in-exercises-13-16-find-a-basis-for-the-solution-space-of-the-difference-equation-prove-that-the-s-3

Question

In Exercises $13-16,$ find a basis for the solution space of the difference equation. Prove that the solutions you find span the solution set. 
$$
y_{k+2}-25 y_{k}=0
$$

Michael Jacobsen · Accepted Answer

in this example, we have a difference equation that&#x27;s provide here. It&#x27;s homogeneous due to the equal zero portion that we see, and we&#x27;re going to let h denote the subset that which contains all solutions to this equation. Now what we want to do is find a basis for age. And in order to do that, we need to convert our difference equation seen here into an auxiliary equation. That auxiliary equation will be R squared, coming from Y to the power of K Plus two minus 25 we set this equal to zero. But this immediately implies that solutions to the AUC&#x27;s early equation R r equals five and negative five. So from this work, we can now go back to the original difference equation and assert that the set, which is going to be five to the power of K negative fighter the power of K contains solutions to the equation. And when we say the equation, we&#x27;re referring to the original difference equation here. So now that we have a set of solutions, were one step closer to determining if we have a basis to do that, let&#x27;s take a look at the form of these solutions notice with five to the power K and a negative on the inside. These are not multiples of each other. If the negative here was on the outside the parentheses, then they would be in multiples of each other. So it&#x27;s a state next that these solutions are not multiples off each other. And so we know whenever we have two vectors coming from a vector space H in this case and we have that they&#x27;re not multiples of each other. We can conclude that they are linearly independent so we can write. And so the set is Lee nearly independent? Well, next before we determining fully determined whether this set is a basis, let&#x27;s consider this difference equation here. The difference equation is of order to where to came from this to that is right here, which gave us our squared in the auxiliary equation. So now that the difference equation is of order of to this implies that the dimension of age is also to But then if we have a set, as we do here, which is already known to be linearly independent and are set, contains two vectors where the dimension of the target spaces to that immediately implies that the SAT must span H But then, with linear independence and spanning, we can conclude that the set fight of the power of K negative fighter the power of K is a basis for the set H and that completes our solution to this problem.

Nikolaos Alexandrakis · Answer

we need to find the dances for the solution. Space off this this critic elation. Now, as usual, we construct for the axillary regulation on Dwayne simplified. And we have are to the gay Times 16 r squared plus a times are miles three equals zero. Now we are interested in the solutions. So we wants to solve basic case on and there&#x27;s a second left active under prosecution. So we will just skip the details and present the two solutions. They know those Yeah, there is no, there&#x27;s no need absolving the execution. Red now equals minus a plus or minus. The square root off the determinant with his 16 squared divided by two times 16. Ah, so we have two solutions After doing the calculations, one is throw 11 over four and the other is are too with a close mind moose three over for. So these two solutions a result in solutions in the solutions one over for the game and minus three over four. So they k no. To verify that the solutions forum the solution set the solution space for this occasion. We just need to tow so that this is linearly independent on DDE taking the castle automatics for K equals zero. And considering that the Taliban and with this according to the Castle Rock Ian results in calculating to the determinant off these metrics by standard theory have you hop, sing the progress and success is then this is something is it will be done. And so we have minus three off four minus one over four sees you except minus one and its other than zero. And so we have Hlynur independence No, we have the The number of solutions is two on DDE echoes the dimension off the solution space given by fear him 17 Thus we&#x27;re done. These two signals form the solution space off these discredit cation.

Nikolaos Alexandrakis · Answer

we need to find a basis for the solution. Space off this disc. Ridicule a shin Andi. So we start by just solving their creation. First we write the absolutely rickerson which is this one, and we simplify it wrote to the K multiplied by row square mindless. So then times are glass 12 a quote zero. We are interested in nontrivial solution. So we need to find the solution. So this second or the regulation with we can easily solved on duh the terminal that two solutions are for in three. That after the solution is we&#x27;re looking for are three toe okay And for so the cake So discreet solutions. And so he&#x27;s the number of solutions equal because the dimension off the solution space by fury 17 Bates to 50. And now we just need to tow dough the casserole to test which we will skip the details Just because we have seen these in previous exercises and way would adjust find the Katsuragi in. See that the technical of the custom mathematics We were evaluated at K equals school and then we will have 11 three for Is that the revenant equals one. It is other than zero. So we have Hlynur independence and we are then these signals three toe again for to the K form the solution space of the situation.

