Question

It can be shown that the algebraic multiplicity of an eigenvalue λ is always greater than or equal to the dimension of the eigenspace corresponding to λ. Find a value of h in the matrix A below such that the eigenspace for λ = -2 is two-dimensional:

Answer 1

by eigen space do u mean the space possible by the eigen vectors

Answer 2

can u define eigenspace for me?

Answer 3

let us ask @ganeshie8

Answer 4

i am not sure,
wolfram defines it the same way you're saying.. and its just adding 0 vector also to make it a subspace

http://mathworld.wolfram.com/Eigenspace.html

Answer 5

its obvious that number of eigenvectors cannot be more than the multiplicity of eigenvalue hmm

Answer 6

so if the OP shows us the matrix A,
we can solve the characteristic equation for h, such that the eigenvalue -2 has a multiplicity of 2 and produces two independent eigenvectors

Answer 7

where is my matrix A T_T

Answer 8

try this


1 2 3
4 5 h
7 8 9

Answer 9

or if u want a simpler matrix

-2 1 2
0 h 3
0 0 1

Answer 10

for 2 dimensions basically u are picking an h such that you are only left with 1 row as long as that row as atleast 3 nonzero components

Answer 11

or well as long as u have 2 free variables to pick from its fine

Answer 12

i believe u dont even need to have 2 row reductions for that u can have just 1 row reduces but u must be left with 2 free variables so u can set one to zero and the otehr one will atomatically then be in its own indepntant space so u will have 2 independant vectors giving u a 2d space

Answer 13

i think that makes sense, its easy for a 2x2 matrix but we know that not all repeated eigenvalues produce independent eigenvectors. so the multiplicity is always greater than the dimension of eigenspace

-2 h
0 -2

Answer 14

guess we want the matrix to be symmetric to have independent eigenvectors

Answer 15

so the only value of `h` for which the dimension of eigenspace for -2 is 2 is h=0