Solving 2x2 Characteristic Equations Over C

From a Matrix to a Quadratic Equation

For a $2\times2$ matrix, finding eigenvalues reduces to solving a quadratic equation. The arithmetic is familiar when the coefficients are real and the roots are real; over $\mathbb{C}$, the same formula also handles negative discriminants and complex coefficients. The key is to perform each operation in the scalar field $\mathbb{C}$, including taking a square root of the discriminant.

Throughout, let $\mathbf{A}\in\mathbb{C}^{2\times2}$. A scalar $\lambda\in\mathbb{C}$ is an eigenvalue of $\mathbf{A}$ if there is a vector $\mathbf{x}\in\mathbb{C}^2$, $\mathbf{x}\ne\mathbf{0}$, such that $\mathbf{A}\mathbf{x}=\lambda\mathbf{x}$. The characteristic polynomial, using the course convention, is $p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_2)$. Its roots are the solutions of the characteristic equation $p_{\mathbf{A}}(\lambda)=0$.

Definition: For a quadratic polynomial $q(\lambda)=a\lambda^2+b\lambda+c$ with $a,b,c\in\mathbb{C}$ and $a\ne0$, its discriminant is $\Delta=b^2-4ac$. A square root of $\Delta$ is any complex number $s$ satisfying $s^2=\Delta$.

Every complex number has a complex square root. For example, if $z=re^{i\theta}$ with $r\ge0$, then $\sqrt r\,e^{i\theta/2}$ is a square root of $z$; when $z=0$, its square root is $0$. If $\Delta\ne0$, its two square roots are $s$ and $-s$. Thus the sign choice in the formula below gives the two roots, with no need to assume that the discriminant is real.

Theorem (Quadratic formula over $\mathbb{C}$): Let $q(\lambda)=a\lambda^2+b\lambda+c$, where $a,b,c\in\mathbb{C}$ and $a\ne0$. If $s^2=b^2-4ac$, then the roots of $q(\lambda)=0$, counted with repetition, are $$ \lambda=\frac{-b+s}{2a} \qquad\text{and}\qquad \lambda=\frac{-b-s}{2a}. $$ If $s=0$, these two expressions give the same root.

Proof. Since $a\ne0$, divide $q(\lambda)=0$ by $a$, then complete the square. Each step is reversible over $\mathbb{C}$, because $2$ and $a$ are nonzero:

\begin{aligned} a\lambda^2+b\lambda+c=0 &\iff \lambda^2+\frac{b}{a}\lambda+\frac{c}{a}=0\\ &\iff \left(\lambda+\frac{b}{2a}\right)^2 =\frac{b^2-4ac}{4a^2}\\ &\iff 2a\lambda+b=s\ \text{or}\ 2a\lambda+b=-s. \end{aligned}

The last equivalence follows because if $y^2=s^2$, then $(y-s)(y+s)=0$; since $\mathbb{C}$ has no zero divisors, $y=s$ or $y=-s$. Solving the two resulting linear equations gives the stated expressions. If $s=0$, both expressions equal $-b/(2a)$; otherwise they are distinct. This also proves that there are no other roots, since every solution had to satisfy one of those two equations. $\square$

Applying the Formula to a 2×2 Matrix

Write the entries of $\mathbf{A}$ as $a,b,c,d\in\mathbb{C}$. Subtracting $\lambda$ from the diagonal entries and using the $2\times2$ determinant formula gives the characteristic polynomial directly:

\mathbf{A}= \begin{bmatrix}a&b\\c&d\end{bmatrix}, \qquad p_{\mathbf{A}}(\lambda) =\det\begin{bmatrix}a-\lambda&b\\c&d-\lambda\end{bmatrix} =\lambda^2-(a+d)\lambda+(ad-bc).

This is always a quadratic: the coefficient of $\lambda^2$ is $1$, including when some entries of $\mathbf{A}$ are zero. Its discriminant, found by using its displayed coefficients, is $(a+d)^2-4(ad-bc)$. To find the eigenvalues, solve this characteristic equation with the quadratic formula.

Theorem (Characteristic roots are eigenvalues): For $\mathbf{A}\in\mathbb{C}^{2\times2}$, a scalar $\lambda\in\mathbb{C}$ is an eigenvalue if and only if $p_{\mathbf{A}}(\lambda)=0$.

