Problems for M 11/2: A =
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1 Math 30 Lesieutre Problem set # November 0 Problems for M /: 4 Let B be the basis given by b b Find the B-matrix for the transformation T : R R given by x Ax where 3 4 A (This just means the matrix for the transformation T but where we use the basis B on both sides) We have a formula to find the matrix for a transformation in a given basis It s For us M [[T (b )] B [T (b )] B ] 3 4 T (b ) 3 4 T (b ) 3 But to get M we need to know the B-coordinates of both of these vectors For that we need to use our older formulas: We have We then obtain [T (b )] B P B (T (b )) [T (b )] B P B (T (b )) P B [T (b )] B 0 [T (b )] B 3 So at the end of the day the B-matrix is given by M 0
2 43 Consider the transformation T : R R given by 0 A 3 4 Find a basis B with respect to which the transformation is diagonal A matrix becomes diagonal when you work in an eigenbasis (ie a basis made up of eigenvectors So we just have to find the eigenvectors and that ll be the basis we want λ det(a λi) det λ 4λ + 3 (λ )(λ 3) 3 4 λ For λ we have and so the eigenvector is and the eigenvector is A λi 3 3 Likewise for λ 3 it s A λi 3 3 So our basis B has two vectors 3 3 Early in the course I mentioned that linear algebra gives you a way to find a formula for the Fibonacci numbers I didn t get to it in lecture so I will let you work it out The Fibonacci numbers are a sequence defined by F and F and F n+ F n + F n They start off You can read lots of fun facts about them on Wikipedia (a) Let A [ 0 ] and x be the vector yourself that A n x is the vector We have It seems to work that A n x [ Fn+ F n+ ] Ax 0 A x A(Ax) A 3 x A(Ax) [ Fn+ F n+ Compute Ax A x A 3 x Convince ] ; every time we apply A the new first entry is the sum of the preceding ones and the new second entry is the old first entry
3 (b) Computing A n x directly is hard but we can do it by working with coordinates in an eigenbasis First find the eigenvectors and eigenvalues of A (hint: your answer will be a little messy involving some s) Then write down a diagonalization of A The characteristic polynomial is det(a λi) det To solve these we need the quadratic formula λ λ λ λ λ b ± b 4ac a ± This number + is going to show up a lot in this problem so let s give it a name: φ (better known as the golden ratio) For the eigenvalue with a instead of a plus write µ We know find the eigenvectors: φ A φi φ φ Using the trick for eigenvectors our eigenvector is (I m using the second row instead of the first here for simplicity; you ll get the same answer either way eventually) The other eigenvector is similarly found via µ A µi µ µ whence we get So the diagonalization is A P DP where φ µ φ 0 P D 0 µ (c) Let B be the basis given by the eigenvectors you found Compute the coordinate vector [x] B To get the coordinate vector we need [ φ µ [x] B P B x [ µ φ µ ] ] φ φ µ µ φ
4 Notice that So φ µ + [x] B µ φ (d) Our formula for linear transformations in an eigenbasis tells us that [A n (x)] B D n [x] B where D is the diagonalization of A Use your answers to the previous questions to find [A n (x)] B Now n φ 0 [A n (x)] B D n µ [x] B 0 µ φ But φµ [ φ n 0 0 µ n ( ] µ φ n µφ n φ φµ n µ n + ) ( ) 4 4 Then µφ n (µφ)φ n φ n and φµ n µ n so [A n (x)] B φ n + φ n µ n µ n (e) Convert this back into regular coordinates to get an expression for A n (x) What is your formula for F n+? Now we change coordinates back to the usual ones A n (x) P B [A n (x)] B So our answer is that [ φ µ ] ( [ φ n + φ n µ n µ n F n+ ( φ n + φ n µ n µ n ) ]) φ n+ + φ n µ n+ µ n φ n + φ n µ n µ n Equivalently F n ( φ n + φ n µ n µ n ) This can be simplified a little bit We have φ n +φ n φ n (+φ) φ n (φ ) φ n using the fact that φ φ 0 Similarly µ n + µ n µ n So ( ) n ( F n φn µ n + ) n Notice that ( )/ < and so when n is very big this is roughly just φ n / This explains why the ratio of consecutive Fibonacci numbers is close to the golden ratio: it s an eigenvalue of the matrix that generates them!
5 Problems for W /4: 48 Define T : R 3 R 3 by T (x) Ax where A is a 3 3 matrix with eigenvalues and Does there exist a basis B such that the B-matrix for T is a diagonal matrix? Discuss There s a matrix with respect to which the transformation is diagonal if the matrix A can be diagonalized They haven t told us much about A Because it s a 3 3 matrix with only two eigenvalues one of them has to be repeated So maybe A is diagonalizable and maybe it isn t; it depends on whether or not the eigenspace corresponding to this eigenvalue is -dimensional or only -dimensional Find the roots of x 4x Use the quadratic formula: Find (3 + 4i)( 6i) Just multiply it out: x 4 ± ± 36 ± 3i (3 4i)( 6i) (3)()+(3)( 6i)+( 4i)()+( 4i)( 6i) 6 8i 8i 4 8 6i 3 Find 3+4i 6i For this one you want to do: 3 + 4i 6i 3 + 4i + 6i 6i + 6i 8 + 6i i 0 4 Write + i in polar form re iθ Use this to compute ( + i) In polar form we have Then r + θ tan π 4 ( + i) ( e iπ/4 ) ( ) e πi/4 4 (cos π 4 + i sin π 4 ) 4 ( i) 4 4i Problems for F /6:
6 Find the (possibly complex) eigenvalues and eigenvectors for the matrix A 3 For this one we have det(a λi) det The quadratic formula gives us λ 3 4λ + λ ( ) λ 4λ + 3 λ λ 4 ± 6 0 ± i Let s use i The complex eigenvector is found by computing + i A ( i)i + i The eigenvector is v + i There are lots of other possible answers; just multiply both entries of this one by any complex number you want The eigenvector for + i is the complex conjugate of this one namely i w 4 Ditto with A 3 This time we get det(a λi) det λ 8λ + λ ( ) λ 8λ λ Let s do this one without the quadratic formula Completing the square this is which is going to be 0 when λ 4 ± i (λ 4) + Let s find the eigenvector for 4 i We have + i A λi + i
7 By the switch the two things in the first row and add a minus sign trick we ve discussed this gives v + i The eigenvector for the eigenvalue 4 + i is just the complex conjugate of this which is w i 9 Find the eigenvalues of A The transformation determined by A is a composition of a rotation and a scaling; give the angle of the rotation and the scaling factor (Hint: look at Example 6) [ ] 3/ / A / 3/ a b Let s follow the hint Example 6 tells us that a matrix has eigenvalues a ± bi b a Our matrix is of this form with a 3/ and b / This has r a + b and θ 7π/6 so it corresponds to a rotation by 7π/6 and a scaling by a factor of (the latter of which doesn t actually do anything) 3 Consider the matrix A 3 a b Find an invertible matrix P and a matrix C of the form (both with real entries) b a such that A P CP You might find your answer to the first question useful We [ have ] a recipe for this We already found the eigenvalue i with eigenvector + i Then A P CP where For our matrix this is a b P [Re(v) Im(v)] C b a P C 0 If you multiply out P CP you ll find that this indeed does the trick
Problems for M 10/26:
Math, Lesieutre Problem set # November 4, 25 Problems for M /26: 5 Is λ 2 an eigenvalue of 2? 8 Why or why not? 2 A 2I The determinant is, which means that A 2I has 6 a nullspace, and so there is an eigenvector
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