MAT224 Practice Exercises - Week 7.5 Fall Comments

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1 MAT224 Practice Exercises - Week 75 Fall 27 Comments The purpose of this practice exercise set is to give a review of the midterm material via easy proof questions You can read it as a summary and you should attempt to prove each question on your own or with your study group Notice that one of the most important theorems which we covered in the previous weeks is that if the dimension of a vector space is n, then it is isomorphic to R n We saw that this allows us to carry all our computations to the world of MAT223 where everything could be computed by a computer In the first term test, you didn t have any computational questions because you needed to understand in which spaces you can do computations In the second term test, you should naturally expect to do some computations However, you should understand the theoretical background in order to apply the computation techniques from MAT223 to our course I did not break this into different sections because I wanted to point out that everything is connected That s why it might seem long and scary, but I believe you could finish this review in 3-4 hours with a good reading Linear Transformations Between Finite Dimensional Vector Spaces Throughout this problem set V and W are finite dimensional vector spaces and f : V W is a linear function We pick a basis α = {v,, v n } for V and a basis {w,, w m } for W Write what it means for f to be linear 2 Here a couple of easy consequences of the definition of linear functions: (a) Let V and W denote the zero vectors of V and W respectively Then, f( V ) = W Note We don t actually need the subscripts and we will not use them from now on We will simply write f() = From the context, you can understand the first zero is the zero vector in V and the second one is the zero vector in W Indeed, the inputs of f are from V and the inputs of f are from W Hint To prove this question, start with + =, apply f to both sides and use linearity (b) For any v V, f( v) = f(v) 3 Show that for any x,, x k V and x,, x k R, we have f(c x + + c k x k ) = c f(x ) + + c k f(x k ) Recall that since α is a basis, we can write any vector v in V as a linear combination So, you proved that v = c v + + c n v n f(v) = f(c v + + c n v n ) = c f(v ) + + c n f(v n ) This means that as long as you know f(v ),, f(v n ), you can compute f(v) for any v in V This is one of the miracles of linear functions

2 4 Recall from MAT223 that if A is the matrix a a m A = a n a mn and if e is the vector then Ae is the first column of A If you don t recall this, convince yourself by computing [ ] [ ] You should get Similarly, you get [ ] [ ] 5 4 = 3 2 [ ] [ ] 5 4 = 3 2 [ ] 5 = 3 [ ] 4 = 4 [ ] 5 + [ ] 4 + [ ] = 5 3 [ ] = 4 2 [ ] + 3 [ ] + 2 [ ] [ ] Check that these computations are correct 5 We will mimick what we did in the previous question to get a matrix which represents f The recipe is as follows: Compute f(v ),, f(v n ) Since these are elements of W, they can be written as a linear combination of w,, w m : f(v ) = a w + + a m w m f(v n ) = a n w + + a mn w m looking back at the previous question, we define the matrix of f with respect to α and β as a a m [f] β α = A = a n a mn Convince yourself that if we choose different bases, the matrix might change So, this matrix depends on the basis you choose This is important 6 The previous question allowed us to store the entire information about the function f using an n m matrix Now, we will store the information about our vectors using vectors To this end, we will define the coordinate vectors For v = c v + + c n v n in V, we will write [v] α = c c n 2