Michael Jacobsen · Answer

for this example, we have a second order difference equation that were considering, and what we&#x27;re going to do here is for the set H, which is the set of all solutions. We want to find a basis. So to do that, we&#x27;re going to convert our difference equation into theocracy Ilary equation, which is R squared minus are plus 2/9 equal zero. To solve this equation, I&#x27;ll take the step of multiplying both sides by nine. So we now have nine R squared minus nine R plus two is zero, and this is becomes an equation which weaken factor pretty quickly. Nine r squared becomes three R and three are in these groups. And if we place a minus one and a minus to the factory, ization is complete. So we now have solutions to the auxiliary equation. They are r equals 1/3 coming from the first factor and 2/3 from the second factor. What this tells us then, is that this set, which is going to be 1/3 to the power K and 2/3 to the power of K, is a set of solutions. Two are provided difference equation. Now, what&#x27;s special about this set of solutions is that it is linearly independent. Weaken say this set is linearly independent in a church, since the signal&#x27;s are not multiples of each other. This method for determining independence works with having to signals. But for three signals we would have to look at a Casarotto matrix and then go from there. So now let&#x27;s pause for a moment. We have a difference equation, which is of order to do to that set our that subscript there. And H is a subspace that contains all the solutions. This tells us that the dimension of H is equal to equal to two, and this implies because this set is linearly independent, that 1/3 to the power of K and 2/3 to the power of K is a basis four h because if the dimension of H is too and we already have a linearly independent set that immediately implies this set spans to form a basis

James Chok · Answer

Okay. In this question, we will find the basis of the subspace span by these vectors. So this is a quiver tow asking what is a column space off this matrix when you just squish all of these colored pictures together? So this column back the girls yet some guys here goes here, goes here, goes here. Does it really matter what water it is? But but I&#x27;ll just take a funeral. So then the first thing that we have to do is to find to convert this into right echelon for so the first couple off steps that will do is take a road to you, add one that she wrote to will take Row three and add two tons of road one. In that interview, I three three and then we&#x27;re full. You was tracked. Why? It has wrote one. Bring that into road for so performing the steps 120 minus 131 plus one gives you zero to my street. Get it minus 10 clothes to give to four months. One gives you three, three months. Eight gives you negative five now very three plus words. You so negative Two plus two gives you zero four minus one gives you 30 My sixties Negative six negative seven miles to give me make you mine and nine plus six kids. A kitty. Okay, now rough Full minus 51 So five minus 50 six minus 10. Giving negative pole eight minus zero gives you eight. Seven plus five. Getting too. No. Sorry. Well, and by minus 15 gives you maybe 10. So now, right oppressed That perform is row three and add three. Title wrote to going down into row three and wrote full will subtract four tunnel road to connect to work for So we get 1 to 0 minus 130 Advice 123 Negative five. And you&#x27;re so right. Three plus road to sew. Three plus three miles st zero to negative six plus six is your negative mind plus nine zero 15 minus 15. Now I wrote for my forward to Is there a negative? Four plus four Q zero eight minus eight is you 0 12 miles. Talk gives you zero, but this is a regular for my sake. Zero on dhe negative. 10 plus 20 gives you 10. So you see, we only have three columns, right? Yeah. So how How homes waste is the space to span by These three letters stood in the basis for this subspace. Ah, one minus one minus 25 23 months. 16 and three. Negative, 89 and five.

Nikolaos Alexandrakis · Answer

So we assume that these three signals So all of these different situation and we need to solve that they form a basis for the solution set off the security firm now by I feel him of 17. We have the dimension off. The solution said off these evacuation is three as it is the order of dedication. Now we present by the castrati test no live to defend him well, that the solutions are linearly independent and we evaluate the casserole automatics. And according to the conservative test, we just need to find one case that&#x27;s that that the turbulent off these Matt Tricks is non zero, meaning that the mother&#x27;s is in vegetable. So we proceed by writing down the metrics for K equals zero. That&#x27;s for simplicity. And we have three. So the power of zero, which is one and then three and then so X squared and then minus five to the power off zero, which is still one and then my miss five and then minus five squared, which is 10 to 5. Hey, no, we do some Lynn around the veterans formations and we just ce obstruct the columns and, uh, find that is the determinant acquires the eminent off this mother tricks with zero here and one here. And, uh, mind most seven here. And, uh, lastly, not this is to square. Leslie would have five in here and, uh, mine on and 21 in here. So these determinant equals the determined these metrics. One timers 75 and 21 wigs, ecu, oz toe into one plus seven, multiplied by five. And, uh, circles 21 class for the plus 35 Hank ones 56 which is positive. And we&#x27;re done. And now we return toe the base, steal him on base on Bales Tau Tau nine. And we have that These three linearly independent solution seems there are three and they are elements of a vector space. We of that thes any solutions, form form a basis for the solution. Set

Problem

In Exercises $13-16,$ find a basis for the soluti…

Problem 15 Hard Difficulty

In Exercises $13-16,$ find a basis for the solution space of the difference equation. Prove that the solutions you find span the solution set.
$$
y_{k+2}-25 y_{k}=0
$$

Answer

the set $\left[ \left\{ 5 ^ { x } , ( - 5 ) ^ { k } \right\} \right]$ forms basis for the solution space of the difference equation.

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

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Catherine R.

Anna Marie V.

Joseph L.

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the set $\left[ \left\{ 5 ^ { x } , ( - 5 ) ^ { k } \right\} \right]$ forms basis for the solution space of the difference equation.

Linear Algebra and Its Applications

Lily A.

Catherine R.

Anna Marie V.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Lily A.

Catherine R.

Anna Marie V.

Joseph L.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications