Hello, in this video we will determine the eigenvalues and eigenvectors of this matrix.
We will perform this in three steps, which I wrote down to the right in short.
I first calculate the characeristic polynomial of the matrix, 
follow by determination of the zeros of the polynomial, providing the eigenvalues,
and the third step is to calculate the corresponding eigenvectors. This is quite a task.
Let's take it easy, we first start with the characteristic polynomial.
For this, I first need to subtract lambda on the diagonal of my matrix.
From that, I want to know the determinant, so I already write down that we want to know the determinant.
I will rewrite the matrix, but on the diagonal I will put lambda.
Like this, 2, another 1, minus lambda, 0, 0, 2 and a 1 - lambda.
I have the diagonal and subtract lambda everywhere, and from this I need to know the determinant.
This I can calculate, so let's do that straight away.
That is 1 - lambda, I first take this term and multiply it by the determinant of this here,
which is the determinant of 1 - lambda, 0, 2, 1 - lambda.
Minus 2 times the determinant of what I am left with after removing this column and this row,
which is the matrix 2, 0, 0, 1 - lambda.
Alright, then I have 1 - lambda, multiplied by, here I have 1 - lambda squared minus 0, 
so I can write down 1 - lambda squared.
Minus 2 times, here I also have only this diagonal, which happens because the other one equals 0.
I obtain 2 times 1 - lambda. This is the characteristic polynomial. 
Before I write out the brackets and those kinds of things,
I look at the second step, which is to determine its zeros.
I need to equal this to 0 and solve it.
It is more convenient to leave the brackets as they are,
and why this is the case we will see immediately.
This was phase 1 so to say, and now we continue with phase 2,
where we set this entire thing equal to 0.
Why did I not remove the brackets? Because in this term I see a term of 1 - lambda,
and there I see a factor 1 - lambda as well. I can simply put it ouside of the brackets!
This is equal to 1 - lambda multiplied by the remainder, 1 - lambda squared,
and over there I had 2 times 2 and a minus sign, makes 4.
Like this, which still equals 0.
I can solve this now, let's continue at the top here, this yields lambda equals 1, 
which would make this term 0, so lambda equals 1,
or, here I obtain 1 - lambda squared minus 4 equals 0, which is the same as 1 - lambda being equal to 4. 
Here again, we see that it is nice to have not worked out the brackets yet,
because from this we can conclude that 1 - lambda is equal to either plus or minus 2.
Before we had a third degree equation, this stated a lambda squared and yet another lambda,
so we already expected 3 solutions. We also have 3 solutions, namely lambda = 1,
or lambda equals, when you take plus 2 this is equal to -1, or when you take -2 then lambda equals 3.
These are the three eigenvalues.
The only problem left is to find for each eigenvalue the corresponding eigenvector. 
Although I say eigenvector, it is not one. 
Keep in mind that there is an infinite amount of corresponding vectors.
Which vectors are they? For that, we solve this equation.
I take my matrix as written there, and I will start with the first case where lambda equals 1.
For lambda equals 1, I have this matrix with 1 filled in everywhere.
I then have 020, 200, and 020. What I need to solve,
I will look for vectors v to obtain the product, I write v as x y z, to obtain 0 0 0.
We can now start sweeping and such, but usually this is not such a good idea.
Namely, the equations you obtain for small matrices are always very clear, and can be solved in one go.
For example, the first row already tells you that 2y should equal 0.
The next row gives 2x = 0, and the last one already tells you that 2y = 0.
In any case, I can already conclude that the solutions are of the form: 0, 0, 
and for z no statements can be made. This is correct, since z can be chosen freely. 
For example, I can choose z to be equal to c.
In any case, this is an eigenvector for lambda = 1.
You can check this with a calculation, of applying this matrix to this eigenvector.
Indeed, this leaves only the last column, because of the 0 0 1, so this indeed works and is an eigenvector.
However, there is an infinite amount of them, so if I want to know the eigenvectors in general,
then we have all multiples of this one. In other words, a constant multiplied by this one.
With the constant simply being a real number. 
Okay, these are the eigenvectors for lambda equals 1.
Let's also take lambda = 2, oh wait we don't have lambda = 2, let's take lambda = -1.
After that, I think the blackboard will be full, so you will get a rain check for the last one.
Lambda = -1 provides me with the matrix: 2 2 0, 2 2 0, and 0 2 2, like this.
It is good to comment that we have two identical equations: we have two identical rows.
In general, you expect the rows to be dependent on each other.
Only when the rows are mutually dependent, then their determinant equals 0.
The determinant detected whether the rows are mutually dependent.
This makes the equations mutually dependent, 
which means that we will always have an infinite amount of solutions, which we illustrated before.
I can translate this into two equations: on the one hand I have 2x + 2y = 0, 
and on the other hand y + 2z = 0. From this, I can conclude that x and y are opposites,
and that y and z are opposites. In any case, I need to find a solution.
Just like before, where we could choose that 1 freely, I can start by choosing something freely here as well.
Let's say I choose x = 1.
I fill in x = 1 here, and let's have a look at the consequences.
When x = 1, then y should equal, for the upper equation to be true, should equal -1.
When y = -1, then this gives -2, and to yield 0 z should equal 1 again.
By this, I have at least one eigenvector.
The nice thing is that all other solutions are an extension of this.
Similar to before, I can state that the general solution is simply a constant,
multiplied by the one eigenvector we had, with the constant a real number.
Unfortunately, lambda = 3 does not fit here anymore.
Perhaps it is a good practice for you at home.
Good luck and this was it!
