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[SOLVED] set of eigenvectors is linearly independent

Fermat

Active member
Nov 3, 2013
188
I know eigenvectors corresponding to different eigenvalues are linearly independent but what about a set ${e_{1},...,e_{n}}$ of eigenvectors corresponding to different eigenvalues?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I don't understand your question because I don't see how the two parts of your question are different.

When you say
I know eigenvectors corresponding to different eigenvalues are linearly independent
Do you mean that two eigenvectors corresponding to two different eigenvalues
but what about a set ${e_{1},...,e_{n}}$ of eigenvectors corresponding to different eigenvalues?
but asking, "what if there are more than two?". One can show generally, "if, in a set of vectors, any two are independent (au+ bv= 0 only if a= b= 0 which is the same as saying that b is NOT a multiple of a and vice-versa) then all the vectors are independent."
One can prove that by first proving 'if [tex]u_1, u_2, ... u_n[/tex] are each independent of v, then so is [tex]a_1u_1+ a_2u_2+ ...+ a_nu_n[/tex] is independent of v for any numbers [tex]a_1[/tex], [tex]a_2[/tex], ..., [tex]a_n[/tex].