It can be shown that the algebraic multiplicity of an eigenvalue lambda is always greater than or equal to the dimension of the eigenspace correspondi

fortdefruitI 2021-01-24 Answered
It can be shown that the algebraic multiplicity of an eigenvalue lambda is always greater than or equal to the dimension of the eigenspace corresponding to lambda. Find h in the matrix A below such that the eigenspace for lambda = 5 is two-dimensional: A=[526103h000540001]
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yagombyeR
Answered 2021-01-25 Author has 92 answers

1) Row reduce the augmented matrix for the equation (A5I)x=0
2) [0261002h000004000040]=[0261002h6100004000040]=[0130000h6000001000000]
3) For a two-dimensional eigenspace, the system above needs two free variables. his happens if and inly if h=6
[01300000000001000000]

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