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eigenvectors (

0	9	3
2	0	4
3	7	0

)

Solution

(

−2

), (

−2(29−4√145)

37−√145

), (

−2(29+4√145)

√145+37

)

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Solution steps

One step at a time

(

0	9	3
2	0	4
3	7	0

)

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 (

0	9	3
2	0	4
3	7	0

): λ=−5, λ=5+√1452 , λ=5−√1452

Eigenvectors for λ=−5: (

−2

)

Eigenvectors for λ=5+√1452 : (

−2(29−4√145)

37−√145

)

Eigenvectors for λ=5−√1452 : (

−2(29+4√145)

√145+37

)

The eigenvectors for (

0	9	3
2	0	4
3	7	0

)

−2

), (

−2(29−4√145)

37−√145

), (

−2(29+4√145)

√145+37

)

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

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