It works just the same if the eigenvalues are complex numbers. As for your second question, the answer is it depends whether or not the geometric multiplicity matches the algebraic multiplicity; this is slightly more technical so I cannot really explain it in a comment, but suffice to say that not all square matrices can be diagonalized.
papi chulo If A^n=PD^nP^-1, invert both sides to see that A^-n=PD^-nP^-1. Since D is a diagonal matrix, D^-n is obtained by inverting the individual eigenvalues along the main diagonal. This is of course based on the assumption that A is invertible and diagonalizable.
i noticed that instead of the usual det(A-lambda*I) you are instead using det(lambda*I - A) which resulted to elements of the matrix having reverse signs. Please clarify...
slcmath@pc If A^n=PD^nP^-1, invert both sides to see that A^-n=PD^-nP^-1. Since D is a diagonal matrix, D^-n is obtained by inverting the individual eigenvalues along the main diagonal. This is of course based on the assumption that A is invertible and diagonalizable.
![Diagonalizing 3x3 Matrix - Full Process [Passing Linear Algebra]](/img/1.gif)
Thank you for this. This is exactly what I was stuck on and found that I only needed to search for how eigenvectors can be applied to find your video.
ture
Very impressive vid
loved the concept and way u xplained
Thank you so much, this was very clear and understandable :)
Thank you, amazing skills
Very nice brother, good concept
Thank you! Interesting.
Perfect!
Thank you
hi is there a method for this where lamda is complex or either where there is only one value of lamda(where the quadratic is a perfect square)
It works just the same if the eigenvalues are complex numbers. As for your second question, the answer is it depends whether or not the geometric multiplicity matches the algebraic multiplicity; this is slightly more technical so I cannot really explain it in a comment, but suffice to say that not all square matrices can be diagonalized.
and for n
papi chulo If A^n=PD^nP^-1, invert both sides to see that A^-n=PD^-nP^-1. Since D is a diagonal matrix, D^-n is obtained by inverting the individual eigenvalues along the main diagonal. This is of course based on the assumption that A is invertible and diagonalizable.
i noticed that instead of the usual det(A-lambda*I) you are instead using det(lambda*I - A) which resulted to elements of the matrix having reverse signs. Please clarify...
Dexter Aparicio This way, the characteristic polynomial is always monic even for a square matrix of odd dimension.
❤
wonderful
thank you so much my exam is after 2 days
Hi, minute 09:24, why did you write the vector (1 -1) and not -1, 1 ??
papi chulo Any nonzero multiple of an eigenvector is also an eigenvector for the same eigenvalue; notice that one vector is the negative of the other.
thanks for the explanation!!! good video.
slcmath@pc If A^n=PD^nP^-1, invert both sides to see that A^-n=PD^-nP^-1. Since D is a diagonal matrix, D^-n is obtained by inverting the individual eigenvalues along the main diagonal. This is of course based on the assumption that A is invertible and diagonalizable.
I hope this works for fractional powers x,o
Have you figured it out? ;-)
massive brain moment
Those are good. :-)
Hello BSCS2C
Aral well :)
Ok so in this video
PxDXPinverse=D