In matrix theory, we often seek ways to simplify matrix forms to make them easier to analyze and compute. Diagonalization is one technique for doing this. Imagine transforming a complex space into a more orderly space where each dimension does not interfere with each other.
The main goal of diagonalization is to find a special basis so that the linear transformation y=A⋅x can be represented through a diagonal matrix B=S−1⋅A⋅S. If this basis is an orthonormal basis, then the transformation matrix has the property .
A matrix A∈Kn×n is called diagonalizable if it is similar to some diagonal matrix Λ∈Kn×n, that is, if there exists an invertible matrix S∈Kn×n such that:
When can a matrix A∈Kn×n be diagonalized? The answer is when we can find a basis of Kn that consists entirely of eigenvectors v1,…,vn∈Kn of A with corresponding eigenvalues λ1,…,λn∈K.
The diagonal matrix Λ is:
Λ=λ1⋱λn=diag(λ1,…,λn)
and S is the matrix with columns:
S=(v1…vn)
If A is diagonalizable, then the columns v1,…,vn of S form a basis of eigenvectors. From Λ=S−1⋅A⋅S we obtain A⋅S=S⋅Λ and thus A⋅vi=λi⋅vi for i=1,…,n.
Conversely, if v1,…,vn is a basis of eigenvectors, then S is invertible and from A⋅vi=λi⋅vi for i=1,…,n we obtain A⋅S=S⋅Λ and thus Λ=S−1⋅A⋅S.
This matrix has eigenvalue λ=1 with algebraic multiplicity μA(1)=2. The eigenspace is the kernel (null space) of A−1⋅I:
EigA(1)=Kern(A−1⋅I)
=Kern(0020)
=Span(10)
which has dimension 1. Since there are no other eigenvalues and eigenvectors, and there is no basis of K2 consisting of eigenvectors of A, then A is not diagonalizable.
If a matrix A∈Kn×n is diagonalizable, then the characteristic polynomial χA(t) of A over K factors into linear factors:
χA(t)=(λ1−t)⋯(λn−t)
where A has n eigenvalues that need not be distinct λi∈K.
When all eigenvalues are distinct, the process becomes simpler. If A∈Kn×n and the characteristic polynomial χA(t) of A over K factors into linear factors:
χA(t)=(λ1−t)⋯(λn−t)
with pairwise distinct eigenvalues λi=λj for i=j with i,j∈{1,…,n}, then A is certainly diagonalizable.
Why is this so? Because eigenvectors for pairwise distinct eigenvalues of A are always linearly independent and form a basis of Kn.
But what if A has repeated eigenvalues? We must check this more carefully. Eigenvalues have algebraic multiplicity μA(λi) and geometric multiplicity dimEigA(λi) with the relationship:
For a matrix A∈Kn×n, the following statements are equivalent:
A is diagonalizable.
Both of the following conditions are satisfied. First, the characteristic polynomial of A must factor into linear factors:
χA(t)=(λ1−t)μA(λ1)⋯(λk−t)μA(λk)
with pairwise distinct eigenvalues λ1,…,λk∈K of A. Second, for all eigenvalues of A, the algebraic multiplicity must equal the geometric multiplicity:
μA(λi)=dimEigA(λi)for i=1,…,k
The direct sum of all eigenspaces is the entire vector space:
EigA(λ1)⊕⋯⊕EigA(λk)=Kn
For each i=1,…,k, let v1(i),…,vdi(i) be a basis of eigenvectors of A for the eigenspace EigA(λi). Then:
