9.E: Exercises
- Last updated
- Feb 27, 2022
- Save as PDF
- Page ID
- 98198
( \newcommand{\kernel}{\mathrm{null}\,}\)
Exercise 9.E.1
Suppose you have R2 and the + operation is as follows: (a,b)+(c,d)=(a+d,b+c). Scalar multiplication is defined in the usual way. Is this a vector space? Explain why or why not.
Exercise 9.E.2
Suppose you have R2 and the + operation is as follows: (a,b)+(c,d)=(0,b+d) Scalar multiplication is defined in the usual way. Is this a vector space? Explain why or why not.
Exercise 9.E.3
Suppose you have R2 and scalar multiplication is defined as c(a,b)=(a,cb) while vector addition is defined as usual. Is this a vector space? Explain why or why not.
Exercise 9.E.4
Suppose you have R2 and the + operation is defined as follows. (a,b)+(c,d)=(a−c,b−d) Scalar multiplication is same as usual. Is this a vector space? Explain why or why not.
Exercise 9.E.6
Denote by RN the set of real valued sequences. For →a≡{an}∞n=1, →b≡{bn}∞n=1 two of these, define their sum to be given by →a+→b={an+bn}∞n=1 and define scalar multiplication by c→a={can}∞n=1 where →a={a+n}∞n=1 Is this a special case of Exercise 9.E.5? Is this a vector space?
Exercise 9.E.7
Let C2 be the set of ordered pairs of complex numbers. Define addition and scalar multiplication in the usual way. (z,w)+(ˆz,ˆw)=(z+ˆz,w+ˆw),u(z,w)≡(uz,uw) Here the scalars are from C. Show this is a vector space.
Exercise 9.E.8
Let V be the set of functions defined on a nonempty set which have values in a vector space W. Is this a vector space? Explain.
Exercise 9.E.9
Consider the space of m×n matrices with operation of addition and scalar multiplication defined the usual way. That is, if A,B are two m×n matrices and c a scalar, (A+B)ij=Aij+Bij,(cA)ij≡c(Aij)
Exercise 9.E.10
Consider the set of n×n symmetric matrices. That is, A=AT. In other words, Aij=Aji. Show that this set of symmetric matrices is a vector space and a subspace of the vector space of n×n matrices.
Exercise 9.E.11
Consider the set of all vectors in R2,(x,y) such that x+y≥0. Let the vector space operations be the usual ones. Is this a vector space? Is it a subspace of R2?
Exercise 9.E.12
Consider the vectors in R2,(x,y) such that xy=0. Is this a subspace of R2? Is it a vector space? The addition and scalar multiplication are the usual operations.
Exercise 9.E.13
Define the operation of vector addition on R2 by (x,y)+(u,v)=(x+u,y+v+1). Let scalar multiplication be the usual operation. Is this a vector space with these operations? Explain.
Exercise 9.E.14
Let the vectors be real numbers. Define vector space operations in the usual way. That is x+y means to add the two numbers and xy means to multiply them. Is R with these operations a vector space? Explain.
Exercise 9.E.15
Let the scalars be the rational numbers and let the vectors be real numbers which are the form a+b√2 for a,b rational numbers. Show that with the usual operations, this is a vector space.
Exercise 9.E.16
Let P2 be the set of all polynomials of degree 2 or less. That is, these are of the form a+bx+cx2. Addition is defined as (a+bx+cx2)+(ˆd+ˆbx+ˆcx2)=(a+ˆa)+(b+ˆb)x+(c+ˆc)x2 and scalar multiplication is defined as d(a+bx+cx2)=da+dbx+cdx2 Show that, with this definition of the vector space operations that P2 is a vector space. Now let V denote those polynomials a+bx+cx2 such that a+b+c=0. Is V a subspace of P2? Explain.
Exercise 9.E.17
Let M,N be subspaces of a vector space V and consider M+N defined as the set of all m+n where m∈M and n∈N. Show that M+N is a subspace of V.
Exercise 9.E.18
Let M,N be subspaces of a vector space V. Then M∩N consists of all vectors which are in both M and N. Show that M∩N is a subspace of V.
Exercise 9.E.19
Let M,N be subspaces of a vector space R2. Then N∪M consists of all vectors which are in either M or N. Show that N∪M is not necessarily a subspace of R2 by giving an example where N∪M fails to be a subspace.
Exercise 9.E.20
Let X consist of the real valued functions which are defined on an interval [a,b]. For f,g∈X,f+g is the name of the function which satisfies (f+g)(x)=f(x)+g(x). For s a real number, (sf)(x)=s(f(x)). Show this is a vector space.
