[BOŞ_SES] Hello.
We solved five examples in the previous session.
The purpose of the selection of these examples was to demonstrate conditions that may arise.
These were examples of simple numerical terms.
First and fifth in two binary matrices,
the other three were also three three-pointers matrix.
In the first example, we found two real roots.
Ie two in a binary matrix.
Again, the two binary complex eigenvalues in a matrix of two valuable
We have seen can be found.
In the second example for each of the three eigenvalues out and we found an eigenvector.
The third and fourth examples of a second discrete eigenvalues found
eigenvalues was a repeated root, but it was a different situation.
In the third example, although one of the two eigenvalues
Repeat if we can find time to opposing the two eigenvalues the eigenvalues.
Therefore, the non-recurring one root
In contrast, we find that the eigenvalues full three independent eigenvectors.
So our ability to build a base emerged.
But in the fourth example of a repeated eigenvalues
We have to find a single one eigenvector.
Therefore, we find only two eigenvectors.
Therefore, a basis vector for a three-dimensional space
We did not have the ability to create a team.
Now there may be a problem with the eigenvalues of leaving the complex.
Because in many cases the eigenvalues of a physical magnitude,
It may indicate a probability value.
For example, a power, a frequency, a probability
With numbers like real value, it is as valuable real definition.
Therefore, to ensure that they are real valued matrix,
We saw it in the matrix in which they must be symmetrical.
So what should be in them as a guarantee
We will ask you out real valued.
Again sufficiently eigenvector
What is the find of the requirements he will ask for it.
Their answer lies in symmetric matrix.
Now, at the beginning we do not know yet, but we will prove the following:
Real roots in the matrix will be symmetrical.
Even if repeated roots, they sufficiently
We'll show would eigenvector
and the eigenvalues of this would be perpendicular to each other, we make happen.
So, there are some supremely superior properties of symmetric matrix.
Now, before you go to the public our önsezgi
In order to strengthen a small matrix start, half a binary matrix.
Here you have no condition on the coefficients.
a 1 1, a 1 2 a 2 1 ,a 2 2.
We are writing to them, this matrix.
We are removing the negative lamdba across the diagonal.
This will give us a determinant eigenvalues.
Indeed, we have here see a single lambda squared term.
Multiply the two numbers on the diagonal,
lambdal the only minus lambda times a 1 1 here,
minus lambda times a second, including two from here, lambdal have these two terms.
And lambda are not available, the absolute numbers
a 1 1 a 2 2 eksi a 1 2 a 2 1 terimi var.
Now we find here the lambda.
These are the roots of what happens when we ask the real answer immediately clear.
This quadratic function, power function
To find the roots of this work A1, B number here,
c squared minus four times, including in a number wherein a,
c plus has to be valuable because it will take the square root.
We are writing here to be square, a 1 1
A 2 minus 2 squared term here four times.
We are witnessing just this: a plus from a 2 1 1 2 1 1 frame work,
2 2 1 1 2 2 squared plus twice that.
But here comes the same term minus four times.
So 4 minus 2 minus 1 will be a twice-to-1 2 2.
Birde a 1 1 kare var a 2 2 kare var.
So when we received the following three terms a 2-to-1 1 minus 2 squared off.
Back at plus 4 times a 1 2 2 1 off.
Now we just ask this question: here we find the roots of the subject,
1 plus 1 plus the square root of minus 2 2 n the D D
We find their roots in the root.
But to be real roots, plus the D's got to be worthwhile.
We have a complex value to be less valuable if D under root.
This D plus the only way to ensure that valuable.
To guarantee in all circumstances
a means is equal to a 1 2 2 1.
Because in this case the first term will be a sum of squared plus another frame,
This will give you an absolute plus valuable.
But this is not necessarily required.
Because a 1 2 2 1 plus can be valuable.
Here, too, there is added value.
So a 1 2 2 1 may be equal to the value without the plus or
there is a surplus.
This product can be the product of two very large negative value.
He took time minus the D.
But in any case, a sufficient condition is equal to a 2 1
If you can not guarantee this in a 1 2 D's plus is worthwhile.
a 2 1 from 1 to 2 means that even here we go to the matrix, they
The diagonal matrix because they are in the equivalent position is meant to be symmetrical.
So here even just half of a binary matrix symmetric matrix
We show that the roots of the property plus enough to be worthwhile.
This does not mean that the roots will be absolutely symmetrical least complex.
Roots again, plus provides, may be real valued.
But if you can not guarantee symmetrical.
But the real roots of symmetric can not guarantee it would be valuable at any time.
We did not prove anything more here in half a duality but also our intuition
developing symmetry means that there will be something she could imagine.
In fact, the next two theorems,
n times the size of the matrix in this
It will indicate validity of this observation.
Theorem as follows: a theorem saying that the real eigenvalues.
A real, symmetric matrix of a matrix that is symmetric and
If the eigenvalues are real valuable real.
Eigenvectors for the eigenvalues be real is real.
Now how do we prove it?
Go with a proven this in reverse.
Please Suppose we say that the roots are not valuable, but the real symmetric matrix.
If a real root valuable time will be the alpha plus beta ii.
Other root must be minus the alpha beta.
Because here we have the account of a negative determinant of lambda i.
A force that numbered function of real coefficients for everything real.
This is a root complex and the other one is surely to be complicated.
We have seen them in such an example in the previous session.
Now that these eigenvalues lambda lambda and a star.
Let us write that equation.
When we found here will be eigen to the lambdayl.
When we found to be the stars to the stars.
Attila AşkarProf. Dr.
