Eigen-everything
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Introduction to Eigenvalues and Eigenvectors
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Proof of formula for determining Eigenvalues
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Example solving for the eigenvalues of a 2x2 matrix
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Finding Eigenvectors and Eigenspaces example
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Eigenvalues of a 3x3 matrix
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Eigenvectors and Eigenspaces for a 3x3 matrix
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Showing that an eigenbasis makes for good coordinate systems
Example solving for the eigenvalues of a 2x2 matrix Example solving for the eigenvalues of a 2x2 matrix
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- In the last video we were able to show that any lambda that
- satisfies this equation for some non-zero vectors, V, then
- the determinant of lambda times the identity matrix
- minus A, must be equal to 0.
- Or if we could rewrite this as saying lambda is an eigenvalue
- of A if and only if-- I'll write it as if-- the
- determinant of lambda times the identity matrix minus A is
- equal to 0.
- Now, let's see if we can actually use this in any kind
- of concrete way to figure out eigenvalues.
- So let's do a simple 2 by 2, let's do an R2.
- Let's say that A is equal to the matrix 1, 2, and 4, 3.
- And I want to find the eigenvalues of A.
- So if lambda is an eigenvalue of A, then this right here
- tells us that the determinant of lambda times the identity
- matrix, so it's going to be the identity matrix in R2.
- So lambda times 1, 0, 0, 1, minus A, 1, 2, 4, 3, is going
- to be equal to 0.
- Well what does this equal to?
- This right here is the determinant.
- Lambda times this is just lambda times all of these
- terms. So it's lambda times 1 is lambda, lambda times 0 is
- 0, lambda times 0 is 0, lambda times 1 is lambda.
- And from that we'll subtract A.
- So you get 1, 2, 4, 3, and this has got to equal 0.
- And then this matrix, or this difference of matrices, this
- is just to keep the determinant.
- This is the determinant of.
- This first term's going to be lambda minus 1.
- The second term is 0 minus 2, so it's just minus 2.
- The third term is 0 minus 4, so it's just minus 4.
- And then the fourth term is lambda minus
- 3, just like that.
- So kind of a shortcut to see what happened.
- The terms along the diagonal, well everything became a
- negative, right?
- We negated everything.
- And then the terms around the diagonal, we've got
- a lambda out front.
- That was essentially the byproduct of this expression
- right there.
- So what's the determinant of this 2 by 2 matrix?
- Well the determinant of this is just this times that, minus
- this times that.
- So it's lambda minus 1, times lambda minus 3, minus these
- two guys multiplied by each other.
- So minus 2 times minus 4 is plus eight, minus 8.
- This is the determinant of this matrix right here or this
- matrix right here, which simplified to that matrix.
- And this has got to be equal to 0.
- And the whole reason why that's got to be equal to 0 is
- because we saw earlier, this matrix has a
- non-trivial null space.
- And because it has a non-trivial null space, it
- can't be invertible and its determinant has
- to be equal to 0.
- So now we have an interesting polynomial
- equation right here.
- We can multiply it out.
- We get what?
- Let's multiply it out.
- We get lambda squared, right, minus 3 lambda, minus lambda,
- plus 3, minus 8, is equal to 0.
- Or lambda squared, minus 4 lambda, minus
- 5, is equal to 0.
- And just in case you want to know some terminology, this
- expression right here is known as the characteristic
- polynomial.
- Just a little terminology, polynomial.
- But if we want to find the eigenvalues for A, we just
- have to solve this right here.
- This is just a basic quadratic problem.
- And this is actually factorable.
- Let's see, two numbers and you take the product is minus 5,
- when you add them you get minus 4.
- It's minus 5 and plus 1, so you get lambda minus 5, times
- lambda plus 1, is equal to 0, right?
- Minus 5 times 1 is minus 5, and then minus 5 lambda plus 1
- lambda is equal to minus 4 lambda.
- So the two solutions of our characteristic equation being
- set to 0, our characteristic polynomial, are lambda is
- equal to 5 or lambda is equal to minus 1.
- So just like that, using the information that we proved to
- ourselves in the last video, we're able to figure out that
- the two eigenvalues of A are lambda equals 5 and lambda
- equals negative 1.
- Now that only just solves part of the problem, right?
- We know we're looking for eigenvalues and
- eigenvectors, right?
- We know that this equation can be satisfied with the lambdas
- equaling 5 or minus 1.
- So we know the eigenvalues, but we've yet to determine the
- actual eigenvectors.
- So that's what we're going to do in the next video.
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