To find the eigenvalues of a matrix, we need a very important mathematical tool in linear algebra. Imagine we want to find all values λ that make the matrix A−λI become singular (not invertible).
Let A∈Kn×n. We can form a special function:
χA(t)=det(A−t⋅I)
χA(t)=an⋅tn+an−1⋅tn−1+⋯+a1t+a0
This function is a polynomial of degree n in t∈K, which we call the characteristic polynomial of A.
with coefficients a0,…,an−1,an∈K.
In fact, χA(t) is indeed a polynomial of degree n for every matrix A∈Kn×n.
Now, what happens if an eigenvalue appears multiple times as a root of the characteristic polynomial? Let A∈Kn×n and λ∈K. The multiplicity of the root t=λ of the characteristic polynomial χA(t) is called the algebraic multiplicityμA(λ) of the eigenvalue λ of A. We say λ is an eigenvalue with multiplicity μA(λ) of A.
Let A=313200−1−41∈K3×3. The characteristic polynomial of A is:
χA(t)=det3−t132−t0−1−41−t
=(3−t)⋅(−t)⋅(1−t)−2⋅(1⋅(1−t)+3⋅4)+1⋅3⋅t
=−3t+3t2+t2−t3−2+2t−24−3t
=−t3+4t2−4t−26
For K=R, χA(t) only has the root λ1≈−1.8003 with algebraic multiplicity μA(λ1)=1.
For K=C, χA(t) has roots λ1≈−1.8003, λ2≈2.9001+2.4559i, and λ3≈2.9001−2.4559i with algebraic multiplicities μA(λ1)=μA(λ2)=μA(λ3)=1 respectively.
Similar matrices have a fascinating property: they have the same characteristic polynomial, and therefore have the same eigenvalues, the same trace, and the same determinant.
Let's see why this is true. Let S∈Kn×n be invertible and B=S−1⋅A⋅S. Then:
Now, what about the eigenvectors of similar matrices? Let A,B∈Kn×n be similar matrices with B=S−1⋅A⋅S and invertible matrix S∈Kn×n. If λ∈K is an eigenvalue of both A and B, and v∈Kn is an eigenvector of A for eigenvalue λ, then w=S−1⋅v is an eigenvector of B for eigenvalue λ.
Let's see why this is true. Let A⋅v=λ⋅v and w=S−1⋅v. Then:
B⋅w=S−1⋅A⋅S⋅S−1⋅v=S−1⋅A⋅v
=S−1⋅λ⋅v=λ⋅S−1⋅v=λ⋅w
This shows that similarity transformation not only preserves eigenvalues, but also provides a systematic way to transform eigenvectors.