Re: fast eigenvalue/eigenvector ?
- From: Igor Khavkine <igor.kh@xxxxxxxxx>
- Date: Sat, 20 Aug 2005 20:01:42 -0400
On 2005-08-20, David Kastrup <dak@xxxxxxx> wrote:
> "kiki" <lunaliu3@xxxxxxxxx> writes:
>
>> Here is a matrix:
>>
>> [4 -1 0;
>> -1 4 0;
>> 0 0 1]
>> how do you identify their eigenvalue and eigenvector fastly using pencil and
>> paper? 30 seconds?
> Another glance spews out an eigenvalue of 3 that makes the first rows
> orthogonal, with eigenvector of [1;1;0], and an eigenvalue 5 with
> eigenvector [1,-1,0].
Any matrix of the form
[ a b ]
[ b a ]
is invariant under the transformation that swaps the first and second
basis vectors. This transformation is
[ 0 1 ]
[ 1 0 ]
By a well known theorem, these two matrices will share eignevectors. For
the second matrix the eigenvectors are [ 1, 1 ] and [ 1, -1 ]. They will
always be eigenvectors of the first matrix. The eigenvalues can then be
read off by pluggin in the eigenvectors, (a+b) and (a-b) respectively.
If you know this little fact, then eigenvectors of the top 2x2 block in
the original post are automatically seen to be [ 1, 1 ] and [ 1, -1 ].
They give the respective eigenvalues 3 and 5.
Igor
.
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