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Linear Algebra For Undergraduate Students

Yilan(Peter) Tan

Section 6.5 Similar Matrices

In this section, we introduce the notion of similar matrices—a relationship between matrices that captures when they represent "the same" transformation viewed from different perspectives. This concept plays a fundamental role in understanding eigenvalue theory and matrix diagonalization in next chapter.
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Subsection 6.5.1 Definition and Basic Properties

Definition 6.5.1. Similar Matrices.

Two \(n \times n\) matrices \(A\) and \(B\) are similar if there exists an invertible \(n \times n\) matrix \(P\) such that:
\begin{equation*} B = P^{-1}AP. \end{equation*}
Equivalently, \(A = PBP^{-1}\text{.}\)
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We write \(A \sim B\) to indicate that \(A\) and \(B\) are similar. The matrix \(P\) is called a similarity transformation or change of basis matrix.
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The operation \(B = P^{-1}AP\) is called a similarity transformation or conjugation by \(P\text{.}\) At first glance, this definition may seem abstract—why should we care about this particular relationship? The answer lies in the remarkable properties that similar matrices share.
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Proof.

  1. Reflexive: Since \(A = I^{-1}AI\) and \(I\) is invertible, we have \(A \sim A\text{.}\)
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  2. Symmetric: Suppose \(A \sim B\text{,}\) so \(B = P^{-1}AP\) for some invertible \(P\text{.}\) Multiplying both sides on the left by \(P\) and on the right by \(P^{-1}\) gives:
    \begin{equation*} PBP^{-1} = PP^{-1}APP^{-1} = A. \end{equation*}
    Therefore \(A = PBP^{-1} = (P^{-1})^{-1}B(P^{-1})\text{,}\) showing \(B \sim A\text{.}\)
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  3. Transitive: Suppose \(A \sim B\) and \(B \sim C\text{,}\) so \(B = P^{-1}AP\) and \(C = Q^{-1}BQ\) for invertible matrices \(P\) and \(Q\text{.}\) Then:
    \begin{align*} C \amp= Q^{-1}BQ = Q^{-1}(P^{-1}AP)Q\\ \amp= (Q^{-1}P^{-1})A(PQ) = (PQ)^{-1}A(PQ). \end{align*}
    Since \(PQ\) is invertible (as the product of invertible matrices), we have \(A \sim C\text{.}\)
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Being an equivalence relation means that similarity partitions the set of all \(n \times n\) matrices into equivalence classes, where matrices in the same class share fundamental properties.
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Proof.

Suppose \(B = P^{-1}AP\) where \(P\) is invertible.
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  1. Determinant: Using the multiplicative property of determinants:
    \begin{align*} \det(B) \amp= \det(P^{-1}AP)\\ \amp= \det(P^{-1})\det(A)\det(P)\\ \amp= \frac{1}{\det(P)}\det(A)\det(P)\\ \amp= \det(A). \end{align*}
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  2. Trace: Using the cyclic property of trace (\(\operatorname{tr}(XY) = \operatorname{tr}(YX)\)):
    \begin{align*} \operatorname{tr}(B) \amp= \operatorname{tr}(P^{-1}AP)\\ \amp= \operatorname{tr}((P^{-1}A)P)\\ \amp= \operatorname{tr}(P(P^{-1}A))\\ \amp= \operatorname{tr}(A). \end{align*}
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  3. Rank: Since \(P^{-1}\) and \(P\) are invertible, multiplication by them preserves rank. Therefore:
    \begin{equation*} \operatorname{rank}(B) = \operatorname{rank}(P^{-1}AP) = \operatorname{rank}(A). \end{equation*}
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  4. Invertibility: This follows immediately from (1): \(A\) is invertible if and only if \(\det(A) \neq 0\text{,}\) which holds if and only if \(\det(B) \neq 0\text{.}\)
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Example 6.5.4. Similar 2×2 Matrices.

