Proof: Diagonalization Is Equivalent to Having an Eigenbasis

From a Diagonalization to an Eigenbasis

Diagonalization connects two descriptions of a linear transformation. One description gives its action through a matrix \(\mathbf{A}\); another uses a basis in which that action is represented by a diagonal matrix. The key is that the vectors in this basis are eigenvectors: each is scaled by the transformation, rather than sent in a new direction. The correspondence works in both directions, but it depends on keeping the eigenvectors and diagonal entries in the same order.

Throughout this tutorial, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). A nonzero vector \(\mathbf{v}\in\mathbb{R}^n\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\in\mathbb{R}\) if \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). An eigenbasis for \(\mathbf{A}\) is an ordered basis of \(\mathbb{R}^n\) whose vectors are all eigenvectors of \(\mathbf{A}\). We say that \(\mathbf{A}\) is diagonalizable over \(\mathbb{R}\) if there is an invertible matrix \(\mathbf{P}\in\mathbb{R}^{n\times n}\) and a diagonal matrix \(\mathbf{D}\in\mathbb{R}^{n\times n}\) such that \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).

In the expression \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), the columns of \(\mathbf{P}\) are the vectors of a basis, and \(\mathbf{D}\) describes how the transformation scales their coordinates. A square matrix is invertible exactly when its columns form a basis of \(\mathbb{R}^n\); the following argument recalls why this fact applies here.

Proof. Write the columns of \(\mathbf{P}\) as \(\mathbf{p}_1,\dots,\mathbf{p}_n\). If they form a basis, every \(\mathbf{y}\in\mathbb{R}^n\) has a unique expression \(\mathbf{y}=c_1\mathbf{p}_1+\cdots+c_n\mathbf{p}_n=\mathbf{P}\mathbf{c}\), where \(\mathbf{c}=(c_1,\dots,c_n)^\top\). Assign to each \(\mathbf{y}\) its unique coefficient vector \(\mathbf{c}\). This assignment is linear, since the coefficients of a sum or scalar multiple are the corresponding sum or scalar multiple of the unique coefficient vectors. Its matrix \(\mathbf{Q}\) satisfies \(\mathbf{Q}\mathbf{P}=\mathbf{I}_n\) and \(\mathbf{P}\mathbf{Q}=\mathbf{I}_n\), so \(\mathbf{Q}=\mathbf{P}^{-1}\). Conversely, if \(\mathbf{P}\) is invertible, each \(\mathbf{y}\) equals \(\mathbf{P}(\mathbf{P}^{-1}\mathbf{y})\), so it is a linear combination of the columns. If \(\mathbf{P}\mathbf{c}=\mathbf{0}\), multiplying by \(\mathbf{P}^{-1}\) gives \(\mathbf{c}=\mathbf{0}\), so the columns are linearly independent. They therefore form a basis. \(\square\)

The Column Equation That Connects the Two Ideas

Let \(\mathbf{P}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_n\end{bmatrix}\) and let \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\), meaning that the \(\lambda_i\) are the diagonal entries and all off-diagonal entries are zero. Multiplying \(\mathbf{P}\) by \(\mathbf{D}\) scales its \(i\)-th column by \(\lambda_i\). Thus, the matrix equation \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) says, column by column, that \(\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\).

Proof. The \(i\)-th column of \(\mathbf{A}\mathbf{P}\) is \(\mathbf{A}\mathbf{v}_i\), by the definition of matrix multiplication. The \(i\)-th column of \(\mathbf{P}\mathbf{D}\) is \(\lambda_i\mathbf{v}_i\), because \(\mathbf{D}\) is diagonal. Two matrices are equal if and only if their corresponding columns are equal. Therefore, the matrix equation holds exactly when the stated column equations all hold. \(\square\)

This proposition does not require \(\mathbf{P}\) to be invertible. Invertibility matters when we want the columns to form a basis, rather than just to be eigenvectors. When \(\mathbf{P}\) is invertible, \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) can be multiplied on the right by \(\mathbf{P}^{-1}\) to give \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). This is the algebraic link behind the equivalence.

The Equivalence

Proof. First suppose \(\mathbf{A}\) is diagonalizable, so \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) for an invertible \(\mathbf{P}\) and a diagonal \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\). Write the columns of \(\mathbf{P}\) as \(\mathbf{v}_1,\dots,\mathbf{v}_n\). Right-multiplying the diagonalization equation by \(\mathbf{P}\) gives \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). By the proposition, \(\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\) for every \(i\). None of these columns is zero, because an invertible matrix has linearly independent columns. The lemma says those columns form a basis. Hence they are an eigenbasis. Conversely, suppose \(\mathbf{v}_1,\dots,\mathbf{v}_n\) is a basis of eigenvectors, and let \(\lambda_i\) satisfy \(\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\) for each \(i\). Set \(\mathbf{P}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_n\end{bmatrix}\) and \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\). The lemma makes \(\mathbf{P}\) invertible, and the proposition gives \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). Multiplying on the right by \(\mathbf{P}^{-1}\) yields \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), so \(\mathbf{A}\) is diagonalizable. Both directions are proved. \(\square\)

The theorem also tells us what the entries of a diagonalizing matrix mean: its columns are an eigenbasis, and the diagonal entries record the corresponding eigenvalues in that same order. If we reorder the basis vectors, we must reorder the diagonal entries to keep the column equations true.

Reading a Diagonalization in Both Directions

Common Pitfall: Matching the Wrong Eigenvalue to a Column

A diagonal matrix \(\mathbf{D}\) must list eigenvalues in the order of the columns of \(\mathbf{P}\). For a concrete check, let \(\mathbf{A}=\begin{bmatrix}2&0\\0&3\end{bmatrix}\), and choose the eigenvectors in the reversed order \(\mathbf{v}_1=(0,1)^\top\), \(\mathbf{v}_2=(1,0)^\top\). A tempting but wrong choice is \(\mathbf{D}=\begin{bmatrix}2&0\\0&3\end{bmatrix}\), copied in the order of the diagonal entries of \(\mathbf{A}\) rather than matched to the columns. With \(\mathbf{P}=\begin{bmatrix}0&1\\1&0\end{bmatrix}\), we have \(\mathbf{P}^{-1}=\mathbf{P}\), and the wrong choice gives \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}=\begin{bmatrix}3&0\\0&2\end{bmatrix}\ne\mathbf{A}\). The correction is to pair \(\mathbf{v}_1\) with its eigenvalue \(3\) and \(\mathbf{v}_2\) with its eigenvalue \(2\). Then \(\mathbf{D}=\begin{bmatrix}3&0\\0&2\end{bmatrix}\), and \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}=\begin{bmatrix}2&0\\0&3\end{bmatrix}=\mathbf{A}\). Equivalently, check each column equation before forming the product.

Why the Equivalence Is Useful

The two directions of the theorem serve different purposes. If a diagonalization is already known, the columns of \(\mathbf{P}\) provide a basis of eigenvectors without a separate basis search. If an eigenbasis has been found, placing it into \(\mathbf{P}\) and recording the matching eigenvalues in \(\mathbf{D}\) constructs a diagonalization. In either case, the column equation \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) is the most direct check.

The examples also show why the word “basis” is essential. Having one or several eigenvectors is not enough: the vectors must span the whole space and be linearly independent. The scalar matrix has a repeated eigenvalue but still has an eigenbasis, while the last matrix has eigenvectors but not enough independent ones. The theorem makes no claim that eigenvalues must be distinct; it asks whether the eigenvectors can be assembled into a basis.