- Answer
-
The axioms of a vector space all hold because they hold for a vector space. The only thing left to verify is the assertions about the things which are supposed to exist. 0 would be the zero function which sends everything to 0. This is an additive identity. Now if f is a function, −f(x)≡(−f(x)). Then (f+(−f))(x)≡f(x)+(−f)(x)≡f(x)+(−f(x))=0 Hence f+−f=0. For each x∈[a,b], let fx(x)=1 and fx(y)=0 if y≠x. Then these vectors are obviously linearly independent.
Exercise 9.E.21
Consider functions defined on {1,2,⋯,n} having values in R. Explain how, if V is the set of all such functions, V can be considered as Rn.
- Answer
-
Let f(i) be the ith component of a vector →x∈Rn. Thus a typical element in Rn is (f(1),⋯,f(n)).
Exercise 9.E.22
Let the vectors be polynomials of degree no more than 3. Show that with the usual definitions of scalar multiplication and addition wherein, for p(x) a polynomial, (ap)(x)=ap(x) and for p,q polynomials (p+q)(x)=p(x)+q(x), this is a vector space.
- Answer
-
This is just a subspace of the vector space of functions because it is closed with respect to vector addition and scalar multiplication. Hence this is a vector space.
Exercise 9.E.23
Let V be a vector space and suppose {→x1,⋯,→xl} is a set of vectors in V. Show that →0 is in span{→x1,⋯,→xk}.
- Answer
-
k∑i=10→xk=→0
Exercise 9.E.24
Determine if p(x)=4x2−x is in the span given by span{x2+x,x2−1,−x+2}
Exercise 9.E.25
Determine if p(x)=−x2+x+2 is in the span given by span{x2+x+1,2x2+x}
Exercise 9.E.27
Show that the spanning set in Exercise 9.E.26 is a spanning set for M22, the vector space of all 2×2 matrices.
Exercise 9.E.28
Consider the vector space of polynomials of degree at most 2, P2. Determine whether the following is a basis for P2. {x2+x+1,2x2+2x+1,x+1} Hint: There is a isomorphism from R3 to P2. It is defined as follows: T→e1=1,T→e2=x,T→e3=x2 Then extend T linearly. Thus T[111]=x2+x+1,T[122]=2x2+2x+1,T[110]=1+x It follows that if {[111],[122],[110]} is a basis for R3, then the polynomials will be a basis for P2 because they will be independent. Recall that an isomorphism takes a linearly independent set to a linearly independent set. Also, since T is an isomorphism, it preserves all linear relations.
Exercise 9.E.29
Find a basis in P2 for the subspace span{1+x+x2,1+2x,1+5x−3x2} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others. Hint: This is the situation in which you have a spanning set and you want to cut it down to form a linearly independent set which is also a spanning set. Use the same isomorphism above. Since T is an isomorphism, it preserves all linear relations so if such can be found in R3, the same linear relations will be present in P2.
Exercise 9.E.30
Find a basis in P3 for the subspace span{1+x−x2+x3,1+2x+3x3,−1+3x+5x2+7x3,1+6x+4x2+11x3} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.31
Find a basis in P3 for the subspace span{1+x−x2+x3,1+2x+3x3,−1+3x+5x2+7x3,1+6x+4x2+11x3} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.32
Find a basis in P3 for the subspace span{x3−2x2+x+2,3x3−x2+2x+2,7x3+x2+4x+2,5x3+3x+2} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.33
Find a basis in P3 for the subspace span{x3+2x2+x−2,3x3+3x2+2x−2,3x3+x+2,3x3+x+2} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.34
Find a basis in P3 for the subspace span{x3−5x2+x+5,3x3−4x2+2x+5,5x3+8x2+2x−5,11x3+6x+5} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.35
Find a basis in P3 for the subspace span{x3−3x2+x+3,3x3−2x2+2x+3,7x3+7x2+3x−3,7x3+4x+3} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.36
Find a basis in P3 for the subspace span{x3−x2+x+1,3x3+2x+1,4x3+x2+2x+1,3x3+2x−1} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.37
Find a basis in P3 for the subspace span{x3−x2+x+1,3x3+2x+1,13x3+x2+8x+4,3x3+2x−1} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.38
Find a basis in P3 for the subspace span{x3−3x2+x+3,3x3−2x2+2x+3,−5x3+5x2−4x−6,7x3+4x−3} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.39
Find a basis in P3 for the subspace span{x3−2x2+x+2,3x3−x2+2x+2,7x3−x2+4x+4,5x3+3x−2} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.40
Find a basis in P3 for the subspace span{x3−2x2+x+2,3x3−x2+2x+2,3x3+4x2+x−2,7x3−x2+4x+4} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.41
Find a basis in P3 for the subspace span{x3−4x2+x+4,3x3−3x2+2x+4,−3x3+3x2−2x−4,−2x3+4x2−2x−4} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.42
Find a basis in P3 for the subspace span{x3+2x2+x−2,3x3+3x2+2x−2,5x3+x2+2x+2,10x3+10x2+6x−6} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.43
Find a basis in P3 for the subspace span{x3+x2+x−1,3x3+2x2+2x−1,x3+1,4x3+3x2+2x−1} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.44
Find a basis in P3 for the subspace span{x3−x2+x+1,3x3+2x+1,x3+2x2−1,4x3+x2+2x+1} If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Exercise 9.E.45
Here are some vectors. {x3+x2−x−1,3x3+2x2+2x−1} If these are linearly independent, extend to a basis for all of P3.