Show that \(A = \begin{bmatrix} 1 \amp 2 \\ 0 \amp 3 \end{bmatrix}\) and \(B = \begin{bmatrix} 5 \amp 6 \\ -2 \amp -1 \end{bmatrix}\) are similar.
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Solution.
We need to find an invertible matrix \(P\) such that \(B = P^{-1}AP\text{.}\) Let’s try
\begin{equation*} P = \begin{bmatrix} 2 \amp -3 \\ 1 \amp -1 \end{bmatrix}. \end{equation*}
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First, check that \(P\) is invertible: \(\det(P) = 2(-1) - (-3)(1) = -2 + 3 = 1 \neq 0\text{.}\) So
\begin{equation*} P^{-1} = \begin{bmatrix} -1 \amp 3 \\ -1 \amp 2 \end{bmatrix}. \end{equation*}
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Now compute:
\begin{align*} P^{-1}AP \amp= \begin{bmatrix} -1 \amp 3 \\ -1 \amp 2 \end{bmatrix} \begin{bmatrix} 1 \amp 2 \\ 0 \amp 3 \end{bmatrix} \begin{bmatrix} 2 \amp -3 \\ 1 \amp -1 \end{bmatrix}\\ \amp= \begin{bmatrix} -1 \amp 7 \\ -1 \amp 4 \end{bmatrix} \begin{bmatrix} 2 \amp -3 \\ 1 \amp -1 \end{bmatrix}\\ \amp= \begin{bmatrix} 5 \amp 6 \\ -2 \amp -1 \end{bmatrix} = B. \end{align*}
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Therefore \(A \sim B\text{.}\) We can verify the invariants: \(\det(A) = 3 = \det(B)\) and \(\operatorname{tr}(A) = 4 = \operatorname{tr}(B)\text{.}\)
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Remark 6.5.5. Necessary but Not Sufficient.

The invariants in Theorem 6.5.3 are necessary conditions for similarity, but not sufficient. For example, the matrices
\begin{equation*} A = \begin{bmatrix} 1 \amp 1 \\ 0 \amp 1 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix} \end{equation*}
have the same determinant (1), trace (2), and rank (2), but they are not similar. We will see in Chapter 8 that eigenvalues and eigenvectors provide the complete criteria for similarity.
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Subsection 6.5.2 Optional: Matrix Representations in Different Bases

We have defined similarity as an algebraic relationship between matrices. But where does this concept come from? The answer lies in the theory of linear transformations. This subsection explores the deep connection between similarity and change of basis—readers may skip this material on first reading, but it provides valuable geometric insight into why similarity is natural and important.
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Recall that we defined the matrix representation of a linear transformation relative to specific bases. What if we choose different bases? How are the resulting matrices related? The answer is: they are similar!
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Proof.

Let \(\mathbf{v} \in V\) be arbitrary. We trace how the coordinates of \(\mathbf{v}\) and \(T(\mathbf{v})\) transform under change of basis.
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By [provisional cross-reference: Change of Basis Formula (Ch. 4)], the transition matrix \(P = P_{\alpha \to \beta}\) relates coordinates in the two bases:
\begin{equation*} [\mathbf{v}]_\alpha = P[\mathbf{v}]_\beta \quad \text{and} \quad [T(\mathbf{v})]_\alpha = P[T(\mathbf{v})]_\beta. \end{equation*}
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By Theorem 6.2.2, we have:
\begin{equation*} [T(\mathbf{v})]_\alpha = [T]_\alpha [\mathbf{v}]_\alpha = [T]_\alpha P[\mathbf{v}]_\beta. \end{equation*}
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But we also have \([T(\mathbf{v})]_\alpha = P[T(\mathbf{v})]_\beta = P [T]_\beta [\mathbf{v}]_\beta\text{.}\) Therefore:
\begin{equation*} [T]_\alpha P[\mathbf{v}]_\beta = P [T]_\beta [\mathbf{v}]_\beta. \end{equation*}
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Multiplying both sides on the left by \(P^{-1}\text{:}\)
\begin{equation*} P^{-1}[T]_\alpha P[\mathbf{v}]_\beta = [T]_\beta [\mathbf{v}]_\beta. \end{equation*}
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Since this holds for all \([\mathbf{v}]_\beta\text{,}\) we conclude that:
\begin{equation*} [T]_\beta = P^{-1} [T]_\alpha P. \end{equation*}
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Insight 6.5.7. Geometric Meaning of Similarity.

Theorem 6.5.6 reveals the geometric origin of similarity: similar matrices represent the same linear transformation expressed in different coordinate systems. The matrix \(P\) performs the change of basis, connecting the two perspectives.
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Think of it this way: if you describe the same geometric transformation using different "languages" (coordinate systems), the matrices will look different but they’re fundamentally the same transformation. They capture the same geometric action, just written in different notation. This is why we call them similar!
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This also explains why the invariants in Theorem 6.5.3—determinant, trace, rank—don’t depend on the choice of basis. They are intrinsic properties of the transformation itself, not artifacts of how we write it down.
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Example 6.5.8. Rotation in Different Bases.

