- #1

Seydlitz

- 263

- 4

My question is this. Is it possible to prove that there exist an eigenvectors for a symmetric matrix without discussing about what eigenvalues are and going into details with characteristic equations, determinants, and so on? This my short proof for that: (The only assumption is ##A## is symmetric)

Suppose there doesn't exist any vector so that ##Av = \lambda v##. This then happens ##Av = b_{1}##. ##A^{T}v = b_{2}.## Clearly ##b_{1} \neq b_{2}## and thus ## Av \neq A^{T}v##. But this implies ##A^{T} \neq A##, whereas we assume ##A## is symmetric. Thus ##v## must exist, as required.

Is this proof legit? It may be too simple but I'm not certain. The alternative would be showing that if the determinant of ##det(A−\lambda I)=0## then ##v## must exist. But then there's no mention of the symmetric property of matrix in this case.