Exercise 9.E.46
Here are some vectors. {x3−2x2−x+2,3x3−x2+2x+2} If these are linearly independent, extend to a basis for all of P3.
Exercise 9.E.47
Here are some vectors. {x3−3x2−x+3,3x3−2x2+2x+3} If these are linearly independent, extend to a basis for all of P3.
Exercise 9.E.48
Here are some vectors. {x3−2x2−3x+2,3x3−x2−6x+2,−8x3+18x+10} If these are linearly independent, extend to a basis for all of P3.
Exercise 9.E.49
Here are some vectors. {x3−3x2−3x+3,3x3−2x2−6x+3,−8x3+18x+40} If these are linearly independent, extend to a basis for all of P3.
Exercise 9.E.50
Here are some vectors. {x3−x2+x+1,3x3+2x+1,4x3+2x+2} If these are linearly independent, extend to a basis for all of P3.
Exercise 9.E.51
Here are some vectors. {x3+x2+2x−1,3x3+2x2+4x−1,7x3+8x+23} If these are linearly independent, extend to a basis for all of P3.
Exercise 9.E.52
Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. {x+1,x2+2,x2−x−3}
Exercise 9.E.53
Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. {x2+x,−2x2−4x−6,2x−2}
Exercise 9.E.54
Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. {[1201],[−72−2−3],[4012]}
Exercise 9.E.55
Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. {[1001],[0101],[1010],[0011]}
Exercise 9.E.56
If you have 5 vectors in R5 and the vectors are linearly independent, can it always be concluded they span R5?
- Answer
-
Yes. If not, there would exist a vector not in the span. But then you could add in this vector and obtain a linearly independent set of vectors with more vectors than a basis.
Exercise 9.E.57
If you have 6 vectors in R5, is it possible they are linearly independent? Explain.
- Answer
-
No. They can't be.
Exercise 9.E.58
Let P3 be the polynomials of degree no more than 3. Determine which of the following are bases for this vector space.
- {x+1,x3+x2+2x,x2+x,x3+x2+x}
- {x3+1,x2+x,2x3+x2,2x3−x2−3x+1}
- Answer
-
- Suppose c1(x3+1)+c2(x2+x)+c3(2x3+x2)+c4(2x3−x2−3x+1)=0 Then combine the terms according to power of x. (c1+2c3+2c4)x3+(c2+c3−c4)x2+(c2−3c4)x+(c1+c4)=0 Is there a non zero solution to the system c1+2c3+2c4=0c2+c3−c4=0c2−3c4=0c1+c4=0, Solution is: [c1=0,c2=0,c3=0,c4=0] Therefore, these are linearly independent.
Exercise 9.E.59
In the context of the above problem, consider polynomials {aix3+bix2+cix+di,i=1,2,3,4} Show that this collection of polynomials is linearly independent on an interval [s,t] if and only if [a1b1c1d1a2b2c2d2a3b3c3d3a4b4c4d4] is an invertible matrix.
- Answer
-
Let pi(x) denote the ith of these polynomials. Suppose ∑iCipi(x)=0. Then collecting terms according to the exponent of x, you need to have C1a1+C2a2+C3a3+C4a4=0C1b1+C2b2+C3b3+C4b4=0C1c1+C2c2+C3c3+C4c4=0C1d1+C2d2+C3d3+C4d4=0 The matrix of coefficients is just the transpose of the above matrix. There exists a non trivial solution if and only if the determinant of this matrix equals 0.
Exercise 9.E.60
Let the field of scalars be Q, the rational numbers and let the vectors be of the form a+b√2 where a,b are rational numbers. Show that this collection of vectors is a vector space with field of scalars Q and give a basis for this vector space.
- Answer
-
When you add two of these you get one and when you multiply one of these by a scalar, you get another one. A basis is {1,√2}. By definition, the span of these gives the collection of vectors. Are they independent? Say a+b√2=0 where a,b are rational numbers. If a≠0, then b√2=−a which can’t happen since a is rational. If b≠0, then −a=b√2 which again can’t happen because on the left is a rational number and on the right is an irrational. Hence both a,b=0 and so this is a basis.