3 Similarly, for w = d w + + d m w m in W, we put [w] β = We have the following very important consequence of this symbolic definition This can be seen by computing f(v) = f(c v + + c n v n ) = c f(v ) + + cf(v n ) d d m [f] β α[v] α = [f(v)] β = c [a w + + a m w m ] + + c n [a n w + + a mn w m ] = (c a + + c n a n )w + + (c a m + + c n a mn )w m So, c c n c a + + c n a n c a m + + c n a mn You should see this on a small example where n = 2 and m = 3, create your own example This is the key observation, so make sure you understand how this works You should do lots of computations of this sort, you should get your hands dirty 7 Most likely you didn t do your own small example which was suggested at the end of the previous question So, here is a small example Let V = P (R) and W = R 3 Define f : V W by f(a + bx) = (a + b, a b, 2a + 3b) Let α = {, + x} and β = {(,, ), (,, ), (,, )} Then, f() = (,, 2) = 2 (,, ) + ( ) (,, ) + (,, ) f( + x) = (2,, 5) = 5 (,, ) + ( 5) (,, ) + 2 (,, ) Hence, 2 5 [f] β α = 5 2 Now, consider 3 + 5x (a) What is [3 + 5x] α? (b) What is f(3 + 5x)? (c) What is [f(3 + 5x)] β? (d) Is it true that [f(3 + 5x)] β = [f] β α[3 + 5x] α? You should check this 3

4 8 We need to make more definitions to be able to simplify what we will say later These words are not new for most of you, as you already heard them in MAT223 We define the kernel of f as ker(f) = {v V such that f(v) = } Here are some easy but important consequences of this definition (a) Show that ker f is a subspace of V (b) Show that if f is one-to-one, then ker(f) = {} (c) Show that if ker(f) = {}, then f is one-to-one (d) Show that if v is in ker(f), then [v] α is in the nullspace of [f] β α (e) Show that if [v] α is in the nullspace of [f] β α, then v is in ker(f) 9 We define the image (range) of f to be im(f) = {w W such that there is a v V with f(v) = w} (a) Show that im(f) is a subspace of W (b) Conclude that dim(im(f)) dim W (c) Show that im(f) = W iff f is onto (Notice that there is nothing to show here, this is actually definition of being onto) (d) Show that w im(f) iff [w] β is in the column space of [f] b αeta (e) Note that Conclude from this that w im(f) f(v) = w for some v V Combining Q8d and Q8e, we can argue that and from 9c, we can conclude that f(c v + + c n v n ) = w for some c,, c n R c f(v ) + + c n f(v n ) = w for some c,, c n R im(f) = span{f(v ),, f(v n )} dim(ker(f)) = nullity([f] β α) dim(im(f)) = rank([f] β α) In MAT223, you learned the Rank-Nullity Theorem which we state as the Dimension Theorem as follows dim(ker(f)) + dim(im(f)) = dim V Now, we take a step back and prove the following facts (a) Let x,, x k V and assume {f(x ),, f(x k )} is a linearly independent subset of W Show that {x,, x k } is a linearly independent subset of V (b) Let {x,, x k } be a linearly independent subset of V and assume f is one-to-one Show that {f(x ),, f(x k )} is a linearly independent subset of W 4

5 (c) Show with a counterexample that if f is not one-to-one, then the consequence of part b is no longer true Hint For many such questions, the zero transformation (which takes all the vectors in V to zero in W ) is a counterexample Try it first (d) Conclude that {f(v ),, f(v n )} is a linearly independent subset of W if and only if f is one-to-one 2 Combine 9b, 9e and d to conclude that if f is one-to-one, then dim V dim W You can also use the Dimension Theorem to prove this Do it both ways 3 Show that the converse of the previous question is not true That is, if dim V dim W, then f does not have to be one-to-one Hint Zero transformation 4 Use 9b, 9c and 9e; or the Dimension Theorem to conclude that if f is onto, then dim V dim W 5 Show that the converse of the previous question is not true That is, if dim V dim W, then f does not have to be onto Hint Zero transformation once again! 6 We will now go back to matrix representations Here is another very important formula whose proof is tedious but quite simple If U is a third vector space with basis γ and if g : W U is another linear transformation, prove the followings (a) g f : V U is a linear transformation (This is straightforward) (b) [g f] γ α = [g] γ β [f]β α This just follows from some calculations Construct your own example to see this in action This is quite useful as we will see soon Note We are aware that there are some tutoring companies out there which sells notes to our students You should know that these notes are freely available and these so called tutoring companies are scams If you need help, we have more than hours of office hours per week available for you with about 6 hours during midterm weeks If you went to all these office hours and you still needed help, us These companies do not know anything about our midterms A good preparation for our midterm is to understand the material here and the computational problems on the problem sets Worry about your knowledge on the material, don t try to guess which questions can be on the midterm You can not learn the material if somebody hands you the solutions without you trying them on your own We care about you Don t be scammed 7 We say that a linear transformation is an isomorphism if it is one-to-one and onto (hence invertible) We say that V is isomorphic to W if there is an isomorphism from V to W (a) Show that the linear transformation T α : V R n defined by the formula T α (v) = [v] α is an isomorphism This is not difficult, but it is very important Make sure you understand how this works (b) Conclude that if dim V = n, then V is isomorphic to R n (c) Show that if f is an isomorphism, then f : W V is also an isomorphism Hint To show that f (x + y) = f (x) + f (y), you can let a = f (x + y), b = f x, f (c) = y Now your purpose is to show that a = b + c Compute f(a) and f(b + c) Then, use the fact that f is one-to-one 5

