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13 Express the solution to the recursion nt+1 Ant in terms of the eigenvalues and eigenvectors ofA. Use DeMoivre' s theorem to simplify the solution if appropr...

Question

13 Express the solution to the recursion nt+1 Ant in terms of the eigenvalues and eigenvectors ofA. Use DeMoivre' s theorem to simplify the solution if appropriate A = L2 and no

13 Express the solution to the recursion nt+1 Ant in terms of the eigenvalues and eigenvectors ofA. Use DeMoivre' s theorem to simplify the solution if appropriate A = L2 and no



Answers

$15-20$ Express the solution to the recursion $\mathbf{n}_{t+1}=A \mathbf{n}_{t}$ in terms of the eigenvectors and eigenvalues of $A,$ assuming that $\mathbf{n}_{0}=\left[ \begin{array}{l}{1} \\ {1}\end{array}\right]$
$A=\left[ \begin{array}{ll}{2} & {0} \\ {0} & {1}\end{array}\right]$

We want to solve the Rikers in and T plus one equals a anti or a equals one minus 1 to 1 and and not equals 10 The solution to this is going to look like this will be on the falling Formenty equals P D to the t p inverse and not where a is P D. P inverse and D is a diagonal matrix. So D since FAA is don't Dagnall Izabal than the diagonal, Matrix D is going to have the Egan values of a on the diagonal. So let's find the Agon values of A by taking the determinant of a minus lambda times the identity. So this gets us one minus lambda squared plus two, which equals Lambda Squared minus to Lambda plus one plus two. So, plus three, if we set the sequel to zero, we can solve for the Agon values by using the quadratic formula. Land equals two plus or minus square root of four minus 12 all over too, which equals so four minus 12. We're going to get minus eight. This is one plus or minus. Hi! Squared of eight over too and squared of eight is really two times the square root of juice This is one plus or minus. I squared two So we found d Then the becomes one. Plus I squared of 21 minus I squared of two 00 And to find e to the T If you exponentially eight a day ago Matrix, you just exponentially eat the terms on the diagonal. So all that's left is to find p and therefore p inverse And, uh, so pee is going to have the Eiken vectors of a as columns. And so let's find the Eiken vectors of a So let's suppose that 12 minus 11 Let's suppose x y z agon the vector with wagon value one plus I squared of two. So on one hand, we have x minus y two x plus y. On the other hand, we have, um one plus I squared of two x one plus I squared of two x So we have x minus. Why equals one Plus I square it up to Oh, sorry. This should be a why x two x plus y equals one. Plus I square it of two. Why? So what shall we do? We can solve for why in the first equation we get, why equals X times I square root of two that's very minus. There should be a minus there, and then we plug that into our second equation. We get to X minus square of two. I x equals one plus I squared of two. It's well, let's plug in for why again, we get minus squared up to I Becks minus. So it's a plus two X. And as we see this equation is just an identity. So doesn't give us any information. So our first Eigen vector will be X could be anything. So let's say one. And then why is minus I square root of two? No, to find the second Eigen vector, we do the same thing except multiplied by one minus. I squared up to So this becomes one minus. I squared of two was becomes one minus size squared up to So what do we get again? We could solve for y in the first equation. Why equals I X square root of two? And if we plugged it into the second equation, we get two x plus I X squared up to equals by X squared of two loss, two x So again we can pick Why x to be anything. So let's say one and why will be I Square two? So now we found P and therefore confined P inverse. We have d to the T and we have a Not so we have all the pieces of information we need to solve this recurring.

We have the Matrix one minus a a one. All this matrix A and we want to solve the Riker, Shin and T plus one equals A and T where and not is the vector 10 The solution is going to be of the Formenty equals P d to the t p and not where a equals p d. I'm sorry for BP Universe PDP in verse. So we solve this. We just need to find d to the t p and p inverse. So we'll start by finding D. D is a diagonal matrix that hey is similar to, which means that it will have the Ivan values of a on its diagonal. Let's find the Eiken values of A by taking the determined of a minus lambda times the identity we get one minus lambda squared, plus a square which equals Lambda squared minus two lamb DM plus one plus a squared and the AG and values of their going to be the roots of this polynomial. So we can find the roots by the quadratic formula. Lambda equals to closer minus square root of four minus four minus for a square all over too, which is we can simplify. Toe one plus or minus four minus four cancels out, uh, before we can bring out to a tube divides. But the divides the cancel about tuna denominator. So we get minus a squared or aye, aye. So these are the two Aiken values. So we found d one plus a I won minus a I. And we also found e to the T because two exponentially eight diagonal matrix, you just exponentially eight the diagonal elements. So all that's left is to find P and Pampers. So P is going to be. It's the change of basis matrix into the basis vibing vectors. So it's going to have the Eiken vectors of a as its columns. So let's find the agon vectors of a Let's start by hoops. Assuming that ex wide is the eye in vector of I again value one plus a I so by matrix multiplication and get X minus a y and a X plus wide. And on the other hand, because this is an icon vector of really one plus a I, we get one plus a I thanks one plus a i. Why? So we get two equations x minus a Why equals X plus a i X And from here we can solve for y we get Let's see what we want to do. Well, the last I don't second equation x close. Why equals why puts a eye wide. So in this case, if we subtract wide from both sides, we get a X equals a I wine and thus we get X equals I. Why? Well, we actually get that only a is not zero. Well, if a so we can break this up with the two cases. If a is zero, then capital a r matrix is just the identity matrix, in which case P d to the t p inverse are all the identity matrix. So that salt, if it is non zero weaken do X equals I. Why? Which means that the first column vector of P we can pick wide to be anything we want to say. What one and then x equals I times that so right for to find the second column vector We just replace these with minus is to look for the other Aiken vector that's becomes minus. This becomes minus a X equals minus a i Y X equals minus I Y. Now our second column vector looks like minus. I won because that's the simplest vectors that simplest rector I can get where X is minus times. Why other than 00? But we never picks 00 because, um, does the column. Vector 00 is not and hide in Vector. Never. Yes, we have. I minus. I won one for pee. All that's left is to find the inverse. So we have p equals. I minus. I want one. So how do we find P inverse? We multiply p and its inverse, which is yet unknown together. And we guess Aye, aye. Plus C minus. Who's that's wrong? Yeah, aye, aye. Minus. See I the eye minus D I Hey plus c b plus de. And this equals the identity because we multiplied to matrix with inverse. So we get for example, we got a plus c equals zero. So a equals negative c We have a I minus c I equals one. If I replace a by minus C, I get minus two c. I a equals one, So looks like C should be. If it's I, then we get minus one times minus to it too. So I over too. Which means that a it's minus I over too. Then we have the eye minus D. I equals zero. So the I equals D I in fact be equals Dean and B plus d equals one and B equals Dean. The nd both equal 1/2 meaning that our foot we have a final piece of information is minus I over too Cia's eye or two be is one goose 1/2 India's 1/2.


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