Proof. The eigenvalue definition says that $\lambda$ is an eigenvalue exactly when $(\mathbf{A}-\lambda\mathbf{I}_2)\mathbf{x}=\mathbf{0}$ for some $\mathbf{x}\ne\mathbf{0}$. For a $2\times2$ matrix $\mathbf{M}=\begin{bmatrix}u&v\\w&z\end{bmatrix}$, there is a nonzero solution to $\mathbf{M}\mathbf{x}=\mathbf{0}$ exactly when $uz-vw=0$. Indeed, if $uz-vw\ne0$, direct multiplication shows that $\begin{bmatrix}z&-v\\-w&u\end{bmatrix}/(uz-vw)$ is an inverse of $\mathbf{M}$, so its null equation has only the zero solution. If $uz-vw=0$ and the first row $(u,v)$ is nonzero, the nonzero vector $(v,-u)^\top$ is in the null space: its first row product is $uv-vu=0$, and its second is $wv-zu=0$, using $uz=vw$. If the first row is zero and the second row is nonzero, $(z,-w)^\top$ is a nonzero null vector; if both rows are zero, every vector is a null vector. Apply this fact to $\mathbf{M}=\mathbf{A}-\lambda\mathbf{I}_2$. Its determinant is $p_{\mathbf{A}}(\lambda)$, proving both directions. $\square$

Worked Roots in Three Situations

Worked Example: Two Distinct Real Roots

Find the eigenvalues of $\mathbf{A}=\begin{bmatrix}2&1\\1&2\end{bmatrix}$. First form the characteristic polynomial from the entries:

p_{\mathbf{A}}(\lambda) =\det\begin{bmatrix}2-\lambda&1\\1&2-\lambda\end{bmatrix} =(2-\lambda)^2-1 =\lambda^2-4\lambda+3.

Here $a=1$, $b=-4$, and $c=3$, so $\Delta=(-4)^2-4(1)(3)=16-12=4$. Choose $s=2$, since $2^2=4$. The formula gives

\lambda=\frac{4\pm2}{2}, \qquad \lambda=3\ \text{or}\ \lambda=1.

Check: $p_{\mathbf{A}}(3)=3^2-4(3)+3=9-12+3=0$, and $p_{\mathbf{A}}(1)=1^2-4(1)+3=1-4+3=0$. Both candidates satisfy the characteristic equation.

Worked Example: A Negative Real Discriminant

Find the eigenvalues of $\mathbf{B}=\begin{bmatrix}0&-2\\1&0\end{bmatrix}$. Its characteristic polynomial is

p_{\mathbf{B}}(\lambda) =\det\begin{bmatrix}-\lambda&-2\\1&-\lambda\end{bmatrix} =\lambda^2+2.

The coefficients are $a=1$, $b=0$, and $c=2$, so $\Delta=0^2-4(1)(2)=-8$. A square root is $s=2i\sqrt{2}$, because $(2i\sqrt{2})^2=4i^2(2)=-8$. Therefore

\lambda=\frac{0\pm2i\sqrt{2}}{2} =\pm i\sqrt{2}.

Check: $(i\sqrt{2})^2+2=-2+2=0$, and $(-i\sqrt{2})^2+2=-2+2=0$. The roots are nonreal, and in this example they are conjugates. No real square root of $-8$ was used.

Worked Example: Complex Matrix Entries

Find the eigenvalues of $\mathbf{C}=\begin{bmatrix}1+i&0\\0&3-2i\end{bmatrix}$. The characteristic polynomial is

\begin{aligned} p_{\mathbf{C}}(\lambda) &=(1+i-\lambda)(3-2i-\lambda)\\ &=\lambda^2-(4-i)\lambda+(1+i)(3-2i)\\ &=\lambda^2-(4-i)\lambda+(5+i). \end{aligned}

In the product term, $(1+i)(3-2i)=3-2i+3i-2i^2=5+i$. The discriminant is

\Delta=(4-i)^2-4(5+i) =(16-8i+i^2)-(20+4i) =15-8i-20-4i =-5-12i.

Since $(2-3i)^2=4-12i+9i^2=-5-12i$, choose $s=2-3i$. Substitution into the formula gives

\lambda=\frac{(4-i)\pm(2-3i)}{2}, \qquad \lambda=3-2i\ \text{or}\ \lambda=1+i.

