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Similar Matrices — Definition, Formula & Examples

Similar matrices are square matrices AA and BB where B=P1APB = P^{-1}AP for some invertible matrix PP. They represent the same linear transformation but expressed in different bases.

Two n×nn \times n matrices AA and BB are similar if there exists an invertible n×nn \times n matrix PP such that B=P1APB = P^{-1}AP. Similarity is an equivalence relation on the set of n×nn \times n matrices: it is reflexive, symmetric, and transitive. Similar matrices share the same eigenvalues, determinant, trace, rank, and characteristic polynomial.

Key Formula

B=P1APB = P^{-1}AP
Where:
  • AA = An $n \times n$ matrix
  • BB = An $n \times n$ matrix similar to $A$
  • PP = An invertible $n \times n$ matrix (the change-of-basis matrix)

How It Works

To show two matrices are similar, you need to find an invertible matrix PP that conjugates one into the other. In practice, you often check necessary conditions first: if AA and BB have different eigenvalues, traces, or determinants, they cannot be similar. If those match, you attempt to find PP by solving AP=PBAP = PB for the columns of PP. Diagonalization is a special case — when you diagonalize AA into D=P1APD = P^{-1}AP, you are showing AA is similar to a diagonal matrix DD.

Worked Example

Problem: Verify that A and B are similar, where A=(2103)A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}, B=(3012)B = \begin{pmatrix} 3 & 0 \\ 1 & 2 \end{pmatrix}, and P=(0110)P = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.
Step 1: Compute P1P^{-1}. Since PP swaps rows, it is its own inverse.
P1=(0110)P^{-1} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
Step 2: Compute APAP.
AP=(2103)(0110)=(1230)AP = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix}
Step 3: Compute P1APP^{-1}AP.
P1AP=(0110)(1230)=(3012)=BP^{-1}AP = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 1 & 2 \end{pmatrix} = B
Answer: Since P1AP=BP^{-1}AP = B, the matrices AA and BB are similar. Notice they share the same trace (55), determinant (66), and eigenvalues (22 and 33).

Why It Matters

Similar matrices appear throughout linear algebra whenever you change basis — for instance, diagonalizing a matrix to compute its powers efficiently. In differential equations and control theory, recognizing similarity lets you simplify a system into a canonical form without changing its fundamental behavior.

Common Mistakes

Mistake: Assuming that equal eigenvalues, trace, and determinant guarantee similarity.
Correction: These are necessary but not sufficient conditions. For example, the 2×22 \times 2 identity matrix and (1101)\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} share eigenvalues, trace, and determinant, but they are not similar.