[PDF] 1 Let T : P 3(R) → P 3(R) be given by T(α3x3 +

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will hopefully give you some more practice with actual examples Give an example of a linear map T R4 → R4 such that null(T) = range(T) Can you

[PDF] 1 Let T : P 3(R) → P 3(R) be given by T(α3x3 + - Sites at Lafayette

(a) Is x3 − 5x2 + 3x − 6 in null (T)? Explain why why not Solution No thus one choice for a basis for null (TM ) is ( −5 2 1 Find an example of a linear transformation T R4 → R4 so that null (T) = range (T) Example For any x

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The linear map T V → W is called injective (one to one) if, for and the rank of T are the dimensions of null(T) and range(T), So the columns of A span R5 Example For a 9 × 12 matrix A, find the smallest possible value of dim Nul A

[PDF] Math 131 - HW 5 - UCR Maths

(a) Find an example of a linear map S R4 → R4 such that range(S) = null(S) (b) Show that there is NO linear map T R5 → R5 such that range(T) = null(T) Give a proof or counterexample If T V → W is a linear map and v1,,vn spans V ,

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Mar 22, 2012 · (b) I must show that for each u 2 R2 there exists v 2 R2 such that T(v) = u The space spanned by the zero vector is zero dimensional so dim(null(T)) = 0 (d) The matrix representation of a linear transformation is the matrix whose columns are the images 64 Give an example of a function f R2 R having

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there exist 1 dimensional subspaces U1,,Un of V so that V = U1 ⊕···⊕ Un 2 B5) Give an example of a linear map T R4 → R4 so that rangeT = nullT 1

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Homework 9 Key

Let T:P3(R)! P3(R) be given by

T(3x3+2x2+1x+0) = 21x3+ (3+2)x+ (1+0):

(a)

Is x35x2+ 3x6 in null(T)? Explain why/why not.

Solution: No, because

T(x35x2+ 3x6) = 6x34x236=0:

(b)

Is 4 x34x2in null(T)? Explain why/why not.

Solution: Yes, because

T(4x34x2) = (44)x=0:

(c)

Is 8 x3x1 in range(T)? Explain why/why not.

Solution: Yes, because

T(x3+ 4x5) = 8x3x1:

(d)

Is 4 x33x2+ 7 in range(T)? Explain why/why not.

Solution: No, because no vector whosex2component has nonzero coecient is in the range ofT. 2.

Giv en

M=3 2 11

2 1 8 deneTM:R3!R2by T

M(v) =Mv:

(a)

Find the rank of M.

Solution: The RREF ofMis1 0 5

0 12 since this matrix has 2 leading 1s, its rank is 2. (b)

Find a basis for the n ullspace of TM.

Solution: Solutions toMx=0may be parameterized as

x=s0 @5 2 11 A thus one choice for a basis for null(TM) is 0 @5 2 11 A 1

Homework 9 Key

(c)

Find a basis for the range of TM.

Solution: Row reducing the augmented matrix for the system, we have

3 2 11jv1

2 1 8jv2

!1 0 5j2v2v1

0 12j2v13v2

This system is always consistent, so the range ofTMis all ofR2; thus we may choose any basis we like forR2, say1 0 ;0 1 (d)

V erifythe F undamentalTheorem for TM.

Solution: The dimension of null(TM) = 1, and dimension of range(TM) = 2; we have dim(null(TM)) + dim(range(TM)) = 3 = dim(R3): 3.

Dene T:M3(R)! M3(R) by

T(X) =XX>:

(a)

Find a b asisfor the n ullspace of T.

Solution: IfXX>=0;then we haveX=X>, that isXis symmetric. Thus one choice of basis for null(T) is 0 @1 0 0 0 0 0

0 0 01

A ;0 @0 1 0 1 0 0

0 0 01

A ;0 @0 0 1 0 0 0

1 0 01

A ;0 @0 0 0 0 1 0

0 0 01

A ;0 @0 0 0 0 0 1

0 1 01

A ;0 @0 0 0 0 0 0

0 0 11

A (b)

Find a basis for the range of T.

Solution: IfV=XX>, it is clear that

V >= (XX>)> =X+X> =(XX>) =V: Thus every vector in range(T) is skew symmetric, and a basis for range(T) is 0 @0 1 0 1 0 0

0 0 01

A ;0 @0 0 1 0 0 0

1 0 01

A ;0 @0 0 0 0 0 1 01 01 A (c)

V erifythe F undamentalTheorem for T.

