Similar Matrices

Similarity is the algebraic engine behind diagonalization: an operator is diagonalizable exactly when its matrix is similar to a diagonal matrix, and eigenvalue algorithms in numerical linear algebra (QR algorithm, Jacobi eigenvalue method) work by iteratively applying similarity transformations to drive a matrix toward a simpler (triangular or diagonal) similar form while provably preserving eigenvalues at every step. In differential equations and dynamical systems, similarity transformations diagonalize or block-diagonalize the coefficient matrix of a linear system $\dot{x} = Ax$, decoupling the system into independent scalar equations solvable in closed form — the qualitative long-term behavior (stability, oscillation) is read off from eigenvalues, which similarity guarantees are basis-independent. In quantum mechanics and physics, changing between position and momentum representations, or diagonalizing a Hamiltonian to find energy eigenstates, are similarity (or unitary conjugation, a refinement of similarity) changes of basis; energy levels are physical quantities and must be basis-independent, which is exactly why they correspond to eigenvalues, invariant under similarity. In computer graphics and robotics, coordinate frames attached to moving bodies are related by change-of-basis matrices, and the same rigid transformation expressed in different frames is a similarity transformation of the transformation matrix. In statistics, principal component analysis diagonalizes a covariance matrix via an orthogonal similarity transformation, exploiting the fact that a symmetric matrix's diagonalizing basis reveals uncorrelated directions of variance while total variance (the trace) remains invariant under the change of coordinates. In cryptography and coding theory, invariant polynomial and rank computations that must remain robust to a change of internal representation (e.g. of a linear feedback shift register's state matrix) rely on similarity invariance of characteristic polynomials to guarantee cycle-length and period properties are representation-independent.

Ask the tutor: How does the Jordan canonical form complete the classification of matrices up to similarity when the characteristic polynomial alone is not enough?

Think of a linear operator $T : V \to V$ on a finite-dimensional vector space as a fixed geometric action — a rotation, a shear, a projection — that exists independently of any coordinate system. To compute with $T$, we choose a basis $\mathcal{B}$ of $V$ and represent $T$ by a matrix $[T]_{\mathcal{B}}$: the columns of this matrix record where each basis vector of $\mathcal{B}$ is sent, expressed in coordinates relative to $\mathcal{B}$ itself. If we instead choose a different basis $\mathcal{C}$, we obtain a possibly very different-looking matrix $[T]_{\mathcal{C}}$. Both matrices describe the same operator $T$; they are two photographs of the same object taken from different angles. The change-of-basis matrix $P$ (whose columns are the $\mathcal{C}$-coordinates of the $\mathcal{B}$-basis vectors, or vice versa depending on convention) is the dictionary translating between the two coordinate systems, and the relation between the two matrices is exactly $[T]_{\mathcal{C}} = P^{-1} [T]_{\mathcal{B}} P$. A good mental picture: imagine an ellipse-shaped trampoline being stretched by a linear map into another ellipse. In the standard basis, the matrix of this stretch might have off-diagonal entries mixing $x$ and $y$. But if we choose a basis along the ellipse's own axes, the same stretch becomes a simple diagonal scaling. Nothing about the physical stretching changed — only our descriptive language did.

Example 1: Verifying similarity directly and confirming the invariants. Let

First check $P$ is invertible: $\det P = 1\cdot 3 - 2 \cdot 1 = 1 \neq 0$. Its inverse (using the $2\times 2$ formula $\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}$) is

Compute $B = P^{-1}AP$ in two steps. First $AP$:

Then $P^{-1}(AP)$:

So $A \sim B$ with $B = \begin{pmatrix}3&3\\0&2\end{pmatrix}$. Note $A$ and $B$ do not share a single common entry, yet:

• $\operatorname{tr}(A) = 2+3 = 5$ and $\operatorname{tr}(B) = 3+2 = 5$. Match.
• $\det(A) = 2\cdot3 - 1\cdot0 = 6$ and $\det(B) = 3\cdot2 - 3\cdot0 = 6$. Match.
• Characteristic polynomial of $A$: $\det(A - \lambda I) = (2-\lambda)(3-\lambda) = \lambda^2 - 5\lambda + 6$.

Characteristic polynomial of $B$: $\det(B - \lambda I) = (3-\lambda)(2-\lambda) - 3\cdot 0 = \lambda^2 - 5\lambda + 6$. Match, as guaranteed by the corollary.

