Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Anyone good at Eigen Vectors.

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
im good at eigen values
ohh finally.. ok i have a question.. i know how to solve.. just for oncae type of case i dont know how to solve.. i will post up the question [ 6 5 2 2 0 -8 5 4 0 ] this is the matrix.
do you have your eigen values ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

yes i have .. its 2
If I am not getting it wrong, then I think we will do like this to find eigen values : A - IX = 0 Something like that... No knowledge... @UnkleRhaukus please help your friend (me)...
Where is lambda?? Oh my God...
\[\textbf Ax=\lambda x\]\[(\textbf A-\lambda \textbf I)x=0\]
Oh yeah.... I forgot all the things... One day, I found a page thrown by someone on the road, I read that.. There on the page, it was shown how to find Eigen values but the question was not complete.. And I have forgot that too...
i have found lammida.. and its 2
there is repeated roots?
yes all 3 values of lamida are 2
I try to find eigen value now... You can carry on @UnkleRhaukus
\[\textbf A=\left[\begin{array}{ccc}6&5&2\\2&0&-8\\5&4&0\end{array}\right]\]\[x=\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]\]\[\lambda=2\] \[\left[\begin{array}{ccc}6&5&2\\2&0&-8\\5&4&0\end{array}\right]\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]=2\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]\]
\[\left[\begin{array}{ccc}6-2&5&2\\2&0-2&-8\\5&4&0-2\end{array}\right]\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]=\left[\begin{array}{ccc} 0\\0\\0\end{array}\right]\]
ok.. got this far...
you have three simultaneous equations
ok so i have to just solve simultaneously to get first eigen vector?
something like that, i always go confused when it comes to eigenvectors
i think first eigen vector is x1= = 2 x2 = -2 x3 = 1
\[4x_1 + 5x_2 + 2x_3 = 0\] \[2x_1 -2x_2 - 8x_3 = 0\] \[5x_1 + 4x_2 - 2x_3 = 0\]
@nubeer , that certainly solves the three equations, how did you get there?
i have found x1 and x2 in terms of x3 while using subsitution method.
ok thanks guys.. i think i have done it... thank you very much espacially @UnkleRhaukus :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question