Exercise 9.E.61
Suppose V is a finite dimensional vector space. Based on the exchange theorem above, it was shown that any two bases have the same number of vectors in them. Give a different proof of this fact using the earlier material in the book. Hint: Suppose {→x1,⋯,→xn} and {→y1,⋯,→ym} are two bases with m<n. Then define φ:Rn↦V,ψ:Rm↦V by φ(→a)=n∑k=1ak→xk,ψ(→b)=m∑j=1bj→yj Consider the linear transformation, ψ−1∘φ. Argue it is a one to one and onto mapping from Rn to Rm. Now consider a matrix of this linear transformation and its reduced row-echelon form.
- Answer
-
This is obvious because when you add two of these you get one and when you multiply one of these by a scalar, you get another one. A basis is {1,√2}. By definition, the span of these gives the collection of vectors. Are they independent? Say a+b√2=0 where a,b are rational numbers. If a≠0, then b√2=−a which can’t happen since a is rational. If b≠0, then −a=b√2 which again can’t happen because on the left is a rational number and on the right is an irrational. Hence both a,b=0 and so this is a basis.
Exercise 9.E.62
Let M={→u=(u1,u2,u3,u4)∈R4:|u1|≤4}. Is M a subspace of R4?
- Answer
-
This is not a subspace. [1111] is in it, but 20[1111] is not.
Exercise 9.E.63
Let M={→u=(u1,u2,u3,u4)∈R4:sin(u1)=1}. Is M a subspace of R4?
- Answer
-
This is not a subspace.
Exercise 9.E.64
Let W be a subset of M22 given by W={A|A∈M22,AT=A} In words, W is the set of all symmetric 2×2 matrices. Is W a subspace of M22?
Exercise 9.E.65
Let W be a subset of M22 given by W={[abcd]|a,b,c,d∈R,a+b=c+d} Is W a subspace of M22?
Exercise 9.E.66
Let W be a subset of P3 given by W={ax3+bx2+cx+d|a,b,c,d∈R,d=0} Is W a subspace of P3?
Exercise 9.E.67
Let W be a subset of P3 given by W={p(x)=ax3+bx2+cx+d|a,b,c,d∈R,p(2)=1} Is W a subspace of P3?
Exercise 9.E.68
Let T: P2→R be a linear transformation such that T(x2)=1;T(x2+x)=5;T(x2+x+1)=−1. Find T(ax2+bx+c).
- Answer
-
By linearity we have T(x2)=1,T(x)=T(x2+x−x2)=T(x2+x)−T(x2)=5−1=5, and T(1)=T(x2+x+1−(x2+x))=T(x2+x+1)−T(x2+x))=−1−5=−6. Thus T(ax2+bx+c)=aT(x2)+bT(x)+cT(1)=a+5b−6c.
Exercise 9.E.69
Consider the following functions T: R3→R2. Explain why each of these functions T is not linear.
- T[xyz]=[x+2y+3z+12y−3x+z]
- T[xyz]=[x+2y2+3z2y+3z+z]
- T[xyz]=[sinx+2y+3z2y+3z+z]
- T[xyz]=[x+2y+3z2y+3z−lnz]
Exercise 9.E.70
Suppose T is a linear transformation such that T[11−7]=[333]T[−106]=[123]T[0−12]=[13−1] Find the matrix of T. That is find A such that T(→x)=A→x.
- Answer
-
[31132333−1][621521611]=[29954613827115]
Exercise 9.E.71
Suppose T is a linear transformation such that T[12−18]=[525]T[−1−115]=[335]T[0−14]=[25−2] Find the matrix of T. That is find A such that T(→x)=A→x.
- Answer
-
[53223555−2][114110411231]=[1093810112351081348]
Exercise 9.E.72
Consider the following functions T: R3→R2. Show that each is a linear transformation and determine for each the matrix A such that T(→x)=A→x.
- T[xyz]=[x+2y+3z2y−3x+z]
- T[xyz]=[7x+2y+z3x−11y+2z]
- T[xyz]=[3x+2y+zx+2y+6z]
- T[xyz]=[2y−5x+zx+y+z]
Exercise 9.E.73
Suppose [A1⋯An]−1 exists where each Aj∈Rn and let vectors {B1,⋯,Bn} in Rm be given. Show that there always exists a linear transformation T such that T(Ai)=Bi.
Exercise 9.E.74
Let V and W be subspaces of Rn and Rm respectively and let T: V→W be a linear transformation. Suppose that {T→v1,⋯,T→vr} is linearly independent. Show that it must be the case that {→v1,⋯,→vr} is also linearly independent.
- Answer
-
If r∑iai→vr=0, then using linearity properties of T we get 0=T(0)=T(r∑iai→vr)=r∑iaiT(→vr). Since we assume that {T→v1,⋯,T→vr} is linearly independent, we must have all ai=0, and therefore we conclude that {→v1,⋯,→vr} is also linearly independent.
Exercise 9.E.75
Let V=span{[1120],[0111],[1101]} Let T→x=A→x where A is the matrix [1111011001211112] Give a basis for Im(T).
