Can anyone please help me in the question no. 25 of exercise 7.3? I am having a lot of trouble in comprehending how to calculate the multiplicity.
The geometric multiplicity of a matrix is directly related to the no. of independent eigenvectors in the eigen basis. Since, the no. of independent vectors is 3 (column wise), therefore, the geometric multiplicity is 3.
Can anyone please explain me how to go about solving Q. 14 Exercise 7.3 ? i guess i'm probably making a mistake somewhere in my calculations ...
Hello, I am getting the lambda values to be 0,0,1 for the question. The determinant value becomes 0 =-(lambda)^3+ 2(labmda)^2 - labmda.Now it should be pretty straight forward to find out the corresponding eigen-vectors.
Hi, for Q. 14 in Section 7.3, I calculated the determinant to be 0 = [-(lambda)][(1-lambda)^2]. Therefore, I got three eigenvalues of: lambda = 0, lambda = 1, lambda = 1. Which led to an eigenbasis of: [0 1 0], [1 -5 0], [0 2 1].
Hey friends, like geometric multiplicity of an eigenvalue is related to the nullity of the matrix (A- λIn), is there a way to relate algebraic multiplicity on similar terms ?
Yea.I meant orthogonal.sorry. Thank you though for the answer.
Review for final Chapter 1 &2 by B Zhou [1]
Review for final Chapter 3&4 By B zhou [2]
Various exercises from Chapters 6 & 7, by Dan Coroian (dcoroian) Final review
Hey.Can anyone please explain me the 20th question of exercise 7.1. I am not able to understand how to interpret the question. Thanks
I believe that there would be no eigenvalue corresponding to the rotation in about e3 in R3 ! However, I would recommend asking the question to Prof. Kummini in this regard !
Hey I think the eigenvalue would be 1 since any vector on the axis spanned by e3 would be an eigen vector !
