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eigenvectors (

6	−1
2	3

)

Solution

(

), (

)

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Solution steps

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(

6	−1
2	3

)

Matrix eigenvectors

Let A be a square matrix, η a vector and λ a scalar that satisfy Aη=λη,
then η is an eigenvector of A and λ is the eigenvalue associated with it

To find the eigenvectors η, solve (A−λ I)η=0 for each eigenvalue λ

Find the eigenvalues for (

6	−1
2	3

): λ=5, λ=4

Eigenvectors for λ=5: (

)

Eigenvectors for λ=4: (

)

The eigenvectors for (

6	−1
2	3

)

), (

)

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6	−1
2	3

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Notebook

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

`a`	1
0	2`a`

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