Similar Matrix

𝐴 and 𝐵 are similar matrices when:

• 𝐴 = 𝑃𝐵𝑃^-1

where:

• 𝑃 - any invertible matrix (𝑃 can be thought of as the change of basis matrix)

Similar Matrices - Properties

Suppose that 𝐴 = 𝑃𝐵𝑃^-1, then:

• 𝐴 and 𝐵 have exactly the same characteristic polynomial

   proof

  Suppose that 𝐴 = 𝑃𝐵𝑃^-1, where 𝐴, 𝐵, 𝑃 are 𝑛×𝑛 matrices. The characteristic polynomial of 𝐴 is 𝑑𝑒𝑡(𝐴 − 𝜆𝐼):

  1. 𝐴 − 𝜆𝐼 = 𝑃𝐵𝑃⁻¹ − 𝜆𝑃𝑃⁻¹
  2. 𝐴 − 𝜆𝐼 = 𝑃𝐵𝑃⁻¹ − 𝑃𝜆𝑃⁻¹
  3. 𝐴 − 𝜆𝐼 = 𝑃𝐵𝑃⁻¹ − 𝑃𝜆𝐼𝑃⁻¹
  4. 𝐴 − 𝜆𝐼 = 𝑃(𝐵 − 𝜆𝐼)𝑃⁻¹
  5. 𝑑𝑒𝑡(𝐴 − 𝜆𝐼) = 𝑑𝑒𝑡(𝑃(𝐵 − 𝜆𝐼)𝑃⁻¹)
  6. 𝑑𝑒𝑡(𝐴 − 𝜆𝐼) = 𝑑𝑒𝑡(𝑃) 𝑑𝑒𝑡(𝐵 − 𝜆𝐼) 𝑑𝑒𝑡(𝑃)⁻¹ # multiplicative property of determinants 𝑑𝑒𝑡(𝑃)⁻¹ = 1/𝑑𝑒𝑡(𝑃)
  7. 𝑑𝑒𝑡(𝐴 − 𝜆𝐼) = 𝑑𝑒𝑡(𝐵 − 𝜆𝐼)

• 𝐴 and 𝐵 have exact same eigenvalues

   proof

  • 𝐴𝑥 = 𝜆𝑥 #eigenvalue/eigenvector equation

  • 𝐴𝑃𝑃^-1𝑥 = 𝜆𝑥 # 𝑃𝑃^-1 = 𝐼
  • 𝑃^-1𝐴𝑃𝑃^-1𝑥 = 𝜆𝑃^-1𝑥
  • 𝐵𝑃^-1𝑥 = 𝜆𝑃^-1𝑥

  So, if 𝑥 is an eigenvector of 𝐴, with eigenvalue 𝜆, then 𝑃^-1𝑥 is an eigenvector of 𝐵 with the same eigenvalue. So, every eigenvalue of 𝐴 is an eigenvalue of 𝐵.
  Since you can interchange the roles of 𝐴 and 𝐵 in the previous calculations, every eigenvalue of 𝐵 is an eigenvalue of 𝐴 too. Hence, 𝐴 and 𝐵 have the same eigenvalues.

  Geometrically, in fact, also 𝑥 and 𝑃^-1𝑥 are the same vector, written in different coordinate systems. Geometrically, 𝐴 and 𝐵 are matrices associated with the same endomorphism. So, they have the same eigenvalues and geometric multiplicities

• 𝐴 and 𝐵 have the same number but different eigenvectors:

  • 𝑥 is an eigenvector of 𝐴 → 𝑃^-1𝑥 is an eigenvector of 𝐵

     proof

    Suppose that 𝑥 is an eigenvector of 𝐴 with eigenvalue 𝜆 (i.e. 𝐴𝑥 = 𝜆𝑥).
    Let's prove that 𝑃^-1𝑥 is also an eigenvector of 𝐵, such that 𝐵𝑃^-1𝑥 = 𝜆𝑃^-1𝑥:

    • 𝐵𝑃^-1𝑥 = 𝐵𝑃^-1𝑥
    • 𝐵𝑃^-1𝑥 = 𝑃^-1𝑃𝐵𝑃^-1𝑥
    • 𝐵𝑃^-1𝑥 = 𝑃^-1𝐴𝑥
    • 𝐵𝑃^-1𝑥 = 𝑃^-1𝜆𝑥
    • 𝐵𝑃^-1𝑥 = 𝜆𝑃^-1𝑥

    So that 𝑃^-1𝑥 is an eigenvector of 𝐵 with eigenvalue 𝜆.

