Similarity To A Rotation-Scaling Form

From a Complex Eigenvector to a Real Basis

A real \(2\times2\) matrix can have eigenvalues that are not real. In that case, a complex eigenvector still contains information about how the matrix acts on the real plane: its real and imaginary parts give two real vectors, and together they form a basis. In that basis the matrix has a compact rotation-scaling form. The key is to choose the sign of the imaginary part consistently.

Let \(\mathbf{A}\in\mathbb{R}^{2\times2}\). When considering complex eigenvalues and eigenvectors of \(\mathbf{A}\), we use the same entries of \(\mathbf{A}\) to act on vectors in \(\mathbb{C}^2\). Suppose \(\lambda=a+bi\), where \(a,b\in\mathbb{R}\) and \(b\ne0\), is an eigenvalue, and let \(\mathbf{z}\in\mathbb{C}^2\) be a corresponding nonzero eigenvector. Write its real and imaginary parts as \(\mathbf{z}=\mathbf{x}+i\mathbf{y}\), where \(\mathbf{x},\mathbf{y}\in\mathbb{R}^2\).

The eigenvector equation is \(\mathbf{A}\mathbf{z}=\lambda\mathbf{z}\). Expanding both sides and equating real and imaginary parts gives

These equations indicate why the ordered basis \((\mathbf{x},-\mathbf{y})\) is convenient. The first equation says that \(\mathbf{A}\mathbf{x}=a\mathbf{x}+b(-\mathbf{y})\); the second says that \(\mathbf{A}(-\mathbf{y})=-b\mathbf{x}+a(-\mathbf{y})\). Thus, if these vectors are a basis, the columns of the matrix in that basis are \((a,b)^\top\) and \((-b,a)^\top\). We must still prove that they really are a basis.

Proof. First suppose \(\mathbf{A}\) has the eigenpair in the statement. We show that \(\mathbf{x}\) and \(-\mathbf{y}\) are linearly independent over \(\mathbb{R}\). If they were dependent, then \(\mathbf{x}\) and \(\mathbf{y}\) would lie in a real subspace of dimension at most one. Since \(\mathbf{z}\ne\mathbf{0}\), there would be a nonzero real vector \(\mathbf{u}\) and a nonzero complex scalar \(c\) with \(\mathbf{z}=c\mathbf{u}\). Substituting this into \(\mathbf{A}\mathbf{z}=\lambda\mathbf{z}\) and cancelling \(c\) gives \(\mathbf{A}\mathbf{u}=\lambda\mathbf{u}\). The left side is real, while the imaginary part of the right side is \(b\mathbf{u}\). Since \(b\ne0\), this forces \(\mathbf{u}=\mathbf{0}\), a contradiction. Therefore \(\mathbf{x}\) and \(-\mathbf{y}\) are independent, so \(\mathbf{P}=\begin{bmatrix}\mathbf{x}&-\mathbf{y}\end{bmatrix}\) is invertible. The real and imaginary part equations give \(\mathbf{A}\mathbf{P}=\mathbf{P}\begin{bmatrix}a&-b\\b&a\end{bmatrix}\); multiplying on the left by \(\mathbf{P}^{-1}\) proves the claimed similarity. For the reverse direction, suppose \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{B}\), where \(\mathbf{B}=\begin{bmatrix}a&-b\\b&a\end{bmatrix}\) and \(b\ne0\). The complex vector \(\mathbf{w}=(1,-i)^\top\) is nonzero, and direct multiplication gives \(\mathbf{B}\mathbf{w}=(a+bi)\mathbf{w}\). Since \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{B}\), we have \(\mathbf{A}(\mathbf{P}\mathbf{w})=(a+bi)\mathbf{P}\mathbf{w}\). Invertibility of \(\mathbf{P}\) ensures \(\mathbf{P}\mathbf{w}\ne\mathbf{0}\). Thus \(a+bi\) is a nonreal eigenvalue of \(\mathbf{A}\). \(\square\)

Constructing and Checking the Similarity

The theorem gives a direct procedure: find a nonreal eigenpair, split the eigenvector into real and imaginary parts, place \(\mathbf{x}\) and \(-\mathbf{y}\) into the columns of \(\mathbf{P}\), and verify \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{B}\). This last equation is often the cleanest check because it verifies the action of \(\mathbf{A}\) on each proposed basis vector. Once \(\mathbf{P}\) is known to be invertible, it is equivalent to \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{B}\).

Why the Sign of the Imaginary Part Matters

The formula depends on the order and signs in the chosen basis. If \(\mathbf{z}=\mathbf{x}+i\mathbf{y}\), then the basis giving the block \(\begin{bmatrix}a&-b\\b&a\end{bmatrix}\) is \((\mathbf{x},-\mathbf{y})\), not \((\mathbf{x},\mathbf{y})\). The latter is still a basis, but its matrix has a different sign pattern. The eigenvector equations make this distinction explicit:

Relative to \((\mathbf{x},\mathbf{y})\), these equations give the columns \((a,-b)^\top\) and \((b,a)^\top\), so the matrix is \(\begin{bmatrix}a&b\\-b&a\end{bmatrix}\). Replacing the second basis vector by its negative changes the signs in the corresponding column and produces the theorem’s stated block. Either calculation can be correct; the error is to use one basis and claim the matrix belonging to the other.

Common Pitfall: Mixing the Basis and the Block

A common incorrect calculation uses \((\mathbf{x},\mathbf{y})\) as the basis but writes down the block for \((\mathbf{x},-\mathbf{y})\). For the eigenpair in the first example, \(a=2\), \(b=3\), \(\mathbf{x}=(1,1)^\top\), and \(\mathbf{y}=(-1,0)^\top\). If one uses \(\mathbf{Q}=\begin{bmatrix}\mathbf{x}&\mathbf{y}\end{bmatrix}\), the columns are independent, but the actual matrix in that basis is \(\begin{bmatrix}2&3\\-3&2\end{bmatrix}\), not \(\begin{bmatrix}2&-3\\3&2\end{bmatrix}\). The correction is to match the basis to the formula: use \(\mathbf{P}=\begin{bmatrix}\mathbf{x}&-\mathbf{y}\end{bmatrix}\) for the latter block, or keep \(\mathbf{Q}\) and use the former block. In either case, verify the claim by multiplying the original matrix by the basis matrix.

The constructed basis is useful because it turns a complex eigenpair into a real description of the action on \(\mathbb{R}^2\). The similarity does not claim that the original matrix equals the block in its original coordinates; it says that the two matrices describe the same linear map in different real bases. The distinction between the matrix and its representation in a chosen basis is exactly what the equation \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{B}\) records.