In linear algebra, we are often interested in special vectors that have unique properties when multiplied by matrices. Imagine vectors that are only "stretched" or "shortened" by the matrix, but their direction remains unchanged.
Let A∈Kn×n be a square matrix. An eigenvectorv∈Kn for an is a non-zero vector that satisfies:
eigenvalue
λ∈K
v=0
A⋅v=λ⋅v
This equation shows that when matrix A operates on vector v, the result is a scalar multiple of the same vector.
By definition, eigenvalues can equal 0, but eigenvectors are always non-zero.
Eigenvectors have fundamental properties that are very useful in various mathematical applications.
Scalar Multiplication: Let A∈Kn×n and v∈Kn with v=0 be an eigenvector of A for eigenvalue λ∈K. Then all multiples t⋅v with t=0 are also eigenvectors of A for the same eigenvalue λ.
Why is this true? A⋅(t⋅v)=t⋅(A⋅v)=t⋅(λ⋅v)=λ⋅(t⋅v)
This property shows that if we find one eigenvector, then all its non-zero multiples are also eigenvectors for the same eigenvalue.
One important result in eigenvector theory is that eigenvectors corresponding to different eigenvalues are always linearly independent.
There's a very important result about linear independence of eigenvectors. Let A∈Kn×n and λ1,…,λk∈K be pairwise distinct eigenvalues of A, that is λi=λj for i=j with i,j∈{1,…,k}. Then the corresponding eigenvectors v1,…,vk∈Kn are linearly independent.
This theorem can be proven using mathematical induction and has the important consequence that an n×n matrix has at most n distinct eigenvalues.
Eigenvalues have a close relationship with the invertibility property of matrices.
Now, let's look at an interesting relationship between invertibility and eigenvalues. Matrix A∈Kn×n is invertible if and only if all eigenvalues λ∈K of A satisfy λ=0.
Why is this true? A is invertible if and only if RankA=n, which means EigA(0)=kerA={0}, so λ=0 is not an eigenvalue of A.
