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Hey everyone.
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So the problem we're looking at today builds off of a previous problem.
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I did a video on that too.
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This is the final screen we got.
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If you guys watch that video, if you didn't, feel free to check it out.
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Or if you're fine on that problem, and you need to help with this.
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Just a little tour of what we have.
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We have right here.
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We have written our eigenvectors and their multiplicity.
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These are all the different permutations, additive permutations, we can get to equal their multiplicity.
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So there's seven, for the fives, two for the negative, seven for the sixes, two for the negative twos.
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And that's just equivalent to the different number of ways we can break up each set of eigenvalues here in our jordan canonical form into different jordan blocks.
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Okay.
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So without further ado, the new question.
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Basically, it's asking us, well, not basically, it is asking us which, which matrices in a set we wrote down, which was, if you remember, the set of all, possible jordan canonical forms for something with this characteristic polynomial.
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So which of those matrices have exactly five linearly independent eigenvectors? now, this is a question about a certain theorem that you have in your textbook, which states that the number of jordan blocks is equal to the number of linearly independent eigenvectors for the matrix.
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So basically rewarded, the question asks us for which matrices in the set we wrote down earlier.
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So which possible jordan canonical forms have exactly five jordan canonical.
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Jordan blocks, sorry, jordan blocks in their jordan canonical form.
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So we can go through our list of possibilities right here for each eigenvalue, right? and we can make combinations that add up to five.
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And all of those are going to be answers to this question.
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All of those matrices are going to have five and exactly five linearly independent eye connectors.
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So the numbers here, remember, are the sizes of the blocks.
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So this is one block of size two, and the amount of things added.
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So this is one block of size one and one block of size one.
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So this is actually two blocks here.
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Yeah, that's two blocks there.
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And over here is going to be one so don't just add two, like you could say that, i don't know, two plus three is five, but this is actually one block, so you need one, and then in the next one you need four blocks.
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Right, so just don't get confused there.
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It's the number of blocks, not the size of the blocks that we're looking at.
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So if we have one block in the twos, and remember that's going to be a block of size two, so just one right there.
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We need how many different ways can we get four blocks in, the sixes...