Exercise 9.E.76
Let V=span{[1001],[1111],[1441]} Let T→x=A→x where A is the matrix [1111011001211112] Find a basis for Im(T). In this case, the original vectors do not form an independent set.
- Answer
-
Since the third vector is a linear combinations of the first two, then the image of the third vector will also be a linear combinations of the image of the first two. However the image of the first two vectors are linearly independent (check!), and hence form a basis of the image. Thus a basis for Im(T) is: V=span{[2013],[4245]}
Exercise 9.E.77
If {→v1,⋯,→vr} is linearly independent and T is a one to one linear transformation, show that {T→v1,⋯,T→vr} is also linearly independent. Give an example which shows that if T is only linear, it can happen that, although {→v1,⋯,→vr} is linearly independent, {T→v1,⋯,T→vr} is not. In fact, show that it can happen that each of the T→vj equals 0.
Exercise 9.E.78
Let V and W be subspaces of Rn and Rm respectively and let T: V→W be a linear transformation. Show that if T is onto W and if {→v1,⋯,→vr} is a basis for V, then span{T→v1,⋯,T→vr}=W.
Exercise 9.E.79
Define T: R4→R3 as follows. T→x=[321822−2611−13]→x Find a basis for Im(T). Also find a basis for ker(T).
Exercise 9.E.80
Define T: R4→R3 as follows. T→x=[120111011]→x where on the right, it is just matrix multiplication of the vector →x which is meant. Explain why T is an isomorphism of R3 to R3.
Exercise 9.E.81
Suppose T: R3→R3 is a linear transformation given by T→x=A→x where A is a 3×3 matrix. Show that T is an isomorphism if and only if A is invertible.
Exercise 9.E.82
Suppose T: R3→R3 is a linear transformation given by T→x=A→x where A is a m×n matrix. Show that T is never an isomorphism if m≠n. In particular, show that if m>n, T cannot be onto and if m<n, then T cannot be one to one.
Exercise 9.E.83
Define T: R2→R3 as follows. T→x=[101101]→x where on the right, it is just matrix multiplication of the vector →x which is meant. Show that T is one to one. Next let W=Im(T). Show that T is an isomorphism of R2 and Im(T).
Exercise 9.E.84
In the above problem, find a 2×3 matrix A such that the restriction of A to Im(T) gives the same result as T−1 on Im(T). Hint: You might let A be such that A[110]=[10],A[011]=[01] now find another vector →v∈R3 such that {[110],[011],→v} is a basis. You could pick →v=[001] for example. Explain why this one works or one of your choice works. Then you could define A→v to equal some vector in R2. Explain why there will be more than one such matrix A which will deliver the inverse isomorphism T−1 on Im(T).
Exercise 9.E.85
Now let V equal span{[101],[011]} and let T: V→W be a linear transformation where W=span{[1010],[0111]} and T[101]=[1010],T[011]=[0111]
Explain why T is an isomorphism. Determine a matrix A which, when multiplied on the left gives the same result as T on V and a matrix B which delivers T−1 on W. Hint: You need to have A[100111]=[10011101]
Now enlarge [101], [011] to obtain a basis for R3. You could add in [001] for example, and then pick another vector in R4 and let A\left[\begin{array}{c}0\\0\\1\end{array}\right] equal this other vector. Then you would have A\left[\begin{array}{ccc}1&0&0\\0&1&0\\1&1&1\end{array}\right]=\left[\begin{array}{ccc}1&0&0\\0&1&0\\1&1&0\\0&1&1\end{array}\right]\nonumber
This would involve picking for the new vector in \mathbb{R}^4 the vector \left[\begin{array}{cccc}0&0&0&1\end{array}\right]^T. Then you could find A. You can do something similar to find a matrix for T^{-1} denoted as B.
Exercise \PageIndex{86}
Let V=\mathbb{R}^3 and let W=span(S),\text{ where }S=\left\{\left[\begin{array}{r}1\\-1\\1\end{array}\right],\:\left[\begin{array}{r}-2\\2\\-2\end{array}\right],\:\left[\begin{array}{r}-1\\1\\1\end{array}\right],\:\left[\begin{array}{r}1\\-1\\3\end{array}\right]\right\}\nonumber Find a basis of W consisting of vectors in S.
- Answer
-
In this case \text{dim}(W) = 1 and a basis for W consisting of vectors in S can be obtained by taking any (nonzero) vector from S.
Exercise \PageIndex{87}
Let T be a linear transformation given by T\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}1&1\\1&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\nonumber Find a basis for \text{ker}(T) and Im(T).
- Answer
-
A basis for \text{ker}(T) is \left\{\left[\begin{array}{r}1\\-1\end{array}\right]\right\} and a basis for Im(T) is \left\{\left[\begin{array}{r}1\\1\end{array}\right]\right\}. There are many other possibilities for the specific bases, but in this case \text{dim}(\text{ker}(T)) = 1 and \text{dim}(Im(T)) = 1.
