You are given the eigenvectors ( mathbfu1=left[eginarrayc-1...

You are given the eigenvectors ( mathbf{u}_{1}=left[egin{array}{c}-1 \ 1 \ 1end{array} ight], mathbf{u}_{2}=left[egin{array}{l}0 \ 1 \ 3end{array} ight] ) and ( mathbf{u}_{3}=left[egin{array}{c}2 \ 2 \ -4end{array} ight] ) of the matrix ( A ) given by [ A=left[egin{array}{ccc} -frac{9}{7} & -frac{3}{7} & frac{1}{7} \ frac{13}{7} & -frac{5}{7} & frac{11}{7} \ 7 & 3 & 3 end{array} ight] ] (a) Find the eigenvalues ( lambda_{1}, lambda_{2} ) and ( lambda_{3} ) of the eigenvectors ( mathbf{u}_{1}, mathbf{u}_{2} ) and ( mathbf{u}_{3} ), respectively. [ lambda_{1}=-1 ] [ lambda_{2}=4 ] [ lambda_{3}=-2 ] (b) The vectors ( mathbf{u}_{1}, mathbf{u}_{2} ) and ( mathbf{u}_{3} ) form a basis ( B ) of ( mathbb{R}^{3} ). Let ( mathbf{v} ) be the vector ( left[egin{array}{c}-7 \ -5 \ 13end{array} ight] ). Find the coordinate vector ( [mathbf{v}]_{B} ) of ( mathbf{v} ) in the basis ( B ). [ [mathbf{v}]_{B}=left[egin{array}{l} 1 \ 0 \ -3 end{array} ight] ] (c) Calculate ( A^{20} mathbf{v} ). Below, you can use exponentials to enter your answer, you don't have to write the full decimal expansion of the entries of the vector below, which could be large numbers. For instance, to write ( 5 cdot 9^{20} cdot 8^{8} ) enter ( 5^{*} 9^{wedge}(20)^{*} 8 ).

Transcript

00:01 We are given matrix a equal negative 9 7th, negative 3 7th, 1 over 7, first row, second row, 13 over 7, negative 5 over 7, and 11 over 7, and the third row, 7, 3, 3.

00:20 So we have also been given eigenvectors of a, u1 equal negative 1, 1, 1, u2 equal 0, 1, 3, and u3 equal 2, 2, negative 4.

00:31 With corresponding eigenvalues lambda1 equal negative 1, lambda2 equal 4, and lambda3 equal negative 2.

00:39 Indeed, these three eigenvalues were calculated in part a, which is already done.

00:47 So we're going to solve only part c here.

00:51 Part b was to calculate the coordinates of this vector b in the basis u1, u2, u3, because it has been said that these three vectors form a basis for r3, which is true because we have three vectors, and they are linearly independent.

01:08 And very important to remark that this eigenvalue, lambda1 equals negative 1, corresponds to this eigenvector u1.

01:18 In a similar way, lambda2 is the eigenvalue corresponding to this eigenvector, and lambda3 corresponds to u3.

01:25 So it's given vector b, negative 7, negative 5, 13, and we want to calculate a raised to the 20th power times vector b.

01:38 Of course, it is not to calculate a to the 20th.

01:42 One way to calculate the power of a matrix when we have this information is to write a diagonalization of a and try to find there what is the expression for a to the 20th.

01:55 But knowing that the base is formed by eigenvectors of a and we know the corresponding eigenvalue so it's very easy to calculate that power of a multiplied by v.

02:10 So let's remember that being lambda 1 and an eigenvalue of a with an eigenvector u1 that means that a times u1 is equal to lambda 1 u1.

02:30 Also a u2 is lambda 2 u2 because lambda 2 is an eigenvalue with corresponding eigenvector u2 and also a times u3 is equal to lambda 3 u3 for the same reason so we have these three equations which are true just by definition of eigenvalues and eigenvectors so we have that so now we're going to work with this one the other two are similar so let's say we want to calculate a square u1 that is a times a the u of u1 that is a times a the u1 is lambda 1 u1 now lambda 1 is a coefficient a scalar so you get out of the product so lambda one times a u1 that is lambda one times and a u1 again is lambda one u1 and so we get the scalar square lambda one square u1 in other words a square u1 is equal to lambda 1 square u1 we can calculate another one a cube u1 will be a times a square u1 by definition of a cube but that is a times and a square u1 we calculated here is lambda 1 square u1 lambda 1 square is a scalar you get out of the product so so lambda 1 squared times a u1.

04:26 So we get lambda 1 squared and a u1 is lambda 1 u1.

04:34 And so we get the scalar cube times u1.

04:37 That is a cube u1 is equal to lambda 1 cube u1.

04:45 So my induction is to verify that that in general any positive integer power of a times u1 that is a to the nth power u1 is this eigenvalue lambda 1 to the nth power times u1 for any or every n in the natural numbers that is n greater than positive integer so we have this formula in that formula of course we can do exactly the same calculation for the other eigenvectors and eigenvalues so we have similarly a to the nth power u2 will be lambda 2 to the nth power u2 and a to the nth power u3 is equal to lambda 3 to the nth power u3 so we have this also for every positive integer n good we have that information now we're going to use parts b of this exercise as something solves and that part was calculated the coordinate vector of vector b in the base form by the eigen sorry and that is vector b was calculated its coordinates in the base form by the eigenvector so in part b it was found that the coordinates of vector b in the base b is equal to 1 0 negative 3 where b is the base of r3 formed by by the eigenvectors u1, u2, and u3.

07:41 And of course, it's in that order.

07:42 There is 1 corresponds to the coefficient of u1, 0 the coefficient of u2, and 83 the coefficient of u3.

07:51 That means that vector p can be written as a linear combination of u1, u2, u3 using these scalars here.

07:59 So v is 1 times u1 plus 0 times u2 minus 3 times u3...

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