A periodic function f(t) of period 2 is defined by f(t) =...

A periodic function f(t) of period 2 is defined by f(t) = egin{cases} 3t & (0 < t < 1)\ 3 & (1 < t < 2) end{cases} f(t+2)=f(t) Sketch a graph of f(t) for -4 le t le 4 and determine a Fourier series expansion for the function. Answer: frac{9}{4} - frac{6}{pi^{2}}sum_{n=1}^{infty}frac{cos((2n-1)pi t)}{(2n-1)^{2}} - frac{3}{pi}sum_{n=1}^{infty}frac{sin(npi t)}{n}

Transcript

00:03 Consider the following 3x3 matrix a, which has linear dependent columns.

00:08 We want to use this information to find one eigenvalue and associated eigenvector of a.

00:15 So first, let's recall how we compute, or how we determine, eigenvalues.

00:23 We find eigenvalues by evaluating the determinant of a minus lambda i, and setting this equal to 0.

00:38 And now with property of determinant, we can write this as determinant of a is equal to the determinant of lambda i.

00:48 Now with the information that a has linear dependent columns, this implies that the determinant of a is equal to zero.

01:16 And determinant of a is equal to zero, well this means that determinant of lambda i is is equal to 0, which will only be the case if lambda is equal to 0, because determinant of an entity is equal to 1.

01:34 So we found our first eigenvalue.

01:40 And how do we find eigenvectors? so an eigenvector is a vector, say, u such that au is equal to lambda u.

01:56 U.

02:05 So in this case, with lambda 1 equal to 0, we want to solve for u such that au is equal to 0, essentially finding its null space, and we find that its null space, u1, i'm not going to give details, i'll just provide the result, is equal to the vector 1 minus 2, 1.

02:41 So we found the first eigenvalue and associated eigenvector.

02:45 Now in the information that a is a markov matrix, we want to find the other eigenvalues.

02:55 So because a is a markov matrix, this implies that it has another eigenvalue equal to 1.

03:20 And now with lambda the 2 equal to 1, we want to solve for vector u2 such that this will give us identity.

03:39 This product must give us an entity.

03:42 And we find that the second associated eigenvector is the vector 3, 3, 4.

03:59 And these all, we don't know it, they're multiplicity.

04:08 So if they're all multiplicity 1, then it's possible that we have a third eigenvalue.

04:15 And if we do go through all the calculation, we do find a third eigenvalue.

04:21 So by developing the determinants of a minus minus lambda i.

04:27 This was tricky.

04:28 I don't know, i can't think of any other trick than to just calculate it from first principles.

04:38 We find a third eigenvector equal to minus 1 over 5, and its associated eigenvector yields the vector minus 1, 1, 0...

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