Recovering Scale And Angle From Trace And Determinant

What Trace and Determinant Reveal

For a real \(2\times2\) matrix with nonreal eigenvalues, the eigenvalues form a conjugate pair. The trace and determinant are enough to recover that pair’s size and direction in the complex plane: the determinant gives the square of the modulus, and the trace fixes the angle. This gives a way to find the scale and angle parameters without first finding an eigenvector or putting the matrix into a particular form.

Let \(\mathbf{A}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\) be a real \(2\times2\) matrix. Its trace is \(\operatorname{tr}\mathbf{A}=a+d\), and its determinant is \(\det\mathbf{A}=ad-bc\). The characteristic polynomial, using the convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_2)\), is

Write \(\tau=\operatorname{tr}\mathbf{A}\) and \(\delta=\det\mathbf{A}\). The eigenvalues are the roots of \(p_{\mathbf{A}}\). When they are nonreal, designate by \(\lambda_+\) the one with positive imaginary part, and let \(\lambda_-\) be its conjugate. We use the principal argument in \((-\pi,\pi]\), so the upper-half-plane eigenvalue has argument between \(0\) and \(\pi\).

Proof. The characteristic polynomial is \(\lambda^2-\tau\lambda+\delta\), whose discriminant is \(\tau^2-4\delta\). The quadratic formula gives roots \(\frac{\tau}{2}\pm\frac{1}{2}\sqrt{\tau^2-4\delta}\). If the discriminant is nonnegative, both roots are real. If it is negative, then \(4\delta-\tau^2>0\), and the square root of \(\tau^2-4\delta\) is \(i\sqrt{4\delta-\tau^2}\); the roots are nonreal and have the displayed form. This proves both directions of the criterion. Finally, \(4\delta>\tau^2\ge 0\), so \(\delta>0\). \(\square\)

Recovering the Modulus and Argument

The modulus of a complex number \(x+iy\) is \(\sqrt{x^2+y^2}\). For the eigenvalue \(\lambda_+\) in the theorem, its real and imaginary parts are \(\tau/2\) and \(\sqrt{4\delta-\tau^2}/2\). Squaring and adding gives \(\delta\), so its modulus is \(\sqrt{\delta}\). To recover its argument, divide each part by that modulus: the resulting cosine is determined by the trace, while the positive sine identifies the upper-half-plane choice.

Proof. By the preceding theorem, \(\delta>0\) and \(\lambda_+=\frac{\tau}{2}+\frac{i}{2}\sqrt{4\delta-\tau^2}\). Therefore $$ |\lambda_+|^2=\left(\frac{\tau}{2}\right)^2+ \left(\frac{\sqrt{4\delta-\tau^2}}{2}\right)^2 =\frac{\tau^2+4\delta-\tau^2}{4} =\delta. $$ Since a modulus is nonnegative and \(\delta>0\), \(|\lambda_+|=\sqrt{\delta}\). The real part divided by the modulus is \(\tau/(2\sqrt{\delta})\). The number lies strictly between \(-1\) and \(1\), because \(\tau^2<4\delta\). Thus its inverse cosine is a unique angle \(\theta\in(0,\pi)\). The imaginary part of \(\lambda_+\) is positive, so its sine is positive; hence its principal argument is this \(\theta\), rather than \(-\theta\). The roots are conjugates, so \(\lambda_-=\overline{\lambda_+}\) has the same modulus and argument \(-\theta\). \(\square\)

In practical terms, first check the strict inequality \(\tau^2<4\delta\). Only when it holds do the formulas for a nonreal pair apply. Then compute \(r=\sqrt{\delta}\) and \(\theta=\arccos(\tau/(2r))\). The positive-imaginary-part eigenvalue is \(r(\cos\theta+i\sin\theta)\); its conjugate is \(r(\cos\theta-i\sin\theta)\). This is a description of the eigenvalues, not a claim that trace and determinant determine the matrix itself.

Worked Recoveries

What the Data Do—and Do Not—Determine

Trace and determinant determine the characteristic polynomial, and therefore the eigenvalues including their multiplicities. In the nonreal case, the trace fixes the real part of each root, and the determinant fixes the product of the conjugate roots. Together these specify the eigenvalue pair. They do not specify eigenvectors, or all the entries of the matrix. Different matrices can have the same trace and determinant and hence the same eigenvalues.

For instance, consider \(\mathbf{E}=\begin{bmatrix}2&-9\\1&-2\end{bmatrix}\) and \(\mathbf{F}=\begin{bmatrix}0&-5\\1&0\end{bmatrix}\). They are different matrices. For \(\mathbf{E}\), the trace is \(2+(-2)=0\), and the determinant is \((2)(-2)-(-9)(1)=-4+9=5\). For \(\mathbf{F}\), the trace is \(0\), and the determinant is \(0-(-5)(1)=5\). Both therefore have characteristic polynomial \(\lambda^2+5\), and both have eigenvalues \(i\sqrt{5}\) and \(-i\sqrt{5}\). The recovered modulus and arguments agree, but the matrices themselves do not.

The condition \(\tau^2<4\delta\) is essential. If it fails, the inverse-cosine expression need not describe a nonreal pair. For example, when \(\tau^2=4\delta\), the discriminant is zero and the root is real and repeated; when \(\tau^2>4\delta\), both roots are real. In the nonreal case, \(\delta\) must be positive, so a zero or negative determinant also rules out nonreal eigenvalues for a real \(2\times2\) matrix.

Common Pitfalls

Using the determinant itself as the modulus. For \(\mathbf{B}=\begin{bmatrix}0&-5\\1&0\end{bmatrix}\), a tempting but incorrect calculation is \(r=\delta=5\). The eigenvalues are \(\pm i\sqrt{5}\), whose modulus is \(\sqrt{5}\), not \(5\). The correct relation is \(r^2=\delta\), so \(r=\sqrt{\delta}\). This also follows directly from multiplying the conjugate eigenvalues: \(\lambda_+\lambda_-=|\lambda_+|^2=\delta\).

Ignoring which argument is being reported. The trace determines \(\cos\theta\), which alone cannot distinguish \(\theta\) from \(-\theta\). The convention here is to take the eigenvalue with positive imaginary part and its argument \(\theta\in(0,\pi)\); its conjugate has principal argument \(-\theta\). The sign is fixed by identifying the eigenvalue, not by the cosine formula alone.

Assuming the recovered values identify every matrix entry. The matrices \(\mathbf{E}\) and \(\mathbf{F}\) above have the same trace, determinant, and eigenvalues, yet different entries. The formulas recover the scale and angle of the eigenvalue pair, not the full matrix or its eigenvectors.