Most of the time when I see someone worry their eigenvectors are not correct, it is just because of a sign error on the eigenvectors. But in this case, your code is incorrect, and I would suggest you have one other subtle issue. First, I would note that your eigenvalues are not coming out in the correct order. The inverse power method should find the smallest eigenvalue first. And that alone tells me there is an issue in your code.
A = [3.6574, -1.9723, 0, 0.8581, 3.0865;
0.5177, 1.4469, 0, 1.1376, -3.7955;
-0.7951, 0.1142, 0.7290, -0.5829, -0.2565;
-0.7273, 2.1819, 0, 2.2816, -6.5275;
0.4301, -1.2904, 0, -0.8602, 3.5590];
[V,D] = eig(A);
D = diag(D)
Now, suppose you find the smallest eigenvalue of A, and then deflate A?
vec = V(:,5);
lambda = D(5);
Ahat = A - lambda*vec*vec'
Now, what are the eigenvalues of Ahat, after deflation?
eig(Ahat)
What is the SMALLEST eigenvalue of Ahat? ZERO. So now what will the inverse power method find when you apply it to the matrix Ahat? Things will now get confusing. The inverse power method will now "think" the smallest eigenvalue of Ahat is ZERO.
This tells us that we cannot use a simple deflation scheme, shifting things so the smallest eigenvalue is zero.
What else can we do? If we are using the inverse power method, one idea might be to shift that eigenvalue to something large instead.
lambdashift = max(sum(abs(A),2))*1.1;
You should be able to show that lambdashift is indeed larger in magnitude than any possible eigenvalue of A. We can try using that as a shift instead.
Ahat2 = A + (lambdashift - lambda)*vec*vec';
eig(Ahat2)
As you should see, this leaves the other small eigenvalues of A unchanged. And now the inverse power method will be able to again find the smallest eigenvalue.
Finally, there is the issue of your code to solve for the smallest eigenvalue/vector, using the inverse power method. That still needs to be fixed...