Exercise \PageIndex{88}
Let T be a linear transformation given by T\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}1&0\\1&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\nonumber Find a basis for \text{ker}(T) and Im(T).
- Answer
-
In this case \text{ker}(T) = \{0\} and Im(T) = \mathbb{R}^2 (pick any basis of \mathbb{R}^2).
Exercise \PageIndex{89}
Let V=\mathbb{R}^3 and let W=span\left\{\left[\begin{array}{c}1\\1\\1\end{array}\right],\:\left[\begin{array}{r}-1\\2\\-1\end{array}\right]\right\}\nonumber Extend this basis of W to a basis of V.
- Answer
-
There are many possible such extensions, one is (how do we know?): \left\{\left[\begin{array}{r}1\\1\\1\end{array}\right],\:\left[\begin{array}{r}-1\\2\\-1\end{array}\right],\:\left[\begin{array}{c}0\\0\\1\end{array}\right]\right\}\nonumber
Exercise \PageIndex{90}
Let T be a linear transformation given by T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&1&1\\1&1&1\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]\nonumber What is \text{dim}(\text{ker}(T))?
- Answer
-
We can easily see that \text{dim}(Im(T)) = 1, and thus \text{dim}(\text{ker}(T)) = 3−\text{dim}(Im(T)) = 3−1 = 2.
Exercise \PageIndex{91}
Consider the following functions which map \mathbb{R}^n to \mathbb{R}^n.
- T multiplies the jth component of \vec{x} by a nonzero number b.
- T replaces the ith component of \vec{x} with b times the jth component added to the ih component.
- T switches the ith and jth components.
Show these functions are linear transformations and describe their matrices A such that T (\vec{x}) = A\vec{x}.
- Answer
-
- The matrix of T is the elementary matrix which multiplies the jth diagonal entry of the identity matrix by b.
- The matrix of T is the elementary matrix which takes b times the jth row and adds to the ith row.
- The matrix of T is the elementary matrix which switches the ith and the jth rows where the two components are in the ith and jth positions.
Exercise \PageIndex{92}
You are given a linear transformation T: \mathbb{R}^n → \mathbb{R}^m and you know that T(A_i)=B_i\nonumber where \left[\begin{array}{ccc}A_1&\cdots&A_n\end{array}\right]^{-1} exists. Show that the matrix of T is of the form \left[\begin{array}{ccc}B_1&\cdots&B_n\end{array}\right]\:\left[\begin{array}{ccc}A_1&\cdots&A_n\end{array}\right]^{-1}\nonumber
- Answer
-
Suppose \left[\begin{array}{c}\vec{c}_1^T \\ \vdots \\ \vec{c}_n^T\end{array}\right]=\left[\begin{array}{ccc}\vec{a}_1&\cdots&\vec{a}_n\end{array}\right]^{-1}\nonumber Thus \vec{c}_i^T\vec{a}_j=\delta_{ij}. Therefore \begin{aligned} \left[\begin{array}{ccc}\vec{b}_1&\cdots&\vec{b}_n\end{array}\right]\: \left[\begin{array}{ccc}\vec{a}_1&\cdots&\vec{a}_n\end{array}\right]^{-1}\vec{a}_i &=\left[\begin{array}{ccc}\vec{b}_1&\cdots&\vec{b}_n\end{array}\right]\:\left[\begin{array}{c}\vec{c}_1^T \\ \vdots \\ \vec{c}_n^T\end{array}\right] \vec{a}_i \\ &=\left[\begin{array}{ccc}\vec{b}_1&\cdots&\vec{b}_n\end{array}\right] \vec{e}_i \\ &=\vec{b}_i\end{aligned} Thus T\vec{a}_i=\left[\begin{array}{ccc}\vec{b}_1&\cdots&\vec{b}_n\end{array}\right]\: \left[\begin{array}{ccc}\vec{a}_1&\cdots&\vec{a}_n\end{array}\right]^{-1}\vec{a}_i=A\vec{a}_i. If \vec{x} is arbitrary, then since the matrix \left[\begin{array}{ccc}\vec{a}_1&\cdots&\vec{a}_n\end{array}\right] is invertible, there exists a unique \vec{y} such that \left[\begin{array}{ccc}\vec{a}_1&\cdots&\vec{a}_n\end{array}\right]\vec{y}=\vec{x} Hence T\vec{x}=T\left(\sum\limits_{i=1}^ny_i\vec{a}_i\right)=\sum\limits_{i=1}^ny_iT\vec{a}_i=\sum\limits_{i=1}^ny_1A\vec{a}_i=A\left(\sum\limits_{i=1}^ny_i\vec{a}_i\right)=A\vec{x}\nonumber
Exercise \PageIndex{93}
Suppose T is a linear transformation such that \begin{aligned}T\left[\begin{array}{r}1\\2\\-6\end{array}\right]&=\left[\begin{array}{c}5\\1\\3\end{array}\right] \\ T\left[\begin{array}{r}-1\\-1\\5\end{array}\right]&=\left[\begin{array}{c}1\\1\\5\end{array}\right] \\ T\left[\begin{array}{r}0\\-1\\2\end{array}\right]&=\left[\begin{array}{r}5\\3\\-2\end{array}\right]\end{aligned} Find the matrix of T. That is find A such that T(\vec{x}) = A\vec{x}.