Check: The factored form of the polynomial is $(1+i-\lambda)(3-2i-\lambda)$. At $\lambda=3-2i$, the second factor is $0$; at $\lambda=1+i$, the first factor is $0$. Hence both values solve $p_{\mathbf{C}}(\lambda)=0$ and are eigenvalues by the characteristic-root theorem. They are not conjugates, which is possible here because the matrix has complex entries.

Repeated Roots and a Reliable Checking Routine

When the discriminant is zero, the quadratic formula produces the same value twice. This is called a repeated root of the characteristic polynomial. It is useful to distinguish this statement about the polynomial from any claim about eigenvectors: this calculation identifies the root and its repetition, but does not by itself describe how many independent eigenvectors the matrix has.

Worked Example: A Repeated Characteristic Root

For $\mathbf{D}=\begin{bmatrix}2&1\\0&2\end{bmatrix}$, compute the characteristic equation and solve it.

\begin{aligned} p_{\mathbf{D}}(\lambda) &=\det\begin{bmatrix}2-\lambda&1\\0&2-\lambda\end{bmatrix}\\ &=(2-\lambda)(2-\lambda)-0\\ &=\lambda^2-4\lambda+4. \end{aligned}

Here $\Delta=(-4)^2-4(1)(4)=16-16=0$, so $s=0$. The quadratic formula gives $\lambda=(4\pm0)/2=2$ twice. Factoring confirms $p_{\mathbf{D}}(\lambda)=(\lambda-2)^2$.

Check: $p_{\mathbf{D}}(2)=2^2-4(2)+4=4-8+4=0$. Since the factor is squared, this is a repeated root, not two distinct roots that happen to have the same numerical value.

A dependable calculation can be organized in four steps:

Build the polynomial.
Compute $\det(\mathbf{A}-\lambda\mathbf{I}_2)$ carefully, retaining the minus signs in the off-diagonal product.

Read the coefficients.
Write the result as $a\lambda^2+b\lambda+c$. For a $2\times2$ characteristic polynomial, $a=1$.

Compute and square-check the discriminant root.
Find $\Delta=b^2-4ac$, then choose $s$ with $s^2=\Delta$. For a negative real $\Delta$, the square root is imaginary, not real.

Apply the formula and substitute back.
Evaluate $(-b+s)/(2a)$ and $(-b-s)/(2a)$, then check each candidate in $p_{\mathbf{A}}(\lambda)$.

Common Pitfall: Ignoring the Complex Square Root

A frequent error is to solve $\lambda^2+2=0$ by writing $\lambda=\pm\sqrt{2}$, as though the equation were $\lambda^2-2=0$. Substitution exposes the error: $(\sqrt{2})^2+2=2+2=4\ne0$, and $(-\sqrt{2})^2+2=2+2=4\ne0$. The correct calculation is $\lambda^2=-2$, so $\lambda=\pm i\sqrt{2}$. Indeed, $(\pm i\sqrt{2})^2+2=-2+2=0$.

Also take care with signs when expanding the characteristic polynomial. For $\begin{bmatrix}u&v\\w&z\end{bmatrix}$, the determinant is $uz-vw$, not $uz+vw$. For instance, in the negative-discriminant example the off-diagonal product is $(-2)(1)=-2$, so subtracting it contributes $+2$ to $\lambda^2+2$. A sign error there changes the equation and therefore its roots.

The method applies uniformly over $\mathbb{C}$: expand the characteristic polynomial, use the quadratic formula, and verify the roots by substitution. When the entries are real, the coefficients are real, so a negative discriminant gives a pair of nonreal conjugate roots, as the worked example illustrates. For complex entries, coefficients and roots need not be real or conjugate; the formula itself requires no change.

Check Your Understanding

For each problem, form the characteristic polynomial before applying the quadratic formula, and check the proposed roots by substitution.

Find the characteristic roots of $\begin{bmatrix}3&0\\0&-1\end{bmatrix}$.
Solve $\lambda^2+9=0$ over $\mathbb{C}$, and verify both roots.
For which $t\in\mathbb{C}$ does $\lambda^2-2t\lambda+t^2=0$ have a repeated root?
Find the characteristic roots of $\begin{bmatrix}i&0\\0&2+i\end{bmatrix}$, showing the discriminant and one of its square roots.
Explain why a candidate $\lambda$ is an eigenvalue of a $2\times2$ matrix exactly when it makes the characteristic polynomial zero.

Solving 2x2 Characteristic Equations Over C

What You'll Learn

From a Matrix to a Quadratic Equation

Applying the Formula to a 2×2 Matrix