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- Thread starter saadsarfraz
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A theorem is that an eigenvalue's geometric multiplicity (the dimension of the eigenspace) is less than or equal to its algebraic multiplicity (its multiplicity in the characteristic equation). For example, if the characteristic equation is (x-1)^2(x+2) then since 1 has algebraic multiplicity 2, you know that there are at most 2 linearly independent eigenvectors with eigenvalue 1. Since -2 has multiplicity 1, there is exactly one eigenvector (up to scale).

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HallsofIvy

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Suppose [itex]Au= \lambda_1 u[/itex] and [itex]Av= \lambda_2 v[/itex] where [itex]\lambda_1\ne\lambda_2[/itex], u and v non-zero. That is, that u and v are eigenvectors of A corresponding to distinct eigenvalues. Let [itex]a_1u+ a_2v= 0[/itex]. Applying A to both sides of the equation, [itex]a_1A(u)+ a_2A(v)= 0[/itex] or [itex]a_1\lambda_1 u+ a_2\lambda_2 v= 0[/itex].

First, if [itex]\lambda_1= 0[/itex], then we have [itex]a_2\lambda_2 v= 0. Further,[itex]\lambd_2[/itex] is non- zero because the eigenvalues are distinct so it follows that [itex]a_2= 0[/itex]. If [itex]\lambda_1\ne 0[/itex], we can divide by it and get

[tex]a_1u+ \frac{\lambda_2}{\lambda_1}a_2 v= 0[/itex]. Since we also have that [itex]a_1u+ a_2v= 0[/itex], it follows that

[tex]\frac{\lambda_2}{\lambda_1}a_2 v= a_2 v[/tex]

again giving [itex]\lambda_2= 0[/itex].

If two eigenvectors correspond to the

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