# eigenvalues and eigenvectors

#### matamat19

Hi,
I have a problem ,
Give an example to the T Ïµ L(R ^ 4) operator with no real eigenvalues. Please explain

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#### romsek

Math Team
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$T$ is a linear transformation $\mathbb{R}^4 \to \mathbb{R}^4$

#### Greens

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:
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#### matamat19

thank you so much

#### matamat19

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