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Mathematics LibreTexts

6.10.2E: Kernel and Image of a Linear Transformation Exercises

( \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.

  1. [121131021320]
  2. [211310311142]
  3. [121312415022]
  4. [210113123036]
Answer
  1. {[3710],[1101]};{[101],[011]};2,2
  2. {[121]};{[1011],[0112]};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.

  1. T:P2R2; T(a+bx+cx2)=(a,b)
  2. T:P2R2; T(p(x))=(p(0),p(1))
  3. T:R3R3; T(x,y,z)=(x+y,x+y,0)
  4. T:R3R4; T(x,y,z)=(x,x,y,y)
  5. T:M22M22; T[abcd]=[a+bb+cc+dd+a]
  6. T:M22R; T[abcd]=a+d
  7. T:PnR; T(r0+r1x++rnxn)=rn
  8. T:RnR; T(r1,r2,,rn)=r1+r2++rn
  9. T:M22M22; T(X)=XAAX, where
  10. A=[0110]
Answer
  1. {x2x};{(1,0),(0,1)}
  2. {(0,0,1)};{(1,1,0,0),(0,0,1,1)}
  3. {[1001],[0100],[0010]};{1}
  4. {(1,0,0,,0,1),(0,1,0,,0,1), ,(0,0,0,,1,1)};{1}
  5. {[0100],[0001]}{[1100],[0011]}
Exercise 6.10.2E.1

Let P:VR and Q:VR be linear transformations, where V is a vector space. Define T:VR2 by T(v)=(P(v),Q(v)).

  1. Show that T is a linear transformation.
  2. Show that ker T=ker Pker Q, the set of vectors in both ker P and ker Q.
Answer

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 Pker 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].

  1. T:R3R4; T(x,y,z)=(xy+2z,x+yz,2x+z,2y3z)
  2. T:R3R4; T(x,y,z)=(x+y+z,2xy+3z,z3y,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=AT2A 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:VW be a linear transformation where V and W are finite dimensional.

  1. If V=W, then ker TimT.
  2. If dimV=5, dimW=3, and dim(ker T)=2, then T is onto.
  3. If dimV=5 and dimW=4, then ker T{0}.
  4. If ker T=V, then W={0}.
  5. If W={0}, then ker T=V.
  6. If W=V, and imTker T, then T=0.
  7. If {e1,e2,e3} is a basis of V and T(e1)=0=T(e2), then dim(imT)1.
  8. If dim(ker T)dimW, then dimW12dimV.
  9. If T is one-to-one, then dimVdimW.
  10. If dimVdimW, then T is one-to-one.
  11. If T is onto, then dimVdimW.
  12. If dimVdimW, then T is onto.
  13. If {T(v1),,T(vk)} is independent, then {v1,,vk} is independent.
  14. If {v1,,vk} spans V, then {T(v1),,T(vk)} spans W.
Answer
  1. Yes. dim(imT)=5dim(ker T)=3, so imT=W as dimW=3.
  2. No. T=0:R2R2
  3. No. T:R2R2, T(x,y)=(y,0). Then ker T=imT
  4. Yes. dimV=dim(ker T)+dim(imT)dimW+dimW=2dimW
  5. No. Consider T:R2R2 with T(x,y)=(y,0).
  6. No. Same example as (j).
  7. No. Define T:R2R2 by T(x,y)=(x,0). If v1=(1,0) and v2=(0,1), then R2=span{v1,v2} but R2span{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:VW is a linear transformation, show that:

  1. If T is one-to-one and {v1,,vn} is independent in V, then {T(v1),,T(vn)} is independent in W.
  2. 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:RnV by T(r1,,rn)=r1v1++rnvn. Show that T is linear, and that:

  1. T is one-to-one if and only if {v1,,vn} is independent.
  2. T is onto if and only if V=span{v1,,vn}.
Answer

b. imT={iriviri in R}=span{vi}.

Exercise 6.10.2E.9

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

(i) T(v)=0 for some v0 in V;

(ii) T(x)=v has a solution x in V for every v in V.

Exercise 6.10.2E.10

Let T:MnnR denote the trace map: T(A)=trA for all A in Mnn. Show that
dim(ker T)=n21.

Answer

T is linear and onto. Hence 1=dimR=dim(imT)=dim(Mnn)dim(ker T)=n2dim(ker T)

Exercise 6.10.2E.11

Show that the following are equivalent for a linear transformation T:VW.

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 rankArankB.

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, rankArankB.

Exercise 6.10.2E.13

Let A be an m×n matrix of rankr. Thinking of Rn as rows, define V={x in RmxA=0}. Show that dimV=mr.

Exercise 6.10.2E.14

Consider

V={[abcd]|a+c=b+d}

  1. Consider S:M22R with S[abcd]=a+cbd. Show that S is linear and onto and that V is a subspace of M22. Compute dimV.
  2. Consider T:VR 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:PnR by T[p(x)]= the sum of all the coefficients of p(x).

  1. Use the dimension theorem to show that dim(ker T)=n.
  2. Conclude that {x1,x21,,xn1} is a basis of ker T.
Answer

b. B={x1,,xn1} 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={AA in Mmn,AB=0} and V={ABA 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:PnPn where T[p(x)]=p(x)p(x).]

Exercise 6.10.2E.19

Show that every polynomial f(x) in Pn1 can be written as f(x)=p(x+1)p(x) for some polynomial p(x) in Pn. [Hint: Define T:PnPn1 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:MnnMnn by T(A)=AAT 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 k1. Show that every matrix in Mnn has the form BAA for some A in Mnn. [Hint: Show that T:MnnMnn is linear and one-to-one where
T(A)=BAA for each A.]

Exercise 6.10.2E.22

Fix a column y0 in Rn and let
U={A in MnnAy=0}. Show that dimU=n(n1).

Answer

Define T:MnnRn 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=n2dim(imT)=n2n). We have T(0)=0. Let y=[y1y2yn]T0 in Rn. If yk0 let ck=y1ky, and let cj=0 if jk. If A=[c1c2cn], 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 MnnBA=0} and W={BAA in Mnn}. Show that dimU=n(nr) 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:VV be a linear transformation where dimV=n. If ker TimT={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.

  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.]


6.10.2E: Kernel and Image of a Linear Transformation Exercises is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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