T(v1)+⋯+rnT(vn).
Exercise 6.10.2E.8
Given {v1,…,vn} in a vector space V, define T:Rn→V by T(r1,…,rn)=r1v1+⋯+rnvn. Show that T is linear, and that:
- T is one-to-one if and only if {v1,…,vn} is independent.
- T is onto if and only if V=span{v1,…,vn}.
- Answer
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b. imT={∑irivi∣ri in R}=span{vi}.
Exercise 6.10.2E.9
Let T:V→V be a linear transformation where V is finite dimensional. Show that exactly one of (i) and (ii) holds:
(i) T(v)=0 for some v≠0 in V;
(ii) T(x)=v has a solution x in V for every v in V.
Exercise 6.10.2E.10
Let T:‖Mnn→R denote the trace map: T(A)=trA for all A in Mnn. Show that
dim(ker T)=n2−1.
- Answer
-
T is linear and onto. Hence 1=dimR=dim(imT)=dim(Mnn)−dim(ker T)=n2−dim(ker T)
Exercise 6.10.2E.11
Show that the following are equivalent for a linear transformation T:V→W.
ker T=V
imT={0}
T=0
Exercise 6.10.2E.12
Let A and B be m×n and k×n matrices, respectively. Assume that Ax=0 implies Bx=0 for every n-column x. Show that rankA≥rankB.
- Answer
-
The condition means ker (TA)⊆ker(TB), so dim[ker(TA)]≤dim[ker(TB)]. Then Theorem [thm:021499] gives dim[im(TA)]≥dim[im(TB)]; that is, rankA≥rankB.
Exercise 6.10.2E.13
Let A be an m×n matrix of rankr. Thinking of Rn as rows, define V={x in Rm∣xA=0}. Show that dimV=m−r.
Exercise 6.10.2E.14
Consider
V={[abcd]|a+c=b+d}
- Consider S:‖M22→R with S[abcd]=a+c−b−d. Show that S is linear and onto and that V is a subspace of M22. Compute dimV.
- Consider T:V→R with T[abcd]=a+c. Show that T is linear and onto, and use this information to compute dim(ker T).
Exercise 6.10.2E.15
Define T:‖Pn→R by T[p(x)]= the sum of all the coefficients of p(x).
- Use the dimension theorem to show that dim(ker T)=n.
- Conclude that {x−1,x2−1,…,xn−1} is a basis of ker T.
- Answer
-
b. B={x−1,…,xn−1} is independent (distinct degrees) and contained in ker T. Hence B is a basis of ker T by (a).
Exercise 6.10.2E.16
Use the dimension theorem to prove Theorem 1.3.1: If A is an m×n matrix with m<n, the system Ax=0 of m homogeneous equations in n variables always has a nontrivial solution.
Exercise 6.10.2E.17
Let B be an n×n matrix, and consider the subspaces U={A∣A in ‖Mmn,AB=0} and V={AB∣A in ‖Mmn}. Show that dimU+dimV=mn.
Exercise 6.10.2E.18
Let U and V denote, respectively, the spaces of even and odd polynomials in Pn. Show that dimU+dimV=n+1. [Hint: Consider T:‖Pn→‖Pn where T[p(x)]=p(x)−p(−x).]
Exercise 6.10.2E.19
Show that every polynomial f(x) in Pn−1 can be written as f(x)=p(x+1)−p(x) for some polynomial p(x) in Pn. [Hint: Define T:‖Pn→‖Pn−1 by T[p(x)]=p(x+1)−p(x).]
Exercise 6.10.2E.20
Let U and V denote the spaces of symmetric and skew-symmetric n×n matrices. Show that dimU+dimV=n2.
- Answer
-
Define T:Mnn→Mnn by T(A)=A−AT for all A in Mnn. Then ker T=U and imT=V by Example [exa:021376], so the dimension theorem gives n2=dimMnn=dim(U)+dim(V).
Exercise 6.10.2E.21
Assume that B in Mnn satisfies Bk=0 for some k≥1. Show that every matrix in Mnn has the form BA−A for some A in Mnn. [Hint: Show that T:Mnn→Mnn is linear and one-to-one where
T(A)=BA−A for each A.]
Exercise 6.10.2E.22
Fix a column y≠0 in Rn and let
U={A in Mnn∣Ay=0}. Show that dimU=n(n−1).
- Answer
-
Define T:‖Mnn→Rn by T(A)=Ay for all A in Mnn. Then T is linear with ker T=U, so it is enough to show that T is onto (then dimU=n2−dim(imT)=n2−n). We have T(0)=0. Let y=[y1y2⋯yn]T≠0 in Rn. If yk≠0 let ck=y−1ky, and let cj=0 if j≠k. If A=[c1c2⋯cn], then T(A)=Ay=y1c1+⋯+ykck+⋯+yncn=y. This shows that T is onto, as required.
Exercise 6.10.2E.23
If B in Mmn has rankr, let U={A in Mnn∣BA=0} and W={BA∣A in Mnn}. Show that dimU=n(n−r) and dimW=nr. [Hint: Show that U consists of all matrices A whose columns are in the null space of B. Use Example 6.10.2E.7.]
Exercise 6.10.2E.24
Let T:V→V be a linear transformation where dimV=n. If ker T∩imT={0}, show that every vector v in V can be written v=u+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.
- Show that T_{f} is linear.
- Show that T_{f} is one-to-one.
Exercise \PageIndex{29}
Let U be a subspace of a finite dimensional vector space V.
- Show that U = \text{ker }T for some linear operator T : V \to V.
-
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.
- 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].]
- 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
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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.]