4 Images, Kernels, and Subspaces

In our study of linear transformations we've examined some of the conditions under which a transformation is invertible. Now we're ready to investigate some ideas similar to invertibility. Namely, we would like to measure the ways in which a transformation that is not invertible fails to have an inverse.

4.1 The Image and Kernel of a Linear Transformation

Denition.Theimageof a function consists of all the values the function assumes. If f:X!Yis a function fromXtoY, then im(f) =ff(x) :x2Xg:

Notice that im(f) is a subset ofY.

Denition.Thekernelof a function whose range isRnconsists of all the values in its domain at which the function assumes the value0. Iff:X!Rnis a function fromXto R n, then ker(f) =fx2X:f(x) = 0g: Notice that ker(f) is a subset ofX. Also, ifT(x) =Axis a linear transformation fromRm toRn, then ker(T) (also denoted ker(A)) is the set of solutions to the equationAx=0. The kernel gives us some new ways to characterize invertible matrices. Theorem 1.LetAbe annnmatrix. Then the following statements are equivalent.

1.Ais invertible.

2. The linear system Ax=bhas a unique solutionxfor everyb2Rn. 3. rref (A) =In. 4. rank (A) =n. 5. im( A) =Rn. 6. k er(A) =f0g. Example 13.(x3.1, Exercise 39 of [1]) Consider annpmatrixAand apmmatrixB. (a) Wha tis the relationship b etweenk er(AB) and ker(B)? Are they always equal? Is one of them always contained in the other? (b) Wh atis the rela tionshipb etweenim( A) and im(AB)? (Solution) 22
(a)Rec allthat k er(AB) is the set of vectorsx2Rmfor whichABx= 0, and similarly that ker(B) is the set of vectorsx2Rmfor whichBx= 0. Now ifxis in ker(B), then Bx= 0, soABx= 0. This means thatxis in ker(AB), so we see that ker(B) must always be contained in ker(AB). On the other hand, ker(AB) might not be a subset of ker(B). For instance, suppose that A=0 0 0 0 andB=1 0 0 1 ThenBis the identity matrix, so ker(B) =f0g. But every vector has image zero under AB, so ker(AB) =R2. Certainly ker(B) does not contain ker(AB) in this case. (b) S upposeyis in the image ofAB. Theny=ABxfor somex2Rm. That is, y=ABx=A(Bx); soyis the image ofBxunder multiplication byA, and is thus in the image ofA. So im(A) contains im(AB). On the other hand, consider A=1 0 0 0 andB=0 0 0 1 Then

AB=0 0

0 0 so im(AB) =f0g, but the image ofAis the span of the vector1 0 . So im(AB) does not necessarily contain im(A). Example 14.(x3.1, Exercise 48 of [1]) Consider a 22 matrixAwithA2=A. (a) If wis in the image ofA, what is the relationship betweenwandAw? (b) Wha tcan y ousa yab outAif rank(A) = 2? What if rank(A) = 0? (c) I fran k(A) = 1, show that the linear transformationT(x) =Axis the projection onto im(A) along ker(A). (Solution) (a) If wis in the image ofA, thenw=Avfor somev2R2. Then

Aw=A(Av) =A2v=Av=w;

sinceA2=A. SoAw=w. 23
(b)If rank( A) = 2, thenAis invertible. SinceA2=A, we see that

A=I2A= (A1A)A=A1A2=A1A=I2:

So the only rank 2 22 matrix with the property thatA2=Ais the identity matrix. On the other hand, if rank(A) = 0 thenAmust be the zero matrix. (c) I frank( A) = 1, thenAis not invertible, so ker(A)6=f0g. But we also know thatA is not the zero matrix, so ker(A)6=R2. We conclude that ker(A) must be a line in R

2. Next, suppose we havew2ker(A)\im(A). ThenAw= 0 and, according to part

(a) ,Aw=w. Sowis the zero vector, meaning that ker(A)\im(A) =f0g. Since im(A) is neither 0 nor all ofR2, it also must be a line inR2. So ker(A) and im(A) are non-parallel lines inR2. Now choosex2R2and letw=xAx. Notice that Aw=AxA2x= 0, sow2ker(A). Then we may writexas the sum of an element of im(A) and an element of ker(A): x=Ax+w: According to Exercise 2.2.33, the mapT(x) =Axis then the projection onto im(A) along ker(A). Example 15.(x3.1, Exercise 50 of [1]) Consider a square matrixAwith ker(A2) = ker(A3).

