Similarity of Matrices

1. Definition

Two square matrices A and B of the same order n are similar if there exists a non-singular matrix P such that

B = P^{-1} A P (equivalently A = P B P^{-1})

P is called the transforming matrix. Similarity is an equivalence relation:

reflexive — A is similar to itself (take P = I);
symmetric — if B = P⁻¹AP then A = Q⁻¹BQ with Q = P⁻¹;
transitive — if A ~ B and B ~ C then A ~ C (multiply the P's together).

Congruence is a related notion, B = PᵀAP, used for quadratic forms (next topic); when P is orthogonal (Pᵀ = P⁻¹) the two ideas coincide, giving orthogonal similarity.

2. Invariants — Properties Preserved by Similarity

If B = P⁻¹AP, then A and B have the same:

Invariant	Why it matters
Determinant	det B = det(P⁻¹)·det A·det P = det A
Trace	trace(P⁻¹AP) = trace(APP⁻¹) = trace A (cyclic property)
Rank	Multiplication by non-singular matrices leaves rank unchanged
Characteristic polynomial and hence the eigenvalues	proof below
Minimal polynomial, nullity, index	same structure of invariants
Invertibility	if A is non-singular, so is B, and B⁻¹ = P⁻¹A⁻¹P

Proof that the characteristic polynomial is preserved (a standard exam question):

det(B - λI) = det(P^{-1}AP - λI) = det(P^{-1}(A - λI)P) = det(P^{-1})·det(A - λI)·det(P) = det(A - λI)

since det(P⁻¹)·det(P) = 1. Hence A and B have the same characteristic equation, the same eigenvalues (with the same multiplicities), the same determinant and the same trace.

3. Eigenvectors Under Similarity

If X is an eigenvector of A for the eigenvalue λ, then Y = P⁻¹X is an eigenvector of B for the same λ:

B Y = P^{-1} A P (P^{-1} X) = P^{-1} A X = P^{-1}(λ X) = λ (P^{-1} X) = λ Y

So similar matrices share eigenvalues but not eigenvectors — the eigenvectors are transformed by P⁻¹. This is exactly the mechanism of diagonalization, where the transforming matrix is built from the eigenvectors themselves.

        [ 4  1 ]                  [ 5  0 ]
   A =  [ 2  3 ]     and    B =   [ 0  2 ]

   A has characteristic equation λ² - 7λ + 10 = 0  ⇒  λ = 5, 2
   B is diagonal, so its eigenvalues (diagonal entries) are 5 and 2  ⇒ same spectrum.

   Eigenvectors of A (from the previous topic):  X₁ = (1,1)ᵀ for λ = 5 , X₂ = (1,-2)ᵀ for λ = 2.
   Take  P = [ 1   1 ]  (columns are the eigenvectors) ; det P = (1)(-2) - (1)(1) = -3 ≠ 0 ✓
             [ 1  -2 ]

   P⁻¹ = (1/-3)[ -2  -1 ] = (1/3)[ 2   1 ]
                [ -1   1 ]        [ 1  -1 ]

   A·P = [ 4  1 ][ 1   1 ] = [ 4+1   4-2 ] = [ 5   2 ]
         [ 2  3 ][ 1  -2 ]   [ 2+3   2-6 ]   [ 5  -4 ]

   P⁻¹·(A·P) = (1/3)[ 2   1 ][ 5   2 ] = (1/3)[ 10+5   4-4 ] = (1/3)[ 15   0 ] = [ 5  0 ] = B ✓
                     [ 1  -1 ][ 5  -4 ]        [  5-5   2+4 ]        [  0   6 ]   [ 0  2 ]
   ⇒  A and B are SIMILAR, with B = P⁻¹AP.

7. Similarity and Diagonalization

A is diagonalizable if and only if it is similar to a diagonal matrix: P⁻¹AP = D = diag(λ₁, …, λₙ), where the columns of P are the eigenvectors of A and the diagonal entries of D are the matching eigenvalues. Every real symmetric matrix is orthogonally similar to a diagonal matrix (P orthogonal, P⁻¹ = Pᵀ) — the basis of the "principal axes" reduction of quadratic forms. Any square matrix is similar to (i) its own transpose, and (ii) an upper triangular matrix with its eigenvalues on the diagonal.

8. Exam Points

Define similar matrices and state the transforming relation B = P⁻¹AP — 5 marks, almost certain.
Prove that similar matrices have the same characteristic polynomial (hence the same eigenvalues) — 5-7 marks, very frequent.
Prove that similar matrices have the same determinant and the same trace — 5 marks.
Verify that two given matrices are similar by finding P with B = P⁻¹AP — 5-7 marks, frequent.
Given A similar to B, find det B / trace B / eigenvalues of B from those of A — 3-5 marks.
Show that two matrices with the same eigenvalues need not be similar (give the 2I vs Jordan block example) — 5 marks.
Explain the connection between similarity, diagonalization and orthogonal similarity — 3-5 marks.

Similarity of Matrices