• Both are upper triangular with eigenvalues $\{2, 3\}$ read off the diagonal — consistent, since similar matrices must have identical eigenvalue multisets, and here the eigenvalues literally appear as diagonal entries in each triangular representative.

Example 2: A basis-change interpretation with an explicit operator on $\mathbb{R}^3$. Let $T : \mathbb{R}^3 \to \mathbb{R}^3$ be the linear operator whose matrix relative to the standard basis $\mathcal{E} = (e_1, e_2, e_3)$ is

Consider the new ordered basis $\mathcal{C} = (w_1, w_2, w_3)$ with

These are linearly independent (the matrix with these columns is lower triangular with nonzero diagonal, hence invertible), so $\mathcal{C}$ is indeed a basis. Let $P$ be the matrix whose columns are the $\mathcal{E}$-coordinates of $w_1, w_2, w_3$ — since $\mathcal{E}$ is the standard basis, coordinates equal components, so

By the theorem, $[T]_{\mathcal{C}} = P^{-1} A P$. Compute $P^{-1}$ by solving $P x = e_i$ for each $i$, or by the standard upper-triangular inverse pattern:

(Check: $PP^{-1}$, first row: $(1)(1)+(1)(0)+(1)(0) = 1$; $(1)(-1)+(1)(1)+(1)(0) = 0$; $(1)(0)+(1)(-1)+(1)(1) = 0$ — matches row 1 of $I$. The remaining rows check similarly.) Now compute $AP$:

Let us redo this multiplication carefully entry by entry. Row 1 of $A$ is $(1,1,0)$:

• column 1 of $P$ is $(1,0,0)^T$: dot product $= 1$.
• column 2 of $P$ is $(1,1,0)^T$: dot product $= 1\cdot1 + 1\cdot1 + 0\cdot0 = 2$.
• column 3 of $P$ is $(1,1,1)^T$: dot product $= 1\cdot1+1\cdot1+0\cdot1 = 2$.

Row 2 of $A$ is $(0,2,1)$:

• vs column 1 $(1,0,0)$: $0$.
• vs column 2 $(1,1,0)$: $0\cdot1+2\cdot1+1\cdot0 = 2$.
• vs column 3 $(1,1,1)$: $0+2+1 = 3$.

Row 3 of $A$ is $(0,0,3)$:

• vs column 1: $0$.
• vs column 2: $0$.
• vs column 3: $3$.

So

Now compute $B = P^{-1}(AP)$, with $P^{-1} = \begin{pmatrix}1&-1&0\\0&1&-1\\0&0&1\end{pmatrix}$. Row 1 of $P^{-1}$ is $(1,-1,0)$, dotted with columns of $AP$:

• column 1 $(1,0,0)$: $1$.
• column 2 $(2,2,0)$: $1\cdot2 + (-1)\cdot2 = 0$.
• column 3 $(2,3,3)$: $1\cdot2+(-1)\cdot3 = -1$.

Row 2 of $P^{-1}$ is $(0,1,-1)$:

• column 1: $0$.
• column 2: $1\cdot2 + (-1)\cdot0 = 2$.
• column 3: $1\cdot3 + (-1)\cdot3 = 0$.

Row 3 of $P^{-1}$ is $(0,0,1)$:

• column 1: $0$.
• column 2: $0$.
• column 3: $3$.

So

Check the invariants: $\operatorname{tr}(A) = 1+2+3 = 6 = \operatorname{tr}(B) = 1+2+3$. Both matrices are triangular, so $\det(A) = 1\cdot2\cdot3 = 6 = \det(B)$, and both display the same eigenvalues $1, 2, 3$ on their diagonals — exactly as guaranteed, since $A$ and $B$ represent the identical operator $T$, merely described in the bases $\mathcal{E}$ and $\mathcal{C}$ respectively.

A frequently made error: concluding $A \sim B$ merely because $A$ and $B$ have the same eigenvalues (even with multiplicity) and the same characteristic polynomial. This is necessary but not sufficient. For instance,

both have characteristic polynomial $(\lambda-2)^2$ and the single eigenvalue $2$ with algebraic multiplicity $2$. But $A = 2I$ is similar only to itself (since $P^{-1}(2I)P = 2P^{-1}P = 2I$ for any invertible $P$), while $B \neq 2I$, so $A \not\sim B$. The characteristic polynomial is an invariant of similarity, not a complete classifier of it; distinguishing matrices with equal characteristic polynomials up to similarity requires finer tools (such as the Jordan form), which lie beyond the present scope.