Just added these guys to both sides of the equation. So let's just graph it just so we have a good visualization of what we're doing. So that tells us that v2 minus v3-- 0 times v1 plus v2 minus v3 is equal to 0. Eigenvectors-Eigenvalues can be defined as while multiplying a square 3x3 matrix by a 3x1 column vector. This is a linear system for which the matrix coefficient is.
This matrix right here-- I've just copied and pasted from above. And then the eigenspace for lambda is equal to minus 3 is a line. And then in the second row is 0, 1, minus 1. Then we just did a little bit of I guess we could call it vector algebra up here to come up with that. In the last video we set out to find the eigenvalues values of this 3 by 3 matrix, A. So this becomes 1, 0, and then 1 minus minus 1 is 2. So that's only one of the eigenspaces.
We can say that the eigenspace for lambda is equal to 3, is the set of all of vectors, v1, v2, v3, that are equal to a times times-- v2 is a, right? This polynomial is called the characteristic polynomial. Because I've officially run out of space. Therefore, all that we need to do here is pick one of the rows and work with it. And then let me do the last row in a different color for fun. So let me move this lower down where I have some free real estate.
Any value of λ for which this equation has a solution is known as an eigenvalue of the matrix A. The 3x3 matrix can be thought of as an operator - it takes a vector, operates on it, and returns a new vector. Eigenvalues and eigenvectors of 3 by 3 matrices Eigenvalues and Eigenvectors of a 3 by 3 matrix Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. For each eigenvalue there will be an eigenvector for which the eigenvalue equation is true. This was lambda is equal to minus 3, just to make us-- it's not related to this stuff right here. This guy's null space is going to be the null space of that guy in reduced row echelon form.
So the null space of this guy-- we could just put in reduced row echelon form-- the null space of this guy is the same thing as the null space of this guy in reduced row echelon form. You can see for an other example. And let me take my second row and divide it by 3. And it has two eigenvalues. Maybe if I draw that vector, that vector might look something like this.
Rewrite the unknown vector X as a linear combination of known vectors. So I'm going to do the same thing. This must be true but this is easier to work with. So this is the eigenspace for lambda is equal to 3. This fact is something that you should feel free to use as you need to in our work. Minus 2 minus 1 is minus 3. Let me move it down here.
Is that definitely the case? Remember that the power on the term will be the multiplicity. I'm going to write a little bit smaller than I normally do because I think I'm going to run out of space. So the eigenspace that corresponds to the eigenvalue 3 is a plane in R3. Since they are eigenvectors of different eigenvalues they are are orthogonal you can verify this calculating the inner product and you can normalize dividing by the modulus. So all of the values that satisfy this make up the eigenvectors of the eigenspace of lambda is equal to 3. In other words, if we know that X is an eigenvector, then cX is also an eigenvector associated to the same eigenvalue.
Now let me replace-- and I'll do it in two steps. I will skip some steps. And it's the span of that guy. EigenValues is a special set of scalar values, associated with a linear system of matrix equations. Let's just divide by 4. Another important use of eigenvalues and eigenvectors is diagonalisation, and it is to this that we now turn. However, each of these will be linearly dependent with the first eigenvector.