Register to reply 
Algebraic Multiplicity and Eigenspace 
Share this thread: 
#1
Apr910, 10:14 AM

P: 17

1. The problem statement, all variables and given/known data
Find h in the matrix A such that the eigenspace for lambda=5 is twodimensional. A= [5,2,6,1] [0,3,h,0] [0,0,5,4] [0,0,0,1] Alambda*I(n) = [0,2,6,1] [0,2,h,0] [0,0,0,4] [0,0,0,4] 2. Relevant equations 3. The attempt at a solution I'm not really sure how to do this. Is there some relation between algebraic multiplicity and the eigenspace of a matrix that would help? I tried solving this by row operations to solve for the eigenvectors in the hope that I would be able to eliminate values of h that would have resulted in more or less than a 2 dimensional eigenspace, but it didn't work out. This is the matrix I got (keep in mind this is after applying the lambda value to the diagonals) A= [0,2,6,1] [0,0, h6,1] [0,0,0,4] [0,0,0,0]. I'm pretty confused about this any help would be greatly appreciated. Thanks. 


#2
Apr910, 11:33 AM

P: 436

Hey there! the eigenspace of having lambda=5 is exactly the nullspace of Alambda I . and since it is 2 dimensional , it suggest that the rank of the matrix is ? 


#3
Apr910, 04:51 PM

P: 17

So the rank = 2 since rank = # columns (4 in this case)  dimNul A (in this case 2). So if the rank is to equal 2 then I will need another free variable, or I need to remove a pivot position. So since h6 is in a pivot position I can easily make it a nonpivot column by setting h=6. This would ensure that the dimension of the null space is 2.
Am I going the right way with this? It makes sense to me if this is right. Thanks for the help! 


#4
Apr1010, 06:27 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

Algebraic Multiplicity and Eigenspace
[tex]\begin{bmatrix}0 & 2 & 6 & 1 \\ 0 & 2 & h & 0 \\ 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & 4\end{bmatrix}\begin{bmatrix}w \\ x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0\end{bmatrix}[/tex]. That gives the equations 2x+ 6y+ z= 0, 2x+ hy= 0, 4z= 0, and 4z= 0. The last two obviously give z= 0 so the first two equations become 2x+ 6y= 0 and 2x+ hy= 0. One obvious eigenvector is (u, 0, 0, 0). There will be another, independent, eigenvector, and so the eigenspace will be two dimensional if and only if there exist nonzero x and y satifying both 2x+ 6y= 0 and 2x+ hy= 0. 


Register to reply 
Related Discussions  
Eigenvalues + Algebraic/Geometric Multiplicity  Calculus & Beyond Homework  2  
Algebraic multiplicity  Calculus & Beyond Homework  1  
Question of algebraic flavor in algebraic topolgy  Differential Geometry  2  
Dimension of eigenspace, multiplicity of zero of char.pol.  Linear & Abstract Algebra  5  
Eigenspace of A?  Calculus & Beyond Homework  1 