1. The problem statement, all variables and given/known data If v and w are eigenvectors with different (nonzero) eigenvalues, prove that they are linearly independent. 2. Relevant equations 3. The attempt at a solution Define an operator A such that a is an nxn matrix, and Av=cIv with c an eigenvalue and v and eigenvector. Define a basis <v1...vn> in that v=vi and w=vk 1<=k<=n and 1<=i<=n, and let ci,i, be an element in A. I is the identity matrix. Consider I*v, a 1xn column matrix with its lonely nonzero (1) at position 1,i. Let the value at ci,i=c.Multiplying I*v by A gives c*Iv . If I*w (another 1xn column matrix) had 1 at position 1,i, it would correspond with ci,i on A and we would get c*Iw. But we assume w has a different eigenvalue. Therefore I*w must have its 1 at a different position to correspond with a different value on A (call it k). Since I*v must have 1 at a row different from I*w, let c*Ia1v+k*Ia2w=0, and since 1 is at different rows, and c and I and k are not zero, a1 and a2 must be 0, so we have c*I*0*v+I*k*0*w=0*v+0*w=0, thus a1 and a2 are trivial so v and w are linearly independent. I kind of have a gut feeling that this may be too wordy.