What is the Best Way to See if Vectors are Equal? Maybe this is a stupid question, but when I started to think about it I started to feel rather unsure. The question is what is the best way to see if vectors, or more specifically eigenvectors are equal? I want to compare two different methods that generates eigenvalues and eigenvectors and I want to show that the eigenvectors I obtained are more or less equal. I know I can compare the norm of the vectors and see if they are equal to each other, but is this really saying that they are equal? They have a direction too! My eigenvectors contains many elements so I can't really put them next to each other and say "Look! They are equal!". What is the best way to tell if two eigenvectors are equal?
 A: You could compute the dot product of the two vectors, and if they are parallel (same direction) their dot product will be equal to the product of their individual norms. Then you can check norms and see if they are equal (same magnitudes).
A: Two vectors are equal if and only if all its components are equal. 
So (1,2,3) is equal to (1,2,3). But (1,2,3) is not
equal to (1, 40, 3) as the 2nd components are different.     
A: One way you could do it is by taking the component-wise difference between the vectors and then checking that the resulting vector is equal to the $0$ vector. 
This method makes it easier to "see" the vectors are the same. For example it is much easier to confirm $$(0,0,12390330)\ne\vec{0}$$
rather than $$(18921049890,128433,352983620)\ne(18921049890,128433,340593290)$$
A: The norm will not tell you much, because if $v$ is an eigenvector, so is any multiple of $v$.
What you can do is to first normalise all your vectors to the same length. Then you can check the norm of the difference of two vectors.
As two vectors are equal if and only if their difference is $0$, and the norm of a vector is $0$ if and only if the vector is $0$.
If you have degenerate eigenvalues, then the corresponding eigenvectors will span a linear subspace. Then you need to check if the subspaces spanned by the eigenvectors found using the two different methods are the same.
A: Similar to Peter's answer above, we must check that each component is equal to its equivalent in the other vector. If your vectors are algebraic not numeric then if one change of variables, re-parametrization or other transformation will take all elements in one vector to the other, then they may be seen as equal or related under the given transformation.
