Choosing and Ordering Eigenvectors in P

Why the Order in \(P\) Matters

To diagonalize a matrix, it is not enough to collect eigenvectors and eigenvalues separately. Their order must match: the first diagonal entry of \(D\) belongs to the first column of \(P\), the second diagonal entry belongs to the second column, and so on. A correct collection with an incorrect pairing generally does not give a diagonalization.

Throughout, matrices and vectors have real entries. Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). An eigenvector of \(\mathbf{A}\) is a nonzero vector \(\mathbf{v}\) for which \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for some scalar \(\lambda\), called its eigenvalue. For a diagonalization, the chosen eigenvectors must form a basis of \(\mathbb{R}^n\). We place that ordered basis into the columns of \(\mathbf{P}\).

The order is encoded in a simple equation. Multiplying \(\mathbf{A}\mathbf{P}\) applies \(\mathbf{A}\) to each column of \(\mathbf{P}\). Multiplying \(\mathbf{P}\mathbf{D}\) scales column \(i\) of \(\mathbf{P}\) by the \(i\)-th diagonal entry of \(\mathbf{D}\). Therefore the correct pairing gives \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). If the columns of \(\mathbf{P}\) form a basis, then \(\mathbf{P}\) is invertible and this equation is equivalent to \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).

The Column-by-Column Test

Proof. The \(i\)-th column of \(\mathbf{A}\mathbf{P}\) is \(\mathbf{A}\mathbf{v}_i\), because matrix multiplication by \(\mathbf{P}\) applies \(\mathbf{A}\) to each column. Since \(\mathbf{D}\) is diagonal, the \(i\)-th column of \(\mathbf{P}\mathbf{D}\) is \(d_i\mathbf{v}_i\). Two matrices are equal exactly when all corresponding columns are equal. Thus \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) holds exactly when \(\mathbf{A}\mathbf{v}_i=d_i\mathbf{v}_i\) for every \(i\). If the columns form a basis, \(\mathbf{P}\) is invertible. Right-multiplying \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) by \(\mathbf{P}^{-1}\) gives \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). Conversely, if that diagonalization equation holds, right-multiplying it by \(\mathbf{P}\) gives \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). \(\square\)

This gives a direct way to check a proposed ordering: compare the columns of \(\mathbf{A}\mathbf{P}\) and \(\mathbf{P}\mathbf{D}\). It also separates two requirements. The column equations check the eigenvalue pairings; the columns must also form a basis for \(\mathbf{P}\) to be invertible and give a diagonalization.

Worked Examples of Pairing and Reordering

Reordering Both Matrices Consistently

The examples illustrate a general fact. Let \(\pi\) be a permutation of the indices \(1,\dots,n\), meaning a rearrangement in which every index appears exactly once. Starting with eigenvectors \(\mathbf{v}_1,\dots,\mathbf{v}_n\) and corresponding eigenvalues \(\lambda_1,\dots,\lambda_n\), form a new ordered list \(\mathbf{v}_{\pi(1)},\dots,\mathbf{v}_{\pi(n)}\). The matching diagonal must then have entries \(\lambda_{\pi(1)},\dots,\lambda_{\pi(n)}\) in that order.

Proof. A permutation changes only the order of the vectors, not which vectors are in the list. Any linear combination of the reordered list can be written as a linear combination of the original list, with the coefficients assigned to the corresponding original indices. Since the original list is a basis, it is independent and spans \(\mathbb{R}^n\); the reordered list therefore has the same properties and is also a basis. For each position \(j\), its \(j\)-th column is \(\mathbf{v}_{\pi(j)}\), and \(\mathbf{A}\mathbf{v}_{\pi(j)}=\lambda_{\pi(j)}\mathbf{v}_{\pi(j)}\). The \(j\)-th diagonal entry of the proposed diagonal matrix is precisely \(\lambda_{\pi(j)}\). The column-by-column theorem now gives \(\mathbf{A}\mathbf{P}_{\pi}=\mathbf{P}_{\pi}\mathbf{D}_{\pi}\). Since the reordered columns form a basis, \(\mathbf{P}_{\pi}\) is invertible, so this equation gives a diagonalization. \(\square\)

When eigenvalues repeat, swapping two eigenvectors with the same eigenvalue leaves the diagonal matrix unchanged, but the columns of \(\mathbf{P}\) have still been swapped. More generally, one may choose a different basis within an eigenspace, provided the complete chosen list remains a basis of \(\mathbb{R}^n\) and each selected vector is paired with its own eigenvalue.

Common Mistake: Reordering Only \(P\)

For the first example, suppose the columns are swapped to form \(\mathbf{P}'=\begin{bmatrix}-1&1\\2&0\end{bmatrix}\), but the original \(\mathbf{D}=\operatorname{diag}(4,2)\) is kept. This incorrectly assigns eigenvalue \(4\) to \(\mathbf{v}_2\) and eigenvalue \(2\) to \(\mathbf{v}_1\). The products expose the mismatch:

They are not equal. The correction is to reorder the diagonal entries as well, using \(\mathbf{D}'=\operatorname{diag}(2,4)\), which gives \(\mathbf{A}\mathbf{P}'=\mathbf{P}'\mathbf{D}'\). A reliable check is to read each column of \(\mathbf{P}\) and ask which eigenvalue multiplies it; those values, in that order, are the diagonal entries of \(\mathbf{D}\).