Monday, April 18, 2016

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 $(2,-1,0),(3,0,-1)$.



No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...