What about the Calculus by Edwin E. Moise? Is it better than Thomas's 3rd ed Calculus? I found two editions of Moise's calculus, one in 1967 and the other in 1972.
The reason for the confusion is that my original idea was to prove that the subspace normal to ##n_3## is a eigenspace with eigenvalue equal to ##\lambda_1##. I'm not taking this as a given premise.
I might know what you mean. If ##v_2=k_1n_1+k_2n_2##, then ##Sv_2=\lambda_1v_2##. I knew that from the beginning. And it all comes with a premise: there are two eigenvectors(##n_1##, ##n_2##) that are orthogonal to each other in the subspace in which any vector is normal to ##n_3## and their...
Here is my idea of proof. First, prove that ##v_2## is an eigenvector. You can see that in #13. And then let's prove that its eigenvalue is ##\lambda_1##. And to prove that, I found that any eigenvector with an eigenvalue of ##\lambda_3## is linearly dependent. So ##\lambda_3## is not the...
I rewrote the proof as follows:
Expand ##n_3## with eigenvalue equal to ##\lambda_3## in an orthogonal base formed by eigenvectors: ##n_1^*,n_2^*,n_3^*##(##n_1^*,n_2^* ##with eigenvalue equal to ##\lambda_1,n_3^*## with eigenvalue equal to ##\lambda_3##)$$n_3=k_1n_1^*+k_2n_2^*+k_3n_3^*,$$...
Expand ##v_2## in an orthogonal base formed by eigenvectors: ##n_1,n_2,n_3,##$$v_2=k_1n_1+k_2n_2+k_3n_3,$$$$Sv_2=\lambda_1\left(k_1n_1+k_2n_2\right)+\lambda_3k_3n_3,$$but the subspace spanned by ##n_1,n_2## and the subspace by ##n_3## do not intersect, so either ##k_1=k_2=0## or ##k_3=0##.
if...
##n_3## is a eigenvector with eigenvalue equal to ##\lambda_3##.Because ##\lambda_1 \ne \lambda_3##, so there is a eigenvector ##n_1## with eigenvalue equal to ##\lambda_1## and ## n_1 \perp n_3##.
Let ##v_2 \perp n_1## and ##v_2 \perp n_3##,next we can prove ##v_2## is a eigenvalue too. For a...