- Answer
-
\left[\begin{array}{rrr}5&1&5\\1&1&3\\3&5&-2\end{array}\right]\left[\begin{array}{ccc}3&2&1\\2&2&1\\4&1&1\end{array}\right]=\left[\begin{array}{ccc}37&17&11\\17&7&5\\11&14&6\end{array}\right]\nonumber
Exercise \PageIndex{94}
Suppose T is a linear transformation such that \begin{aligned}T\left[\begin{array}{r}1\\1\\-8\end{array}\right]&=\left[\begin{array}{c}1\\3\\1\end{array}\right] \\ T\left[\begin{array}{r}-1\\0\\6\end{array}\right]&=\left[\begin{array}{c}2\\4\\1\end{array}\right] \\ T\left[\begin{array}{r}0\\-1\\3\end{array}\right]&=\left[\begin{array}{r}6\\1\\-1\end{array}\right]\end{aligned} Find the matrix of T. That is find A such that T(\vec{x}) = A\vec{x}.
- Answer
-
\left[\begin{array}{rrr}1&2&6\\3&4&1\\1&1&-1\end{array}\right]\left[\begin{array}{ccc}6&3&1\\5&3&1\\6&2&1\end{array}\right]=\left[\begin{array}{ccc}52&21&9\\44&23&8\\5&4&1\end{array}\right]\nonumber
Exercise \PageIndex{95}
Suppose T is a linear transformation such that \begin{aligned}T\left[\begin{array}{r}1\\3\\-7\end{array}\right]&=\left[\begin{array}{c}-3\\1\\3\end{array}\right] \\ T\left[\begin{array}{r}-1\\-2\\6\end{array}\right]&=\left[\begin{array}{4}1\\3\\-3\end{array}\right] \\ T\left[\begin{array}{r}0\\-1\\2\end{array}\right]&=\left[\begin{array}{r}5\\3\\-3\end{array}\right]\end{aligned} Find the matrix of T. That is find A such that T(\vec{x}) = A\vec{x}.
- Answer
-
\left[\begin{array}{rrr}-3&1&5\\1&3&3\\3&-3&-3\end{array}\right]\left[\begin{array}{ccc}2&2&1\\1&2&1\\4&1&1\end{array}\right]=\left[\begin{array}{rrr}15&1&3\\17&11&7\\-9&-3&-3\end{array}\right]\nonumber
Exercise \PageIndex{96}
Suppose T is a linear transformation such that \begin{aligned}T\left[\begin{array}{r}1\\1\\-7\end{array}\right]&=\left[\begin{array}{c}3\\3\\3\end{array}\right] \\ T\left[\begin{array}{r}-1\\0\\6\end{array}\right]&=\left[\begin{array}{c}1\\2\\3\end{array}\right] \\ T\left[\begin{array}{r}0\\-1\\2\end{array}\right]&=\left[\begin{array}{r}1\\3\\-1\end{array}\right]\end{aligned} Find the matrix of T. That is find A such that T(\vec{x}) = A\vec{x}.
- Answer
-
\left[\begin{array}{rrr}3&1&1\\3&2&3\\3&3&-1\end{array}\right]\left[\begin{array}{ccc}6&2&1\\5&2&1\\6&1&1\end{array}\right]=\left[\begin{array}{ccc}29&9&5\\46&13&8\\27&11&5\end{array}\right]\nonumber
Exercise \PageIndex{97}
Suppose T is a linear transformation such that \begin{aligned}T\left[\begin{array}{r}1\\2\\-18\end{array}\right]&=\left[\begin{array}{c}5\\2\\5\end{array}\right] \\ T\left[\begin{array}{r}-1\\-1\\15\end{array}\right]&=\left[\begin{array}{c}3\\3\\5\end{array}\right] \\ T\left[\begin{array}{r}0\\-1\\4\end{array}\right]&=\left[\begin{array}{r}2\\5\\-2\end{array}\right]\end{aligned} Find the matrix of T. That is find A such that T(\vec{x}) = A\vec{x}.
- Answer
-
\left[\begin{array}{rrr}5&3&2\\2&3&5\\5&5&-2\end{array}\right]\left[\begin{array}{ccc}11&4&1\\10&4&1\\12&3&1\end{array}\right]=\left[\begin{array}{ccc}109&38&10 \\112&35&10\\81&34&8\end{array}\right]\nonumber
Exercise \PageIndex{98}
Consider the following functions T: \mathbb{R}^3 → \mathbb{R}^2. Show that each is a linear transformation and determine for each the matrix A such that T(\vec{x}) = A\vec{x}.
