Tuesday, April 3, 2018

linear algebra - Eigenvalues of a nxn matrix without calculations


I have a question about the following matrix:


$$ \begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end{bmatrix} $$


Find the eigenvalues without calculations and define your answer. Now, I was thinking about this problem. And I thought, yeah ok if you try the vector (1,1,1), you can find 6 as one eigenvalue (and I know you have a double multiplicity 0 too). But than you are doing sort of guessing/calculation work.



I see that the columns are linearly dependant. So I know the dimension of the column space and of the null space.


Thank you in advance.


EDIT: follow up question:


Ok, so you find that the dimension of the null space is 2, so there are 2 eigenvectors when the eigenvalue is 0. Now my question is, can the dimension of the eigenspace be bigger than the amount of eigenvalues? I guess not. I know it can be smaller


Answer



Notice that rank=1 and hence $0$ is an eigenvalue of multiplicity $2$. Then trace=sum of eigenvalue and hence the last eigenvalue is $6$.


It is also rather easy to find all eigenvectors without a lot of work. For $6$ the vector is $(1,1,1)$. For $0$ you can take basis $(2,-1,0),(3,0,-1)$.


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