Solution: We have dim(null(T)) = 6, dim(range(T)) = 3, and dim(M3(R)) = 9. So clearly dim(null(T)) + dim(range(T)) = dim(M3(R)): 2

Homework 9 Key

4. Find an example of a l ineartransformation T:R4!R4so that null(T) = range(T). Example: For anyx2R4,T(x)2range(T). Thus the stipulation null(T) = range(T) implies that

T(T(x)) =0

for allx2R4. One possible way to build such an operator is T 0 B B@x 1 x 2 x 3 x 41
C CA =0 B B@0 0 x 1 x 21
C CA: It is clear thatTis indeed a linear transformation, and x2range(T)()x=0 B B@0 0 v w1 C

CA()T(x) =0:

A linear transfor mationT:P2(R)! P2(R) has matrix

A=A(B;C)=0

@13 0 4 131

825 21

A with respect to some basesBandCofP2(R). (a)

Is Tinjective? Explain why/why not.

Solution: IfTis injective, then (T) =f0g, that isT(v) =0()v=0.

NowT(v) =0()(T(v))C=0, and since

i.A(v)B= (T(v))C;and ii. ev eryx2R3is the coordinate vector for somep2 P2, we may translate the observation into one on matrix equations:Ax=0has only the trivial solution if and only if det(A)6= 0. Now the determinant ofAis det(A) = 1, so the only solution toA(v)B=0is the trivial one, (v)B=0. ThusTis indeed injective (b)

Is Tsurjective? Explain why/why not.

Solution: A linear operator is injective if and only if it is also surjective, soTis surjective. 6. Supp osethat R,S, andTare linear operators onVso thatRSTis surjective. Prove thatS is injective. Solution: SinceRSTis surjective, we know that for everyw2V, there is av2Vso that RST(v) =w. SinceRST(v) =R(S(T(v))), there is also a vector (namelyu:=S(T(v))) so 3

Homework 9 Key

thatR(u) =w. Thus the operatorRis surjective as well; this is equivalent toRinjective. Thus for everyw2V, there is precisely onev2Vso thatR(v) =w. ThusSTmust be a surjective map, and using the same reasoning as above,Sis surjective as well, and thus injective. 7. ( Deleted) Recall Theorem 5:10: LetT:V!Vbe a linear operator on the nite dimensional vector spaceV. If1, ..., naredistincteigenvalues ofT, and ifv1, ...,vnare vectors so thatviis an eigenvector associated withi, then the list (v1; v2; :::; vn) is an independent list. Fill in the details of the following sketch of the proof: Proceed by induction: show that, ifv1andv2are eigenvectors forTand are also dependent, then they must be associated with the same eigenvalue. For the inductive hypothesis, letv1, ...,vnbe any eigenvectors associated with unique eigenvalues, so that (v1; :::; vn) is an independent list. Letvn+1be any eigenvector of Tin span(v1; :::; vn), and show thatvn+1must be associated with one of the eigenvalues

1, ...,n.

4quotesdbs_dbs11.pdfusesText_17

[PDF] [PDF] 1 Let T : P 3(R) → P 3(R) be given by T(α3x3 + - Sites at Lafayette

Homework 9 Key

Let T:P3(R)! P3(R) be given by

T(3x3+2x2+1x+0) = 21x3+ (3+2)x+ (1+0):

Is x35x2+ 3x6 in null(T)? Explain why/why not.

Solution: No, because

T(x35x2+ 3x6) = 6x34x236=0:

Is 4 x34x2in null(T)? Explain why/why not.

Solution: Yes, because

T(4x34x2) = (44)x=0:

Is 8 x3x1 in range(T)? Explain why/why not.

Solution: Yes, because

T(x3+ 4x5) = 8x3x1:

Is 4 x33x2+ 7 in range(T)? Explain why/why not.

Giv en

M=3 2 11

M(v) =Mv:

Find the rank of M.

Solution: The RREF ofMis1 0 5

Find a basis for the n ullspace of TM.

Solution: Solutions toMx=0may be parameterized as

Homework 9 Key

Find a basis for the range of TM.

3 2 11jv1

2 1 8jv2

0 12j2v13v2

V erifythe F undamentalTheorem for TM.

Dene T:M3(R)! M3(R) by

T(X) =XX>:

Find a b asisfor the n ullspace of T.

0 0 01

0 0 01

1 0 01

0 0 01

0 1 01

0 0 11

Find a basis for the range of T.

Solution: IfV=XX>, it is clear that

0 0 01

1 0 01

V erifythe F undamentalTheorem for T.

Homework 9 Key

T(T(x)) =0

CA()T(x) =0:

A linear transfor mationT:P2(R)! P2(R) has matrix

A=A(B;C)=0

825 21

Is Tinjective? Explain why/why not.

NowT(v) =0()(T(v))C=0, and since

Is Tsurjective? Explain why/why not.

Homework 9 Key

1, ...,n.