when you compute eigenvalues, the part that you need to form a base . Does the order matter? because i mostly get contradictionary results like
a(1,0,-1) while in the textbook it says a(-1,01) because i extracted an other variable. Does this make a difference? because the next step is combining those bases to do a diagonalization

- anonymous

- jamiebookeater

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- Loser66

The order of eigenvalues doesn't matter!! but the order of theirs eigenvectors is important!!

- ganeshie8

If \(\large \mathrm{e}\) is an eigenvector for eigenvalue \(\lambda\), then any scalar multiple of it, \(\large a \mathrm{e}\) is also an eigenvector of the same eigenvalue \(\lambda\)

- ganeshie8

It helps to keep in mind that there is no "the basis" for a vector space
any linearly independent vectors are fine for "a basis"

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## More answers

- Loser66

|dw:1437765054056:dw|

- Loser66

|dw:1437765276787:dw|

- Loser66

got what I mean?

- anonymous

so this is the same base?

##### 1 Attachment

- ganeshie8

what is your actual matrix that you're trying to diagonalize ?

- anonymous

1 3 3
-3 -5 -3
-3 3 1

- anonymous

everything goes perfect until i need to make bases for my found eigenvalues

- Loser66

Post the original problem, please, by snapshot or scanning!!

- anonymous

##### 1 Attachment

- ganeshie8

1, -2, -2 are your eigenvalues ?

- anonymous

and this is the characteristic equation : |dw:1437765738465:dw|

- anonymous

yes

- ganeshie8

and the corresponding eigenvector matrix is
\[\begin{bmatrix}
1&-1&-1\\
-1&1&0\\
1&0&1
\end{bmatrix}\]
?

- anonymous

yes but i got this:
-1 1 1
1 -1 0
1 0 -1

- ganeshie8

doesn't matter,
both your matrix and mine are same up to a scalar multiple, which is fine.

- anonymous

thank you! and the multiplicity is 2 which implies that you can make a variable 0 ? because it's a linear combination of the other ?

- ganeshie8

what do you mean by making a variable 0 ?

- ganeshie8

this is a nice matrix because the geometric multiplicity for eigenvalue -2 is also 2 :
you got two linearly independent eigenvectors for the eigenvalue -2!

- anonymous

well i find it a bit confusing . so your eigenvalue is -2
so you get after all one equation out of the matrix
x+y+z=0
which gives x=-y-z
and then they say :
we make two bases , one where z is zero and one where y is zero
which gives us
(-1,1,0) and (-1,0,1)

- ganeshie8

Exactly! the nullspace of \(A-(-2)I = 0\) has dimension of \(2\),
thats the reason you got 2 independent eigenvectors for the eigenvalue \(-2\)

- ganeshie8

so as you can see, repeated eigenvalues aren't bad always

- anonymous

well it would be even more fun if we get complex forms with a bigger multiplicity than 1 :p i guess hha

- anonymous

that's what proven here indeed

- ganeshie8

you can get multiple "linearly independent" eigenvectors for the same repeated eigenvalue

- ganeshie8

Yea..

- anonymous

well u really helped me a lot! and loser66 you too :) ! Thank you!!

- ganeshie8

np

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