00:01
Consider this linear dynamic system.
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Okay, we are given a matrix a, which is defined as 0 .1, 0 .2, 0 .4, and 0 .3.
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For the first question, we're required to compute the eigenvalue for a.
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And for the second question, we're required to compute the eigenvector.
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And it's corresponding eigenspace.
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So in fact, we only need to compute the eigenvector because each eigenspace can be spanned by a eigenvector.
00:55
So we are actually required to compute the eigenvector for a.
01:02
And in fact, we could do those two questions altogether.
01:07
And we only need to compute its eigenvalues and the corresponding eigenvectors.
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For the convenience of our computation, let's first define something called a prime.
01:20
Okay, i mean, we define a to be 1 over 10 times a prime.
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Here, a prime will be just equal to 1, 2, 4, 3.
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And it's very easy for us to see that all the eigenvalues for a prime will be 10 times the eigenvalues for a.
01:41
Each eigenvalue of a will be just 1 .0 times the eigenvalue of a prime.
01:46
And a and a prime, and once more, a and a prime share the same eigenvectors because there is only some scaling between them.
01:57
So to compute the eigenvalue of a, we only need to compute the eigenvalues of a prime and then multiplying 1 over 10.
02:07
So to compute the eigenvalues of a prime, we know we need to solve this eigenequation, determine lambda times i minus a prime, right? this will be just equal to determine lambda minus a minus 2 minus 4 lambda minus 3.
02:28
And it's easy to say it will be just equal to lambda square minus 4 lambda plus 3 minus x.
02:37
Which will be just equal to lambda minus 5 times lambda plus 1.
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So we know the eigenvalues for a prime, which can be written as lambda prime 1 will be 5 and lambda prime 2 will be minus 1.
02:59
Okay, this tells us the eigenvalues for a, which can be written as lambda, lambda 1 will be 0 .5 and lambda 2 will be minus 0 .1.
03:10
Okay, now we get the eigenvalues for a.
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Now let's consider its eigenvalues.
03:16
Just as what i said, we only need to consider the eigen...
03:19
Now we want to compute the eigenvectors of a.
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As i said before, we only need to compute the eigenvectors for a prime because they share the same eigenvectors, right? okay, let's compute the eigenvectors of a prime.
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Let's call it x prime.
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1 is the first eigenvector for a prime.
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By the definition, these eigenvectors will satisfy 5i minus a prime times x prime, which will be equal to 0.
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This 0 is the 0 vectors.
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And plug 5 into our expression, we know we'll get...
04:08
Okay, now we get 5 minus 1, which is 4.
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This is equivalent to 4 minus 2, 4 minus 2, times x.
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1 prime is equal to 0.
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This tells us x1 prime can be written as 1, 2.
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Okay, this is the eigenvector corresponding to 5.
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About the second one, do the same thing.
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We know it will be corresponding to minus i minus a prime...