Complex Diagonalization Of Real 2x2 Matrices

Diagonalization Over the Complex Numbers

A real \(2\times2\) matrix can have eigenvalues that are not real. In that case, it cannot be diagonalized using a real diagonal matrix: the diagonal entries of a real diagonal matrix are real eigenvalues. But over \(\mathbb{C}\), the two nonreal eigenvalues occur as a conjugate pair, and their eigenvectors can be placed as the columns of an invertible complex matrix. The resulting diagonalization is a change of coordinates that makes the action of the matrix simple: each eigenvector is scaled by its corresponding eigenvalue.

Here, \(\mathbf{A}\in\mathbb{R}^{2\times2}\) acts on complex columns in \(\mathbb{C}^2\) using the same matrix entries. An eigenvalue \(\lambda\in\mathbb{C}\) has an eigenvector \(\mathbf{z}\in\mathbb{C}^2\) when \(\mathbf{z}\ne\mathbf{0}\) and \(\mathbf{A}\mathbf{z}=\lambda\mathbf{z}\). A diagonalization over \(\mathbb{C}\) is a factorization \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), where \(\mathbf{P}\) is an invertible complex matrix and \(\mathbf{D}\) is diagonal.

In this construction, the bar denotes complex conjugation, applied entry by entry to a vector or matrix. It is essential to pair each column of \(\mathbf{P}\) with its own eigenvalue in \(\mathbf{D}\). The first column \(\mathbf{z}\) is scaled by \(\lambda\), and the second column \(\overline{\mathbf{z}}\) is scaled by \(\overline{\lambda}\).

Why the Construction Works

We first justify the two facts the construction needs: conjugating an eigenpair of a real matrix gives another eigenpair, and eigenvectors for distinct eigenvalues are linearly independent.

Proof. Since every entry of \(\mathbf{A}\) is real, conjugating each entry of \(\mathbf{A}\mathbf{z}=\lambda\mathbf{z}\) gives \(\mathbf{A}\overline{\mathbf{z}}=\overline{\lambda}\,\overline{\mathbf{z}}\). Both vectors are nonzero: \(\mathbf{z}\ne\mathbf{0}\), so its conjugate is nonzero as well. Also \(\lambda\ne\overline{\lambda}\), because \(\lambda=a+bi\) with \(b\ne0\). Suppose the two vectors were linearly dependent. Since both are nonzero, one would be a nonzero scalar multiple of the other, so \(\overline{\mathbf{z}}=c\mathbf{z}\) for some \(c\ne0\). Applying \(\mathbf{A}\) to this equation gives \(\mathbf{A}\overline{\mathbf{z}}=c\lambda\mathbf{z}\), while the conjugate eigenpair equation gives \(\mathbf{A}\overline{\mathbf{z}}=\overline{\lambda}c\mathbf{z}\). Thus \((\lambda-\overline{\lambda})c\mathbf{z}=\mathbf{0}\). Each of \(\lambda-\overline{\lambda}\), \(c\), and \(\mathbf{z}\) is nonzero, a contradiction. Therefore the vectors are independent. \(\square\)

For a \(2\times2\) matrix, two independent columns form an invertible matrix. Thus \(\mathbf{P}\) in the construction is invertible. The key multiplication is then short: its \(j\)th column says that \(\mathbf{A}\) applied to the \(j\)th column of \(\mathbf{P}\) equals that column multiplied by the \(j\)th diagonal entry of \(\mathbf{D}\).

Proof. The proposition shows that \(\overline{\lambda}\) is an eigenvalue with eigenvector \(\overline{\mathbf{z}}\), and that \(\mathbf{z}\) and \(\overline{\mathbf{z}}\) are independent. Hence \(\mathbf{P}\) is invertible. By the two eigenvector equations, the columns of \(\mathbf{A}\mathbf{P}\) are \(\lambda\mathbf{z}\) and \(\overline{\lambda}\,\overline{\mathbf{z}}\), respectively. These are precisely the columns of \(\mathbf{P}\mathbf{D}\), so \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). Multiplying on the right by \(\mathbf{P}^{-1}\) gives \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). \(\square\)

This proof explains why the procedure works rather than just supplying a recipe. The eigenvector equations establish \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\); independence ensures the inverse exists, so the equation can be rearranged into the desired factorization.

Worked Diagonalizations

Common Pitfalls and What the Diagonalization Means

Do not repeat the same eigenvector in both columns. In the second example, taking \(\mathbf{P}=\begin{bmatrix}\mathbf{z}&\mathbf{z}\end{bmatrix}\) gives two identical columns, so \(\det\mathbf{P}=0\) and \(\mathbf{P}^{-1}\) does not exist. The second column must be the conjugate eigenvector \(\overline{\mathbf{z}}\), not another copy of \(\mathbf{z}\).

Do not repeat the same eigenvalue in both diagonal entries. For that example, \(\mathbf{A}\overline{\mathbf{z}}=(2-3i)\overline{\mathbf{z}}\), not \((2+3i)\overline{\mathbf{z}}\). Pairing \(\overline{\mathbf{z}}\) with \(2+3i\) would make the second columns of \(\mathbf{A}\mathbf{P}\) and \(\mathbf{P}\mathbf{D}\) unequal. Conjugate both the eigenvector and the eigenvalue for the second column.

The construction is a diagonalization over \(\mathbb{C}\), even though \(\mathbf{A}\) has real entries. The entries of \(\mathbf{P}\) and \(\mathbf{D}\) need not be real. In fact, a real diagonal matrix can have only real eigenvalues: if a real diagonal matrix is \(\operatorname{diag}(d_1,d_2)\), its characteristic equation is \((d_1-\lambda)(d_2-\lambda)=0\), so its eigenvalues are \(d_1,d_2\in\mathbb{R}\). It therefore cannot be similar to a real matrix whose eigenvalues are nonreal.

There is also a useful real-coordinate interpretation. Writing \(\mathbf{z}=\mathbf{x}+i\mathbf{y}\) with real columns \(\mathbf{x},\mathbf{y}\), the eigenvector equation for \(\lambda=a+bi\) gives, by equating real and imaginary parts, \(\mathbf{A}\mathbf{x}=a\mathbf{x}-b\mathbf{y}\) and \(\mathbf{A}\mathbf{y}=b\mathbf{x}+a\mathbf{y}\). The complex diagonalization records these same paired relationships using conjugate columns and diagonal entries. The complex form is particularly useful when calculations involve eigenvalues or powers of \(\mathbf{A}\): in the eigenvector coordinates, each diagonal entry acts separately.