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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
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 T Ïµ L(R ^ 4)
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April 15th, 2019, 01:16 PM   #3
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$T$ is a linear transformation $\mathbb{R}^4 \to \mathbb{R}^4$

 April 15th, 2019, 02:50 PM #4 Member     Joined: Oct 2018 From: USA Posts: 76 Thanks: 51 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 Thanks from topsquark and matamat19 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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