Charakteristisches Polynom | Mathe by Daniel Jung

Yes, a mathematical vocabulary characteristic polynomial. I link the video where this is explained in detail here. You have multiplied a matrix with a vector in the approach equal to a multiple of the vector. Here now lambda by conversion you come to this expression. E stands for example for the unit matrix. And it must be noted that the determinant of this expression here is equal to zero. Here is no longer a zero vector, but really the number equals zero. As an example, you have here a 2 x 2 matrix with four coefficients. And it is then that determinant 1 minus lambda 2 4 1 minus lambda must be equal to zero. Here again the note in the playlist matrix algebra I also have other examples. For example, if we have a 3 x 3 matrix, that is, which rule can we apply here? Here, for example, the Kramers rule, then we would have cross-multiplied 1 minus lambda times 1 minus lambda. I can also directly write up to 2. Minus again cross-multiplied 4 times 2 is 8 equals 0. And now you have your characteristic equation. That's why I always talk about mathematical vocabulary. Now you unpack a solution. You can look at which scheme you go ahead. Some now multiply out here first. Then we would have here now 1 squared is 1 minus 2 lambda plus lambda squared minus 8 equals 0. 1 minus 8, then we would have minus 7. Sorted written down minus 2 lambda minus 7 equals 0. And now you can think about which solution you take. If, for example, you take the p-q formula, then you have accordingly, it is normalized. You have the p-value here, the q-value here. Minus p half plus minus root out p half raised to 2 minus q should lead you here to lambda 1 and lambda 2. With minus p half 1 times plus times minus and then root out 8. That's what it should be. Calculate in peace where I wanted to go. On the vocabulary. Some only need the vocabulary characteristic polynomial. You will also find in other areas. Characteristic equation. You have to have a solution. And then you should get through easily.