- T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}x+2y+3z \\ 2y-3x+z\end{array}\right]
- T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}7x+2y+z \\ 3x-11y+2z\end{array}\right]
- T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}3x+2y+z \\ x+2y+6z\end{array}\right]
- T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}2y-5x+z \\ x+y+z\end{array}\right]
Exercise \PageIndex{99}
Consider the following functions T: \mathbb{R}^3 → \mathbb{R}^2. Explain why each of these functions T is not linear.
- T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}x+2y+3z+1 \\ 2y-3x+z\end{array}\right]
- T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}x+2y^2+3z \\ 2y+3x+z\end{array}\right]
- T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}\sin x+2y+3z \\ 2y+3x+z\end{array}\right]
- T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}x+2y+3z \\ 2y+3x-\ln z\end{array}\right]
Exercise \PageIndex{100}
Suppose \left[\begin{array}{ccc}A_1&\cdots&A_n\end{array}\right]^{-1}\nonumber exists where each A_j ∈ \mathbb{R}^n and let vectors \{B_1,\cdots ,B_n\} in \mathbb{R}^m be given. Show that there always exists a linear transformation T such that T(A_i) = B_i.
Exercise \PageIndex{101}
Find the matrix for T (\vec{w}) = \text{proj}_{\vec{v}} (\vec{w}) where \vec{v}=\left[\begin{array}{ccc}1&-2&3\end{array}\right]^T.
- Answer
-
Recall that \text{proj}_{\vec{u}}(\vec{v}) = \frac{\vec{v}\bullet\vec{u}}{||\vec{u}||^2}\vec{u} and so the desired matrix has ith column equal to \text{proj}_{\vec{u}} (\vec{e}_i). Therefore, the matrix desired is \frac{1}{14}\left[\begin{array}{rrr}1&-2&3\\-2&4&-6\\3&-6&9\end{array}\right]\nonumber
Exercise \PageIndex{102}
Find the matrix for T (\vec{w}) = \text{proj}_{\vec{v}} (\vec{w}) where \vec{v}=\left[\begin{array}{ccc}1&5&3\end{array}\right]^T.
- Answer
-
\frac{1}{35}\left[\begin{array}{ccc}1&5&3\\5&25&15\\3&15&9\end{array}\right]\nonumber
Exercise \PageIndex{103}
Find the matrix for T (\vec{w}) = \text{proj}_{\vec{v}} (\vec{w}) where \vec{v}=\left[\begin{array}{ccc}1&0&3\end{array}\right]^T.
- Answer
-
\frac{1}{10}\left[\begin{array}{ccc}1&0&3\\0&0&0\\3&0&9\end{array}\right]\nonumber
Exercise \PageIndex{104}
Let B=\left\{\left[\begin{array}{r}2\\-1\end{array}\right],\:\left[\begin{array}{c}3\\2\end{array}\right]\right\} be a basis of \mathbb{R}^2 and let \vec{x}=\left[\begin{array}{r}5\\-7\end{array}\right] be a vector in \mathbb{R}^2. Find C_B(\vec{x}).
Exercise \PageIndex{105}
Let B=\left\{\left[\begin{array}{r}1\\-1\\2\end{array}\right],\:\left[\begin{array}{c}2\\1\\2\end{array}\right],\:\left[\begin{array}{r}-1\\0\\2\end{array}\right]\right\} be a basis of \mathbb{R}^3 and let \vec{x}=\left[\begin{array}{r}5\\-1\\4\end{array}\right] be a vector in \mathbb{R}^2. Find C_B(\vec{x}).
- Answer
-
C_B(\vec{x})=\left[\begin{array}{r}2\\1\\-1\end{array}\right].
Exercise \PageIndex{106}
Let T: \mathbb{R}^2\mapsto \mathbb{R}^2 be a linear transformation defined by T\left(\left[\begin{array}{c}a\\b\end{array}\right]\right)=\left[\begin{array}{c}a+b\\a-b\end{array}\right].
Consider the two bases B_1=\{\vec{v}_1,\vec{v}_2\}=\left\{\left[\begin{array}{c}1\\0\end{array}\right],\:\left[\begin{array}{r}-1\\1\end{array}\right]\right\}\nonumber and B_2=\left\{\left[\begin{array}{c}1\\1\end{array}\right],\:\left[\begin{array}{r}1\\-1\end{array}\right]\right\}\nonumber Find the matrix M_{B_2,B_1} of T with respect to the bases B_1 and B_2.
- Answer
-
M_{B_2B_1}=\left[\begin{array}{rr}1&0\\-1&1\end{array}\right]