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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. $$(-1)^{k}, 3^{k} ; y_{k+3}+y_{k+2}-9 y_{k+1}-9 y_{k}=0$$

Since, by the theorem, we have the set H of all solution of the $n ^ { \text {th } }$ order homogeneous lineardifference equation $y _ { t + n } + a _ { 1 } y _ { k + n - 1 } + \ldots , \ldots \ldots \ldots \ldots , x _ { n - 1 } y _ { k + 1 } + a _ { n } y _ { k } = 0$ For all $k$ is an $n$ - dimensional vector space, the two signals $( - 1 ) ^ { x }$ and $3 ^ { k }$ cannot possibly span a three dimensional space and so cannot be a basis for H.

Calculus 3

Chapter 4

Vector Spaces

Section 8

Applications to Difference Equations

Vectors

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in this example, we have 1/3 order difference equation that'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'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'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's explain why that'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'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.

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