in-exercises-7-12-assume-the-signals-listed-are-solutions-of-the-given-difference-equation-determi-2

Question

In Exercises $7-12$ , assume the signals listed are solutions of the given difference equation. Determine if the signals form a basis for the solution space of the equation. Justify your answers using appropriate theorems. 
$$
2^{k}, 4^{k},(-5)^{k} ; y_{k+3}-y_{k+2}-22 y_{k+1}+40 y_{k}=0
$$

Nikolaos Alexandrakis · Accepted 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

Michael Jacobsen · Answer

in this example. We have a difference equation that&#x27;s provided here. And we&#x27;re told that one k to the power k and negative to to the para que are all solutions to this homogeneous difference equation. Next, we&#x27;re going to let h do you know this up space of s that contains all solutions and determine if this set is a basis for age. To do this, we need to determine the Katsuragi Matrix for this particular difference equation. It&#x27;s going to be a matrix. We can do no by see if we like. And in the first column we have one to the Power K, then one of the power of K plus one, then one of the power of K plus two. So, likewise, for the second column, we go to the second solution and right in to the power K to the Power K plus one, and then to the power of K plus two. Likewise, the less column is negative to power of K that negative two to the power of K plus one and then finally negative two to the power of K plus two. So for this particular matrix, we want to determine if it is potentially in vertebral for some choice of K, and the most nicest choice to work with is the following. Let&#x27;s let K be equal to zero, then our cause. Roddy Matrix C will be equal to the following. Placing K a zero on the first row gives us 11 and one. If K is zero, then on the second row we have the bases all to power one. So we have 12 negative to then raise thes same basis to the power of to we have 14 and four. So this is our matrix, and our next goal is to determine if it&#x27;s in vertebral. Well, this can be assessed through either taking a determinant or through row reduction. Going the road reduction route. I found that this matrix ISRO equivalent to 100 going column wise 11 zero one negative three and 12. And because there is a pivot in every room of this matrix, it follows that see inverse exists. So it&#x27;s pause for a moment. We found that if we take K two b zero, then the Katsuragi matrix becomes in vertebral. That tells us that this set is linearly independent. Let&#x27;s write that down as our next step so we can say this shows that let&#x27;s call this set Maybe well, we could leave it without a name. It&#x27;s one of the K to the power of K and negative to the power of K is leaned early independent. We also know that the dimension off the solution set H is equal to three and that becomes from the order of this equation which is order three. So we now know that we have three linearly independent solutions and the dimension of the spaces h Then we found a basis. So our conclusion is that this set meaning that one right above forms a basis four h and this completes our solution.

Nikolaos Alexandrakis · Answer

we assume that these two signals solve these discreet against on this lunar disc articulation, which is also home agendas. And we need to determine whether they form a basis for that solution. Little space off this occasion? No, by film, 17 on pages 2 50 We have that The dimension off the solution vector space off. The situation equals the number of the order off the Gerson, which is four. And this is greater than two. So which is the number off solutions we have? So these two solutions do not for the basis for all of the solution vector space of this occasion seems there to end their list them. But then the dimension of the

Nikolaos Alexandrakis · Answer

we will. So that the history signals are Who are you with? Something. So all of these things Critical ation forum and basis for the solution set of this occasion. Now we first keep in mind the theorem 17 from the book Bates Toe 50. Who says that the solution said off these discreet occasions Since it is off their three, it is three dimensional vector space. If it were so for their end, it would be an n dimensional vector space. So us in the biggest precise is we proceed the with psoriatic test and evaluate like a certain Patrick&#x27;s with this this one for Katie, plus one and keep lasto. And now we just need to find one. Okay, Uh, what was the best day? Who substitute was one in the bar? You? Because our young, which is the pregnant of these metrics and see if is not zero with me equals zero. He is inevitable, Andi. Then we&#x27;ll have. Then we will have dinner. Independence by the castle. Atticus gets on the phone at the end of hates to 47 eso week of one minus one on one, in here on zero and one and minus one. And, uh, to the five and five square, which is five. All right, so we just do some basic literal transformations, and, uh, we see that these metrics, we will put the equivalent in a parentis. Is this that land is These are them. Equals the government off the equivalent Mataric switches. If with is if we substructure the first column from the second, that is one minus one and one on the first column and then zero and then one minus one equals zero. And then we fell. Um, I miss fund doing here. Three on duh. Then five and five squared my loose minus one with vibrates at six. And mindless one, which is five minus one. It&#x27;s going for and solve this determined evaluates to the government off, please. Too happy to Matt. Tricks on this is minus 24 minus 12. It&#x27;s evaluates of minors 36. Now this is other than zero. And so we are done on the Mavericks is convertible for K equals zero. So by fear him by the basis field among aids stool drinking nine. We thought that see, these three function seems there&#x27;ll in early independent and, uh, they solve this this good system, they for use for basis for the solution set off this occasion