6 (d) Conclude that if V is isomorphic to W, then W is isomorphic to V (e) Show that if f is an isomorphism and g : W U is an isomorphism, then g f is also an isomorphism 8 Conclude using 7b, 7d and 7e that if dim V = dim W, then V is isomorphic to W 9 Use 2 and 4 to conclude that if V is isomorphic to W, then dim V = dim W 2 Combine the previous two questions to conclude that V is isomorphic to W if and only if dim V = dim W Notice that we did not construct an explicit isomorphism to prove this result, we used abstract linear algebra If we want to construct an isomorphism, here is a recipe: send v to w, v 2 to w 2 and so on until v n to w n Show that this recipe actually defines an isomorphism We can choose any other bases, and this recipe gives an isomorphism 2 Be careful We can have dim V = dim W and a linear map between them which is not an isomorphism Hint Zero transformation once again! 22 Show that if I : V V is the identity transformation on V then I is a linear transformation Show that [I] α α is the n n identity matrix I n 23 Assume f is an isomorphism Using the fact that f f = I, and using 6 with f : W V and f : V W and with γ = α, conclude Using 22, conclude that [f ] α β [f]β α = [I] α α [f ] α β = ([f] β α) 24 In 22, you showed that [I] α α is the identity matrix Show with a counterexample that if α is another basis, then [I] α α is not the identity matrix This matrix has a special name It is called the change of basis matrix from α to α It is the matrix of the identity transformation with respect to bases α and α 25 Using 6, 22 and the fact that I I = I, conclude ( [I] α α = [I] α α ) 26 Now, we will need some subscripts Let s denote by I V the identity transformation on V and by I W the identity transformation on W Let α and β be bases for V and W respectively (a) Convince yourself that f = I W f I V (b) Conclude using 6 that You just proved the change of basis formula [f] β α = [I] β β [f]β α [I] α α 27 NOTE From now on, we will assume that W = V So, we have a linear operator f : V V (Operator is just a fancy name to say that the domain and the codomain are the same, a nice property of an operator is that we can compose it with itself, so it allows us to model some dynamical systems) 6