Consider the counterclockwise rotation by \(90°\) in \(\mathbb{R}^2\text{,}\) \(T(\mathbf{x}) = R_{90°}\mathbf{x}\text{.}\) In the standard basis \(\mathcal{E} = \{\mathbf{e}_1, \mathbf{e}_2\}\text{,}\) the matrix is:
\begin{equation*} [T]_{\mathcal{E}} = \begin{bmatrix} 0 \amp -1 \\ 1 \amp 0 \end{bmatrix}. \end{equation*}
Find the matrix representation in the basis \(\mathcal{B} = \left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}\text{.}\)
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Solution.
The transition matrix from \(\mathcal{E}\) to \(\mathcal{B}\) is:
\begin{equation*} P = \begin{bmatrix} 1 \amp 1 \\ 1 \amp -1 \end{bmatrix}, \quad P^{-1} = \frac{1}{2}\begin{bmatrix} 1 \amp 1 \\ 1 \amp -1 \end{bmatrix}. \end{equation*}
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By Theorem 6.5.6:
\begin{align*} [T]_{\mathcal{B}} \amp= P^{-1}[T]_{\mathcal{E}}P\\ \amp= \frac{1}{2}\begin{bmatrix} 1 \amp 1 \\ 1 \amp -1 \end{bmatrix} \begin{bmatrix} 0 \amp -1 \\ 1 \amp 0 \end{bmatrix} \begin{bmatrix} 1 \amp 1 \\ 1 \amp -1 \end{bmatrix}\\ \amp= \frac{1}{2}\begin{bmatrix} 1 \amp -1 \\ -1 \amp -1 \end{bmatrix} \begin{bmatrix} 1 \amp 1 \\ 1 \amp -1 \end{bmatrix}\\ \amp= \frac{1}{2}\begin{bmatrix} 0 \amp 2 \\ -2 \amp 0 \end{bmatrix} = \begin{bmatrix} 0 \amp 1 \\ -1 \amp 0 \end{bmatrix}. \end{align*}
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The matrices \([T]_{\mathcal{E}}\) and \([T]_{\mathcal{B}}\) look different, but they represent the same rotation—they are similar matrices. Note that both have determinant 1 and trace 0, as predicted by Theorem 6.5.3.
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Subsection 6.5.3 The Significance of Similarity

We conclude this section by reflecting on why similarity is such an important concept in linear algebra.
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Insight 6.5.9. Why Similarity Matters.

The intellectual origin: Similarity arises naturally from a fundamental principle: the same object viewed from different perspectives should be recognized as equivalent. When we change coordinate systems (bases) to describe a linear transformation, the matrix changes, but the transformation itself remains the same. Similarity is the mathematical formalization of this equivalence.
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The computational power: Not all matrix representations are equally convenient. Some matrices are easier to work with than others. The central problem of diagonalization (Chapter 8) asks: given a matrix \(A\text{,}\) can we find a similar matrix \(D\) that is diagonal? If so, computations with \(A\)—such as computing \(A^n\) or \(e^{At}\)—become trivial, because we can work with \(D\) instead.
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The theoretical importance: Similarity defines equivalence classes of matrices that share fundamental properties. These properties—determinant, trace, rank, and (as we’ll see) eigenvalues—are invariants that don’t depend on the choice of basis. They represent intrinsic features of the underlying linear transformation. Understanding what properties are preserved under similarity helps us identify which matrix features are "real" and which are merely artifacts of notation.
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Looking ahead: In Chapter 8, we will see that eigenvalues and eigenvectors provide the complete theory of similarity. Two matrices are similar if and only if they have the same eigenvalues with the same multiplicities and the same structure of generalized eigenvectors. This culminates in the Jordan canonical form, which classifies all matrices up to similarity. But the conceptual foundation—that similar matrices represent the same transformation in different coordinates—is what we’ve established here.
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The study of similarity transforms the question "What is this matrix?" into the deeper question "What does this transformation do?" By recognizing that many different matrices can represent the same transformation, we shift our focus from the notation to the underlying geometric or algebraic structure. This shift in perspective is one of the key insights of linear algebra.
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