Fast method for finding Eigen values for a 2x2 Matrix

Feb 16, 2026

Introduction

The prusuit for finding eigen values, and the eigen vector for matrices can be very complicated under the scaling of dimensions, however, such as any probles, the lower the dimensions, the simpler they are to solve.

The issues of dimensions can be seen when it comes to topics such as determinants, which in a 2×22 \times 2 case is very simply to solve, since it’s simply given by some matrix AA

A=[abcd]A = \left[\begin{array}{cc} a & b\\ c & d\\ \end{array}\right]

We then have that

∣A∣=a×d−b×c| A | = a \times d - b \times c

However, when we scale the dimensions of the matrix up, we require more complex methods to solve such cases.

The generalized method

It can be noted, however, that just like the case of the determinant, that in the lower dimension case, there are easier way to solve for the. In the general case case, we have that the eigenvector is given by the eigen values, of which it satify the following equation

Ax=λxAx = \lambda x

This is then solved by the following equation

A−λI=[abcd]−λ[1001]=[a−λbcd−λ]\begin{aligned} A - \lambda I &= \left[\begin{array}{cc} a & b\\ c & d\\ \end{array}\right] - \lambda \left[\begin{array}{cc} 1 & 0\\ 0 & 1\\ \end{array}\right]\\ &= \left[\begin{array}{cc} a - \lambda & b\\ c & d - \lambda\\ \end{array}\right] \end{aligned}

Example

An example of this, can be the following case

A−λI=[7−152−4]−λ[1001]=[7−λ−172−4−λ]\begin{aligned} A - \lambda I &= \left[\begin{array}{cc} 7 & -15\\ 2 & -4\\ \end{array}\right] - \lambda \left[\begin{array}{cc} 1 & 0\\ 0 & 1\\ \end{array}\right]\\ &= \left[\begin{array}{cc} 7 - \lambda & -17\\ 2 & -4 - \lambda\\ \end{array}\right] \end{aligned}

There is then solved for the characteristic polynomial, which is given by

∣A−λI∣=∣[7−152−4]−λ[1001]∣=∣[7−λ−152−4−λ]∣=(7−λ)(−4−λ)−(2×−15)=−28−7λ+4λ+λ2+30=λ2−3λ+2\begin{aligned} \left| A - \lambda I \right| &= \left|\left[\begin{array}{cc} 7 & -15\\ 2 & -4\\ \end{array}\right] - \lambda \left[\begin{array}{cc} 1 & 0\\ 0 & 1\\ \end{array}\right]\right|\\ &= \left|\left[\begin{array}{cc} 7 - \lambda & -15\\ 2 & -4 - \lambda\\ \end{array}\right]\right|\\ &= \left(7-\lambda\right)\left(-4-\lambda\right) - \left(2 \times -15\right)\\ &= -28 - 7\lambda + 4 \lambda + \lambda^{2} + 30\\ &= \lambda^{2}-3\lambda +2 \end{aligned}

Given that we now know the polynomial, we then wish to find the roots for it, which is the case of

λ2−3λ+2=0\lambda^{2}-3\lambda +2 = 0

If we inspect the terms, we know that we’re dealing with λ2\lambda^2 (obviously, it is a polynomial), and we have that there will be two cases, so we apply the quadratic formula, which is given by

−b±b2−4ac2a\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In this case, we have that

a=1b=−3c=2\begin{aligned} a &= 1\\ b &= -3\\ c &= 2 \end{aligned}

Giving us

3±−32−4(1)(2)2(1)=3±9−82=3±12\begin{aligned} \frac{3 \pm \sqrt{-3^2 - 4(1)(2)}}{2(1)} &= \frac{3 \pm \sqrt{9 - 8}}{2}\\ &= \frac{3 \pm 1}{2} \end{aligned}

So the two cases we have is

λ1=3+12λ2=3−12\begin{aligned} \lambda_1 &= \frac{3 + 1}{2}\\ \lambda_2 &= \frac{3 - 1}{2} \end{aligned}

Which gives us

λ1=42=2\begin{aligned} \lambda_1 &= \frac{4}{2}\\ &= 2 \end{aligned} λ2=22=1\begin{aligned} \lambda_2 &= \frac{2}{2}\\ &= 1 \end{aligned}

The fast method

For the fast method, we apply the average value of the diagonal, and the determinant of the matrix.

Average value of diagonal

avgdiag⁡(A)=1n∑n=1Nann\operatorname{avgdiag}(A) = \frac{1}{n} \sum^{N}_{n = 1} a_{nn}

for n∈Nn \in N where NN is is the dimension of a square matrix in RN×N\mathbb{R}^{N \times N}

We then have that the eigen values are given by

λ=avgdiag⁡(A)±(avgdiag⁡(A))2−∣A∣\lambda = \operatorname{avgdiag}(A) \pm \sqrt{\left(\operatorname{avgdiag}(A)\right)^2 - | A |}

Example

We can use the example from before, we have that the matrix AA is given by

A=[7−152−4]A = \left[\begin{array}{cc} 7 & -15\\ 2 & -4\\ \end{array}\right]

So we have that, since the matrix is in RN×N\mathbb{R}^{N \times N} that

avgdiag⁡(A)=12(7−4)=1.5\begin{aligned} \operatorname{avgdiag}(A) &= \frac{1}{2}\left(7-4\right)\\ &= 1.5 \end{aligned}

We have that the determinant is given by

∣A∣=(7)(−4)−(−15)(2)=−28−(−30)=2\begin{aligned} |A| &= (7)(-4) - (-15)(2) \\ &= -28 - (-30) \\ &= 2 \end{aligned}

So we then get

λ=1.5±(1.5)2−2=1.5±2.25−2=1.5±0.25=1.5±0.5\begin{aligned} \lambda &= 1.5 \pm \sqrt{ (1.5)^2 - 2 } \\ &= 1.5 \pm \sqrt{ 2.25 - 2 } \\ &= 1.5 \pm \sqrt{ 0.25 } \\ &= 1.5 \pm 0.5 \end{aligned}

Which gives us

λ1=1.5+0.5=2\begin{aligned} \lambda_1 &= 1.5 + 0.5\\ &= 2 \end{aligned}

and

λ2=1.5−0.5=1\begin{aligned} \lambda_2 &= 1.5 - 0.5\\ &= 1 \end{aligned}

So, as we can see, like the regular method, we end up with exactly the same results.

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