7 28 Let β be a basis for V Use 26 to conclude that [f] α α = [I] α β [f]β β [I]β α Use 25 to argue that [f] α α = ( ) [I] β α [f] β β [I]β α 29 Suppose that there is a basis α = {v,, v n } of V such that for some λ,, λ n Compute [f] α α f(v ) = λ v, f(v 2 ) = λ 2 v 2,, f(v n ) = λ n v n 3 If you actually computed [f] α α in the previous question, you see that you have a diagonal matrix If you didn t, do it now This is important The nice thing about diagonal matrices is that they are so much easier to make computations with Here is an example Compute the 7th power of the following two matrices Measure time while doing it [ ] 2 A =, B = 2 3 [ ] 2 Can you compute the th powers? When a process is modeled by a linear operator and you want to understand what happens in the th step or th day, then you want to be able to compute these powers When you have larger matrices, even wolframalpha can t save you You need matrices which are as close as to a diagonal matrix If there is a basis as in 29, you are lucky You end up with a diagonal matrix If there is no such basis, then we will see what happens at the end of the course 3 Since we don t want to keep refering to 29, we should make a definition and give names to such vectors Let f : V V be a linear operator A nonzero vector v V is called an eigenvector of f if there is a λ such that f(v) = λv Such a λ is called an eigenvalue Show that is a subspace of V E λ = {v V : f(v) = λv} 32 Show that for any v E λ, we have f(v) E λ Conclude from this that we can restrict our function f to E λ and its image lies in E λ : f Eλ : E λ E λ Pick a basis β for E λ and show that [f Eλ ] β β is a diagonal matrix with λs on the diagonal 33 Note that f(v) = λv if and only if f(v) λv = This is equivalent to saying that f(v) (λi)(v) =, which is equivalent to saying that (f λi)(v) = From this conclude that 34 Note that E λ = ker(f λi) f(v) = λv [f(v)] α = [λv] α [f(v)] α = λ[v] α [f] α α[v] α = λ[v] α 7

8 From here, conclude that v is an eigenvalue of f if and only if [v] α is an eigenvalue of [f] α α You learned in MAT223 how to find eigenvalues and eigenvectors of matrices So, finding eigenvalues and eigenvectors of operators is something you already learned All you need to do is to pick a basis and write the corresponding matrix Then, you are back in MAT223 and you can find the eigenvalues 35 If you don t remember how to find eigenvalues of a matrix A, here is a reminder Ax = λx Ax = λix Ax λix = (A λi)x = x null(a λi) In MAT223, you learned that there is a nonzero vector in the nullspace of a matrix if and only if that matrix is not invertible And one of the big theorems said that this is if and only if the determinant of this matrix is zero Hence, we conclude that λ is an eigenvalue of A if and only if det(a λi) = Note that det(a λi) is a polynomial in λ, if you don t believe this, see it on a couple of examples to convince yourself 36 There is a very important point here Go back to 5 and convince yourself that that if we choose two different bases α and β, then [f] α α and [f] β β are different matrices 28 tells you the relation between them Here is a nice observation: Even though they are different matrices, they have the same characteristic polynomial! Here is how you can prove it To make things simpler, let s make a definition here Let A and B be two n n matrices We say that A is similar to B if there is an invertible matrix S such that A = S BS (a) Show that if A is similar to B, then B is similar to A (You can show B = SBS Remember S = S ) (b) Show that if A is similar to B, then for any λ R, A λi is similar to B λi X(Y + Z)W = XY W + XZW (c) Show that if A is similar to B, then det(a) = det(b) Hint det(xy ) = det(x) det(y ) Hint (d) Use the previous two parts to conclude that if A is similar to B, then they have the same characteristic polynomial Use this exercise and 28 to conclude what we wanted to prove at the beginning of this exercise Since it doesn t matter which basis we choose we define the characteristic polynomial of f to be the characteristic polynomial of [f] α α for some basis α 37 We say that a linear operator f : V V is diagonalizable if there is a basis β of V consisting of eigenvectors of f See 29 to understand why this name makes sense 38 Suppose that T : R n R n is a linear operator and suppose that T is diagonalizable Let std be the standard basis of R n and let α be a basis of R n consisting of eigenvectors of T Then, write the change of basis matrix [I] α std from std to α and the change of basis matrix [I]std α from α to std Observe from 25 that these matrices are inverses of each other What does [I] std α look like? What are its columns? Recall [T ] α α is a diagonal matrix and ( ) [T ] α α = [I] std α [T ] std std [I] std α This is how you diagonalize your matrix 39 Happy studying! Come to office hours for questions and/or a chat Good luck in your exam 8