  • 𝑥 is an eigenvector of 𝐵 → 𝑃𝑥 is an eigenvector of 𝐴

     proof

    Suppose that 𝑥 is an eigenvector of 𝐵 with eigenvalue 𝜆 (i.e. 𝐵𝑥 = 𝜆𝑥).
    Let's prove that 𝑃𝑥 is also an eigenvector of 𝛢, such that 𝛢𝑃𝑥 = 𝜆𝑃𝑥:

    • 𝛢𝑃𝑥 = 𝛢𝑃𝑥
    • 𝛢𝑃𝑥 = 𝑃𝐵𝑃^-1𝑃𝑥
    • 𝛢𝑃𝑥 = 𝑃𝐵𝐼𝑥
    • 𝛢𝑃𝑥 = 𝑃𝐵𝑥
    • 𝛢𝑃𝑥 = 𝑃𝜆𝑥
    • 𝛢𝑃𝑥 = 𝜆𝑃𝑥

    So that 𝑃𝑥 is an eigenvector of 𝛢 with eigenvalue 𝜆.

Other

• Let 𝐴 = 𝑃𝐵𝑃^-1, then for any 𝑛≥1, we have 𝐴^𝑛 = 𝑃𝐵^𝑛𝑃^-1

   proof

  First, note that:

  • 𝐴² = 𝐴𝐴 = (𝑃𝐵𝑃^-1)(𝑃𝐵𝑃^-1) = 𝑃𝐵(𝑃^-1𝑃)𝐵𝑃^-1 = 𝑃𝐵𝐼𝐵𝑃^-1 = 𝑃𝐵²𝑃^-1

  Next, we have:

  • 𝐴² = 𝐴²𝐴 = (𝑃𝐵²𝑃^-1)(𝑃𝐵𝑃^-1) = 𝑃𝐵²(𝑃^-1𝑃)𝐵𝑃^-1 = 𝑃𝐵³𝑃^-1

  The pattern is clear

Similar Matrices - Computing 𝐴𝑥 in Terms of 𝑃𝐵𝑃^-1𝑥

Suppose that 𝐴 = 𝑃𝐵𝑃^-1, where 𝑃 is an invertible matrix with columns 𝑣₁, 𝑣₂, ..., 𝑣_𝑛. Let 𝔹 = {𝑣₁, 𝑣₂, ..., 𝑣_𝑛}, a basis for ℝ^𝑛. Let 𝑥 be a vector in ℝ^𝑛. To compute 𝐴𝑥, one does the following:

1. Multiply 𝑥 by 𝑃^-1, which changes to the 𝔹-coordinates: [𝑥]_𝔹 = 𝑃^-1𝑥
2. Multiply this by 𝐵: 𝐵[𝑥]_𝔹 = 𝐵𝑃^-1𝑥
3. Interpreting this vector as a 𝔹-coordinate vector, we multiply it by 𝑃 to change back to the usual coordinates: 𝐴𝑥 = 𝑃𝐵𝑃^-1𝑥 = 𝑃𝐵[𝑥]_𝔹

To summarize: if 𝐴 = 𝑃𝐵𝑃^-1, then 𝐴 and 𝐵 do similar transformations, only in different coordinate systems.

Similar Matrices - Equivalence Relation

The similarity follows the equivalence relation. Let 𝐴, 𝐵, and 𝐶 be 𝑛×𝑛 matrices, then:

1. reflexivity - 𝐴 is similar to itself
2. symmetry - if 𝐴 is similar to 𝐵, then 𝐵 is similar to 𝐴
3. transitivity - if 𝐴 is similar to 𝐵 and 𝐵 is similar to 𝐶, then 𝐴 is similar to 𝐶

Similar Matrices - 𝜆-Eigenspace

If 𝐴 = 𝑃𝐵𝑃^-1, then:

• 𝑃^-1 takes the 𝜆-eigenspace of 𝐴 to the 𝜆-eigenspace of 𝐵
• 𝑃 takes the 𝜆-eigenspace of 𝐵 to the 𝜆-eigenspace of 𝐴

To see why, refer to the properties section above.

Similar Matrices - Identity Matrix

The only matrix similar to the identity matrix (𝐼) is itself

Similar Matrices - In Relation to Eigendecomposition/Diagonalization

Given the following equation 𝐴 = 𝑃^-1𝐵𝑃: if 𝐵 is a diagonal matrix, then 𝑃^-1𝐵𝑃 is an eigendecomposition of matrix 𝐴

Resources

• https://textbooks.math.gatech.edu/ila/similarity.html#similarity-eq-reln