Is ker(A3) = ker(A4)? Justify your answer.

(Solution)Su pposex2ker(A3). ThenA3x=0, so A

4x=A(A3x) =A0=0;

meaning thatx2ker(A4). So ker(A3) is contained in ker(A4). On the other hand, suppose x2ker(A4). ThenA4x=0, soA3(Ax) =0. This means thatAxis in the kernel ofA3, and thus in ker(A2). So A

3x=A2(Ax) =0;

meaning thatx2ker(A3). So ker(A4) is contained in ker(A3). Since each set contains the other, the two are equal: ker(A3) = ker(A4).

4.2 Subspaces

Denition.A subsetWof the vector spaceRnis called asubspaceofRnif it (i) c ontainsthe zero v ector; (ii) is closed under v ectorad dition; (iii) is closed under sca larm ultiplication. 24
One important observation we can immediately make is that for anynmmatrixA, ker(A) is a subspace ofRmand im(A) is a subspace ofRn. Denition.Suppose we have vectorsv1;:::;vminRn. We say that a vectorviisre- dundantifviis a linear combination of the preceding vectorsv1;:::;vi1. We say that the set of vectorsv1;:::;vmislinearly independentif none of them is redundant, and linearly dependentotherwise. If the vectorsv1;:::;vmare linearly independent and span a subspaceVofRn, we say thatv1;:::;vmform abasisofV. Example 16.(x3.2, Exercise 26 of [1]) Find a redundant column vector of the following matrix and write it as a linear combination of the preceding columns. Use this representation to write a nontrivial relation among the columns, and thus nd a nonzero vector in the kernel ofA. A=2

41 3 6

1 2 5

1 1 43

5 (Solution)First w enotice that 3 2 41
1 13 5 +2 43
2 13 5 =2 46
5 43
5 meaning that the third vector ofAis redundant. This allows us to write a nontrivial relation 3 2 41
1 13 5 + 12 43
2 13 5 12 46
5 43
5 =2 40
0 03 5 among the vectors. Finally, these coecients give us a nonzero element of ker(A), since 2

41 3 6

1 2 5

1 1 43

52
43
1 13 5 = 32 41
1 13 5 + 12 43
2 13 5 12 46
5 43
5 =2 40
0 03 5 Example 17.(x3.2, Exercise 53 of [1]) Consider a subspaceVofRn. We dene theorthog- onal complementV?ofVas the set of those vectorswinRnthat are perpendicular to all vectors inV; that is,wv= 0, for allvinV. Show thatV?is a subspace ofRn. (Solution)W eha vet hreeprop ertiesto c heck:that V?contains the zero vector, that it is closed under addition, and that it is closed under scalar multiplication. Certainly0v= 0 for everyv2V, so02V?. Next, suppose we have vectorsw1andw2inV?. Then (w1+w2)v=w1v+w2v= 0 + 0; 25
sincew1v= 0 andw2v= 0. Sow1+w2is inV?, meaning thatV?is closed under addition. Finally, suppose we havewinV?and a scalark. Then (kw)v=k(wv) = 0; sokw2V?. SoV?is closed under scalar addition, and is thus a subspace ofRn. Example 18.(x3.2, Exercise 54 of [1]) Consider the lineLspanned by2 41
2 33
5 inR3. Find a basis ofL?. See Exercise 53. (Solution)Su pposev, with componentsv1;v2;andv3, is inL?. Then 0 = 2 4v 1 v 2 v 33
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