Michael Jacobsen · Answer

in this example, we have 1/3 order difference equation that&#x27;s provided its homogeneous since the right hand side is zero and were also provided with two different solutions Negative one of the K and three of the Power K. So what we&#x27;re going to do is let h denote the subspace of all signals s that contains a solution to this difference equation. And our objective is to determine if this will form a basis for H. Let&#x27;s start off with the following observation. The difference equation is off order three due to the three here by the sub script K plus three since its of order three. This implies that the dimension of the subspace h is three. So now we can say that are set negative one to the power of K and three to the power of K cannot form a basis. Four h. Let&#x27;s explain why that&#x27;s the case. We know since the dimension is equal to three, every basis of age must have exactly three vectors. But this particular set, whether it&#x27;s linearly, independent or not, only contains two different vectors, and so it could never form a basis. So we have to extend this set by finding 1/3 solution, and then and only then can we determine whether this is really a basis.

Michael Jacobsen · Answer

in this video. We&#x27;re looking at an example where 1/3 order difference equation is provided and we&#x27;re told that one to parquet three, The parquet Co sign of Cape I divide by two and similar with sign our solutions to this particular difference equation. Next, we&#x27;re going to let h do note the subspace off all signals and determine if we really have a basis for this set age. The first step is to determine if this is a set is linearly independent. And so to that end, we have that Cassara T matrix that&#x27;s defined below. It&#x27;s a rather clunky matrix because of one to the para que then eventually we have powers of K plus warning K plus two in the second and third grows. But what we&#x27;re going to do next is let K b any number of whatsoever to simplify our matrix toe work with. And the most convenient number would be K equals zero. So if K X equals zero, then this matrix C is going to be equal to the following first. All values of K go to zero, so these air effectively wiped out and we just get ones in her first column Next. When we place zero here, we know that CO Siam zero is a one. So we have a three, but sign of zero is zero. So that puts a zero in this row. So next, if we have zero here, then we have co sign pi over to that we&#x27;re dealing with. And that produces a zero placing zero here gives a sign, Piper to which is a one, then a zero here, it gives us three times one. So we have a three here now off to placing a zero here in here. So we&#x27;ll have three squared, which is a nine a co sign of pie altogether, which is negative one. So we have a negative nine. But if we place a zero here, we have signed up I once again, and that&#x27;s equal to zero. So this is our matrix. We need to next determine if this matrix isn&#x27;t vertebral and what we can do is take a determinant, since there&#x27;s plenty of zeros in the set of the third column. So let&#x27;s say that the determinant of C we&#x27;re going to do a co factor expansion down the third column and the signs will be positive here. Negative. Positive, then negative. To take a sign of negative three form a two by two determines found by deleting Row two and column three. So will next take the determinant of 13 one negative nine All together this is going to be equal to 30 which most importantly, is not zero. And that tells us that see, inverse exists and this is when K equals zero. But finding one value of K for which the inverse exists tells us the following this shows that are set one to the power K three of the power of K Times Co sign of Pai Que Divide by two and three The power of K times sine of pike A divide by two is linearly independent. So now we have linear independence off this set that&#x27;s provided through the Consort Casarotto matrix. The next thing we&#x27;re going to do is note that the order of this equation is order three. That means that the dimension of H, the set of all solutions is also three but are set here contains three different vectors which are linearly independent. That means the following set is a basis four h and this complex completes our check

Problem

In Exercises $7-12$ , assume the signals listed a…

Problem 8 Medium Difficulty

In Exercises $7-12$ , assume the signals listed are solutions of the given difference equation. Determine if the signals form a basis for the solution space of the equation. Justify your answers using appropriate theorems.
$$
2^{k}, 4^{k},(-5)^{k} ; y_{k+3}-y_{k+2}-22 y_{k+1}+40 y_{k}=0
$$

Answer

Since by the basis theorem, we have "If a linearly independent set of n-vectors in an n- dimensional space is automatically a basis".

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

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Heather Z.

Kristen K.

Vectors Intro

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Since by the basis theorem, we have "If a linearly independent set of n-vectors in an n- dimensional space is automatically a basis".

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Heather Z.

Kristen K.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Anna Marie V.

Heather Z.

Kristen K.

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

Linear Algebra and Its Applications