Diagonalizability as Similarity to a Diagonal Matrix

What Diagonalizability Means

A diagonal matrix acts in a particularly simple way: each coordinate is multiplied by its own scalar, with no mixing between coordinates. A matrix that is not diagonal in the standard coordinates may still describe an action of this kind when the same linear map is written in a different basis. Diagonalizability gives a precise language for that simplification: it asks whether a matrix is similar to a diagonal matrix.

Throughout, matrices and vectors have real entries, and vectors are columns. Two \(n\times n\) matrices are called similar when one is obtained from the other by an invertible change of coordinates. The matrix that performs the coordinate change must be invertible: without an inverse, the new coordinates could not be converted back uniquely.

The order of the factors matters. The matrix \(\mathbf{P}\) converts coordinates in the chosen basis to standard coordinates, while \(\mathbf{P}^{-1}\) converts standard coordinates back to the chosen basis. Thus \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) describes the same action as \(\mathbf{A}\), but using the new coordinates. A diagonal result means that the action separates into coordinate-by-coordinate scalings in that basis.

How a Change of Basis Changes the Matrix

Let \(T:\mathbb{R}^n\to\mathbb{R}^n\) be a linear map, and let \(\mathbf{A}\) be its matrix in standard coordinates. An ordered basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) determines a matrix \(\mathbf{P}\) whose \(j\)th column is \(\mathbf{b}_j\). If \(\mathbf{x}_{\mathcal{B}}\) is the coordinate column of a vector in this basis, then the vector in standard coordinates is \(\mathbf{P}\mathbf{x}_{\mathcal{B}}\). The matrix \(\mathbf{P}\) is invertible because the basis vectors are linearly independent and span \(\mathbb{R}^n\).

Proof. Let \(\mathbf{x}_{\mathcal{B}}\) be the coordinate vector of any input. Its standard-coordinate vector is \(\mathbf{P}\mathbf{x}_{\mathcal{B}}\). Applying \(T\), whose standard-coordinate matrix is \(\mathbf{A}\), gives the standard-coordinate output \(\mathbf{A}\mathbf{P}\mathbf{x}_{\mathcal{B}}\). To express that output in basis \(\mathcal{B}\), multiply by \(\mathbf{P}^{-1}\). The output coordinates are therefore \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\mathbf{x}_{\mathcal{B}}\). Since this holds for every input coordinate vector, the matrix of \(T\) in \(\mathcal{B}\) is \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\). \(\square\)

This formula explains the definition: \(\mathbf{A}\) is diagonalizable exactly when there is a basis in which its matrix is diagonal. In using this statement, the basis—and therefore \(\mathbf{P}\)—must be the same on both sides of \(\mathbf{A}\). The left factor changes the output back to the chosen coordinates; the right factor first converts the input to standard coordinates.

Proof. In row \(i\) of the product \(\mathbf{D}\mathbf{x}\), the only possibly nonzero entry of \(\mathbf{D}\) is \(d_i\) in column \(i\). Thus the \(i\)th output entry is \(d_i x_i\). This holds for each \(i=1,\dots,n\), so the product has the stated form. Taking \(\mathbf{x}=\mathbf{e}_i\), the \(i\)th standard basis vector, gives \(\mathbf{D}\mathbf{e}_i=d_i\mathbf{e}_i\): that coordinate direction is scaled by \(d_i\), with no contribution from other directions. \(\square\)

Worked Similarity Calculations

Similarity and Common Errors

Similarity is an equivalence relation on the \(n\times n\) real matrices: every matrix is similar to itself, reversing a similarity gives another similarity, and successive similarities can be combined. These facts explain why a matrix can have more than one useful coordinate description.

Proof. For reflexivity, \(\mathbf{A}=\mathbf{I}_n\mathbf{A}\mathbf{I}_n^{-1}\), so \(\mathbf{A}\) is similar to itself. For symmetry, suppose \(\mathbf{B}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) for an invertible \(\mathbf{P}\). Rearranging gives \(\mathbf{A}=\mathbf{P}\mathbf{B}\mathbf{P}^{-1}\), so \(\mathbf{A}\) is similar to \(\mathbf{B}\). For transitivity, suppose \(\mathbf{B}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) and \(\mathbf{C}=\mathbf{Q}^{-1}\mathbf{B}\mathbf{Q}\), where \(\mathbf{P}\) and \(\mathbf{Q}\) are invertible. Substitution gives \(\mathbf{C}=(\mathbf{P}\mathbf{Q})^{-1}\mathbf{A}(\mathbf{P}\mathbf{Q})\), because \((\mathbf{P}\mathbf{Q})^{-1}=\mathbf{Q}^{-1}\mathbf{P}^{-1}\). The product \(\mathbf{P}\mathbf{Q}\) is invertible, with inverse \(\mathbf{Q}^{-1}\mathbf{P}^{-1}\). Thus \(\mathbf{C}\) is similar to \(\mathbf{A}\). \(\square\)

Common pitfall: reversing the factors. In the first worked example, a tempting but incorrect calculation is to use \(\mathbf{P}\mathbf{A}\mathbf{P}^{-1}\) as the matrix in the new coordinates. In fact,

which is not the diagonal matrix \(\mathbf{D}\). The correct formula for the matrix in the new coordinates is \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\), as follows from first converting the input to standard coordinates and then converting the output back. The equation \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) is equivalent, but its factors are arranged to reconstruct the standard-coordinate matrix from the diagonal one.

Another useful distinction is between a matrix and the linear map it represents. Changing basis does not alter the underlying map; it changes the coordinate description of that map. Diagonalization matters because a diagonal description makes repeated applications straightforward: in the diagonal coordinates, applying \(\mathbf{D}\) twice multiplies coordinate \(i\) by \(d_i^2\), and applying it \(k\) times multiplies that coordinate by \(d_i^k\), for any positive integer \(k\). The same map may look less simple in standard coordinates, but the change-of-basis formula keeps the two descriptions linked.