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April 14th, 2019, 01:11 PM  #1 
Newbie Joined: Apr 2019 From: turkey Posts: 3 Thanks: 0  eigenvalues and eigenvectors
Hi, I have a problem , Give an example to the T Ïµ L(R ^ 4) operator with no real eigenvalues. Please explain 
April 15th, 2019, 01:00 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,732 Thanks: 689  Quote:
 
April 15th, 2019, 01:16 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,408 Thanks: 1310  
April 15th, 2019, 02:50 PM  #4 
Newbie Joined: Oct 2018 From: USA Posts: 19 Thanks: 13 Math Focus: Algebraic Geometry 
So, where $\lambda$ is the eigenvalue, and $A$ is the transform matrix: $\displaystyle AX=\lambda IX$ $\displaystyle (A\lambda I)X = 0$ Since X can only be 0, we know that $(A\lambda I)$ must be rank 4 and therefore $(A\lambda I)$ is row equivalent to $I$. Therefore you need $A$ to be some matrix that stays rank 4 no matter what you add/subtract from the diagonals. $A= \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 4 & 0 & 0 & 0 \end{bmatrix} $ Works since no matter what lambda is the matrix won't have any empty rows or columns. In this case $T = AX$ EDIT: Made a mistake in the original, if there are only $1$'s vectors with all equal elements will be eigenvectors. Apologies Last edited by Greens; April 15th, 2019 at 03:20 PM. Reason: typos. Mistake Fixed 
April 15th, 2019, 03:00 PM  #5 
Newbie Joined: Apr 2019 From: turkey Posts: 3 Thanks: 0 
thank you so much

April 15th, 2019, 03:51 PM  #6 
Newbie Joined: Apr 2019 From: turkey Posts: 3 Thanks: 0 
I have one more question... Find the T is a linear transformation C3â†’C3 operator with eigenvalues 6 and 7 so that no matrix is diagonal 

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