Examples of using The eigenvalue in English and their translations into Polish
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The eigenvalue is off.
Each of them correspond to one of the eigenvalues.
The Eigenvalues in bloom.
Or, the eigenspace that corresponds to the eigenvalue 5.
It's going to be the eigenvalue for the nth eigenvector.
Since Heisenberg's matrices are Hermitian, the eigenvalues are real.
Caluclate the eigenvalues and eigenvectors of this gaussian.
So the eigenspace that corresponds to the eigenvalue 3 is a plane in R3.
Is coming up. The eigenvalue is off. I'm not satisfied with the way this.
So in this case, this would be an eigenvector of A, andthis would be the eigenvalue associated with the eigenvector.
So we know the eigenvalues, but we have yet to determine the actual eigenvectors.
Or we could say that the eigenspace for the eigenvalue 3 is the null space of this matrix.
We figured out the eigenvalues for a 2 by 2 matrix, so let's see if we can figure out the eigenvalues for a 3 by 3 matrix.
Lines, and corners anddots can be detected by the eigenvalues and determinant of the Hessian respectively.
He also showed, in 1829, that the eigenvalues of symmetric matrices are real.
Those are all of the eigenvectors that satisfy-- that work for the equation where the eigenvalue is equal to 5.
In the next video, we will actually solve for the eigenvectors,now that we know what the eigenvalues are.
So lambda is the eigenvalue of A, if and only if, each of these steps are true.
Any vector that when you draw in standard position lies, or points to, points on this line,will be an eigenvector for the eigenvalue minus 1.
And we call their scaling factors the eigenvalues associated with this transformation and that eigenvector.
Automatic determination of the ideal elastic critical moment Mcrfor each member or set of members on every x-location according to the Eigenvalue Method or by comparing moment diagrams.
We have not only figured out the eigenvalues for a 3 by 3 matrix, we now have figured out all of the eigenvectors.
Any vector that satisfies this right here is called an eigenvector for the transformation T. And the lambda,the multiple that it becomes-- this is the eigenvalue associated with that eigenvector.
So the eigenspace that corresponds to the eigenvalue minus 1 is equal to the null space of this guy right here.
If the eigenvalues are all positive, then the frequencies are all real and the stationary point is a local minimum.
So we could write that the eigenspace for the eigenvalue 5 is equal to the span of the vector 1/2 and 1.
That is, if one of the eigenvalues of the system is not both controllable and observable, this part of the dynamics will remain untouched in the closed-loop system.
And what we have learned now is that when you look at the eigenvectors and the eigenvalues, you can change your bases so that you can solve your problems in much simpler ways.