Quick Eigenvectors Question
Hi
The following question ask to find eigenvector:
![\left[ \begin{matrix} 3 & 2 & 0 \\ 2 & 3 & 0 \\ 0 & 0 & 1 \end{matrix} \right]](http://latex.codecogs.com/png.latex?\left[ \begin{matrix} 3 & 2 & 0 \\ 2 & 3 & 0 \\ 0 & 0 & 1 \end{matrix} \right])
Now the answer i got was correct according to book, however i don't understand why they choose these two values instead of the other one, which is the repeated eigenvalue 1:
eigenvalue 5
![\left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right]](http://latex.codecogs.com/png.latex?\left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right])
eigenvalue 1 repeated value
![\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right]](http://latex.codecogs.com/png.latex?\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right])
![\left[ \begin{matrix} -1 \\ 1 \\ 0 \end{matrix} \right]](http://latex.codecogs.com/png.latex?\left[ \begin{matrix} -1 \\ 1 \\ 0 \end{matrix} \right])
not included in answer
![\left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right]](http://latex.codecogs.com/png.latex?\left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right])
P.S
Re: Quick Eigenvectors Question
Quote:
Originally Posted by
Paymemoney
Hi
The following question ask to find eigenvector:
Now the answer i got was correct according to book, however i don't understand why they choose these two values instead of the other one, which is the repeated eigenvalue 1:
eigenvalue 5
eigenvalue 1 repeated value
not included in answer
P.S
Let
.
![A\left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right]=5\left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right]](http://latex.codecogs.com/png.latex?A\left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right]=5\left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right])
![A\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right]=1\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right]](http://latex.codecogs.com/png.latex?A\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right]=1\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right])
![A\left[ \begin{matrix} -1 \\ 1 \\ 0 \end{matrix} \right]=1\left[ \begin{matrix} -1 \\ 1 \\ 0 \end{matrix} \right]](http://latex.codecogs.com/png.latex?A\left[ \begin{matrix} -1 \\ 1 \\ 0 \end{matrix} \right]=1\left[ \begin{matrix} -1 \\ 1 \\ 0 \end{matrix} \right])
Re: Quick Eigenvectors Question
maybe my question is not clear. Is
an acceptable answer?
Re: Quick Eigenvectors Question
is (1,1,0) an eigenvector? yes, corresponding to the eigenvalue 5 (see post #2).
Re: Quick Eigenvectors Question
Quote:
Originally Posted by
Paymemoney
Hi
The following question ask to find eigenvector:
Now the answer i got was correct according to book, however i don't understand why they choose these two values instead of the other one, which is the repeated eigenvalue 1:
eigenvalue 5
eigenvalue 1 repeated value
not included in answer
P.S
I certainly don't understand your question. You say that "i don't understand why they choose these two values instead of the other one, which is the repeated eigenvalue 1". Which two "values" are you saying they give?
![\left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right]](http://latex.codecogs.com/png.latex?\left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right])
certainly is included, as an eigenvector of eigenvalue 5, not 1.