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I need a bit of explanation on the conditions under which there is an eigenvalue that is equal to zero and what it's "physical" meaning.
Thanks in advance.
Thanks in advance.
It takes the eigenvectors and maps in it to the 0 vector. So, I guess you can say the "collaspe" to the 0 vector. That's probably all the "physical" meaning you can get.um, actually no physics, just math class, I was trying to get a better understanding of eigenvalues and how those two say something about each other, matrices and eigenvalues/vectors that is....
Really?By definition, an eigenvalue c will be a solution to det(A-cI)=0. If c=0, then det(A)=0.
ok, what I am going to say is not much of a "physical explanation" (it will really depend on the situation and application), but it may help you visualise what is going on.um, actually no physics, just math class, I was trying to get a better understanding of eigenvalues and how those two say something about each other, matrices and eigenvalues/vectors that is....
What does that have to do with anything? You might want to reread what I said. In particular, what was the constraint on c?Really?
Let A = cI, then det(A-cI)=0, but det(A) is not equal to 0. How did you deduce that conclusion?
By definition, an eigenvalue c will be a solution to det(A-cI)=0. If c=0, then det(A)=0.
?? If det(A) is not equal to 0, then c= 0 cannot be an eigenvalue! Ziox did say "If c= 0".Really?
Let A = cI, then det(A-cI)=0, but det(A) is not equal to 0. How did you deduce that conclusion?