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Old 23rd February 2013, 03:54 PM   #1 (permalink)
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Finding Eigenvector!

i need help with just 1 question...

given a matrix of:
A = {{1,0},{0,1}}

how do i find my eigenvector???

finding the eigenvalue is easy, lambda = 1... but when try to sub, (lambda* I - A) x = 0,
lambda * I = I (identity matrix)
with "I - A", i would get a value of 0?
wouldn't tat make my eigenvector [1 1]?

how did they get the algebraic and geometric multiplicities of these values to be 2 (AM =2, GM=2)....

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Old 23rd February 2013, 06:04 PM   #2 (permalink)
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Re: Finding Eigenvector!

A is the identity matrix right? So wouldn't the eigenvector be any vector of the 2nd dimension? The eigenspace would be span{ [1 0], [0 1]}

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Old 23rd February 2013, 06:06 PM   #3 (permalink)
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Re: Finding Eigenvector!

Originally Posted by megaman123 View Post
A is the identity matrix right? So wouldn't the eigenvector be any vector of the 2nd dimension? The eigenspace would be span{ [1 0], [0 1]}
sorry. i don't really get it.. care to elaborate?

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Old 23rd February 2013, 06:14 PM   #4 (permalink)
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Re: Finding Eigenvector!

Okay, Ax=b*x, here the eigenvalue is b and egienvector is x.
Now, Ax=x as A is the identity matrix. So b=1.
Note that x can be any vector and we will still get Ax=x. So any vector can be the eigenvector of A associated with the eigenvalue 1.

Erm... maybe i am not clear enough... i go type out a working for this problem...


Last edited by megaman123; 23rd February 2013 at 06:39 PM.
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Old 23rd February 2013, 08:22 PM   #5 (permalink)
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Re: Finding Eigenvector!

Here is my solution to it. Hope it is better explained here.

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Old 24th February 2013, 02:44 PM   #6 (permalink)
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Re: Finding Eigenvector!

Sorry.. But how is it that u know that a(1,0) and b(0,1)....

I was also searching online abit. But y is it that if the identity matrix is a 3x3, the last row "c" would be zero.

PS: maybe I know y it's a(1,0) and b(0,1).. Coz x is also equals to a identity matrix rite??
However y is the solution different for 3x3


Last edited by KiLlErDeViL; 24th February 2013 at 03:03 PM.
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Old 24th February 2013, 07:45 PM   #7 (permalink)
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Re: Finding Eigenvector!

Hope this clarify things further.
oh ya... x is a vector not a matrix...
i don't understand what you mean by:
"But y is it that if the identity matrix is a 3x3, the last row "c" would be zero."
Anywayz i included a possible solution to a 3x3 identity matrix.

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Old 24th February 2013, 10:43 PM   #8 (permalink)
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Re: Finding Eigenvector!

i think i get it... Sorry, i saw the wrong question when i talked abt 3x3... i get it now.

the qn i saw was a 3x3, upper triangular matrix. (but i see wrongly and tot tat it was a identity matrix)


PS: nvr in my life would i have ever thought that matrix can get so complex (span, nullspace, etc).. sigh.. so many things to rmb...

THANK YOU VERY MUCH!!!


Last edited by KiLlErDeViL; 24th February 2013 at 10:45 PM.
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