00:01
This question is about eigenvalues and eigenvectors.
00:05
Here we want to find the eigenvalues of this matrix, and for each eigenvalue find a basis for the eigenspaced.
00:17
Recall that to find the eigenvalues, we have to solve the characteristic equation, which is characteristic polynomial equals zero.
00:29
Now what is the characteristic polynomial again? it is, in this case, a cubic polynomial of the scalar lambda defined by the determinant of a minus lambda i, where i is the identity matrix.
00:56
This determinant is the determinant of this matrix here, which is a minus lambda i, or just a with negative lambda is added on to the middle or the diagonal entries.
01:19
Now i'm going to use the co -factor expansion definition of the determinant.
01:25
To find the 3x3 determinant here.
01:29
1 minus lambda, or i'm going to choose this column, take the co -factor expansion of the column, which only has one non -zero entry, the one corresponding to 1 -mast lambda, and that will be 1 -mast lambda times the co -factor of 1 -m minus lambda, which is positive 1 times the determinant of the sub -matrix we get by deleting a column and row of this entry.
01:57
So that's the co -factor expansion of the middle column, 1 minus lambda times its co -factor.
02:20
Now we have to evaluate this 2 by 2 determinant, which is just ad minus b, c.
02:46
So here i've expanded the 4 minus lambda times 1 minus lambda, and once we factor that completely, we end up with our three eigenvalues, 1, 2, and 3.
03:08
So our matrix a has these three eigenvalues, 1, 2, and 3.
03:14
Now we want to find eigenvectors.
03:21
So an eigenvector corresponding to the first eigenvalue is a solution of this system.
03:30
This system is homogeneous, so i must have at least one solution, which is the zero solution, trivial solution, in other words.
03:43
But since we said that, or since lambda equals 1 was a solution to the characteristic equation, a minus i has a determinant of zero and thus must be uninvertible.
04:06
And so there is not just one solution in this case, but infinitely many.
04:11
And in fact, we go about the process of finding eigenvalues like this, specifically so that we would have infinitely many solutions and not just the zero vector.
04:29
We want non -zero vectors, non -zero vector solutions to this.
04:33
And the only way to have that is this way.
04:39
By the way, eigenvectors are defined as being non -zero.
04:43
So how do we find the solutions? well, one way is to row reduce this matrix to row reduced echelon form.
05:09
So this is my row reduced echelon form, which, well, from here to here, i haven't written down my steps, but all i did was subtract this row times three from this row, and then switch the rows...