I need aid finding an eigenspace matching to each eigenvalue of A = $\beginbmatrix 1 & -1 & 0 \\ 2 & 4 & 0 \\ 9 & 5 & 4 \endbmatrix$ ?

I adhered to standard eigen-value detect procedures, and also I was able to find that $\lambda = 4, 2, 3$. I was also able to uncover the basis corresponding to $\lambda = 4$:

$\beginbmatrix 0\\ 0\\ 1 \endbmatrix$

However, ns am can not to uncover the basis corresponding to $\lambda = 2, 3$. I would certainly really appreciate it if someone might please aid me through this.

You are watching: Find a basis for the eigenspace corresponding to the eigenvalue of a given below.

$\lambda = 2$:

First us compute A - I$\lambda$ come get:

$\beginbmatrix -1 & -1 & 0 \\ 2 & 2 & 0 \\ 9 & 5 & 2 \endbmatrix$

linear-algebra eigenvalues-eigenvectors
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edited Apr 4 "17 at 3:29
request Apr 4 "17 in ~ 3:25
user400359user400359
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$\begingroup$ i did that, yet I to be still unable to discover the basis. I will update my article with this computed. Have the right to you please help me through the problem? $\endgroup$
–user400359
Apr 4 "17 at 3:28
$\begingroup$ i updated mine post. $\endgroup$
–user400359
Apr 4 "17 at 3:29
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$\beginbmatrix -1 & -1 & 0 \\ 2 & 2 & 0 \\ 9 & 5 & 2 \endbmatrix\beginbmatrix x_1\\ x_2 \\ x_3 \endbmatrix=\thedesigningfairy.combf 0$

$x_1 + x_2 = 0\\x_2 = -x_1\\9x_1 + 5(-x_1) + 2x_3 = 0\\x_3 = -2x_1$

$\thedesigningfairy.combf x = \beginbmatrix 1\\ -1\\ -2 \endbmatrix$

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reply Apr 4 "17 in ~ 3:38

Doug MDoug M
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Remember that the eigenspace of an eigenvalue $\lambda$ is the vector an are generated through the matching eigenvector.

So, every you should do is compute the eigenvectors and also check how many linearly independent facets you can form from calculating the eigenvector.

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answer Apr 4 "17 in ~ 3:41

Aspiringthedesigningfairy.comematicianAspiringthedesigningfairy.comematician
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You have actually computed $A - 2 I$. If friend cannot view what the nullspace is immediately, you deserve to do heat operations to gain it into RREF form.

\beginbmatrix1 & 1 & 0\\2 & 0 & 1\\0 & 0 & 0\endbmatrix

Then the is much easier to view that the nullspace is composed of vectors the the type $(x, -x, -2x)$.

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reply Apr 4 "17 in ~ 3:43

angryavianangryavian
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