Explainer: Eigenvalues and Eigenvectors

In this explainer, we will learn how to find the eigenvalues of a matrix and the corresponding eigenvectors of a matrix.

It is a little exaggeration to say that eigenvalues and eigenvectors are among the most widely used concepts in linear algebra. The development of this vast field of abstract mathematics was piecemeal and nonlinear in the sense that it took several millennia for the different components of linear algebra to be sufficiently developed to the point that they could be incorporated into an overarching, more complete subject. Unsurprisingly, the development of eigenvalues and eigenvectors was similarly nonlinear and piecemeal.

The study of eigenvalues and eigenvectors is largely considered to have begun properly with the seemingly distant topic of differential equations, when the esteemed mathematician Leonhard Euler was looking to solve certain rigid body problems. The concept of eigenvalues and eigenvectors was further developed by some of the most recognized mathematicians who followed Euler, including Laplace, Hermite, Cauchy, and Liouville. Only in the 20th century did these disparate and nonuniform topics become drawn together in a formal pattern of study by the eminent mathematician David Hilbert.

Especially since the age of contemporary science and computing, eigenvalues and eigenvectors have become versatile and pervasive ideas, offering crucial insights in fields such as computer graphics, quantum mechanics, acoustics, thermodynamics, and many, many others. When dealing with a linear operator (in our case, this will be a matrix), calculation of the eigenvalues and eigenvectors is tantamount to having a full understanding of spatial properties of the operator and the ways in which calculations can be significantly simplified.

It would be very easy to list all of the many significant advances that have resulted from the study of eigenvalues and eigenvectors, but these will ultimately be meaningless until we have defined these concepts properly. In this explainer, we will focus only on calculating the eigenvalues of a matrix, which is ordinarily a fairly straightforward process that involves the factorization of a very particular polynomial called the characteristic polynomial. First, we define what we mean by β€œeigenvalues” and β€œeigenvectors.”

Definition: Eigenvalues and Eigenvectors of a Square Matrix

Consider a square matrix 𝐴 with order 𝑛×𝑛. Then, an β€œeigenvector” v is a nonzero vector with order 𝑛×1 such that 𝐴v=𝑑v.

The scaling constant 𝑑 is referred to as the β€œeigenvalue” corresponding to the eigenvector v.

One way of interpreting this definition is that the eigenvectors of a matrix 𝐴 are those vectors which are unchanged after being operated on by the original matrix using matrix multiplication, except for the scaling constant 𝑑. We will demonstrate this first by an example.

Consider the following square matrix and vector: 𝐴=2130,v=22.

We will illustrate the calculation of the matrix product 𝐴v using the known rules of matrix multiplication. Since 𝐴 is a matrix with order 𝑛×𝑛 and v is a matrix with order 𝑛×1, then we know that the matrix product 𝐴v is well defined and will have order 𝑛×1. This means that 213022=ο’βˆ—βˆ—οž, where the βˆ— are entries that we need to calculate. To calculate the unknown entry in the first row and first column of the rightmost matrix, we would need to highlight the first row of 𝐴 and the first column of v, as shown: 213022=ο’βˆ—βˆ—οž.

We combine the highlighted elements in order to find 2Γ—2+1Γ—2=6, meaning that 213022=6βˆ—οŸ.

We must now calculate the remaining unknown entry in the second row and first column of the rightmost matrix. Accordingly, we highlight the second row of 𝐴 and the first column of v: 213022=6βˆ—οŸ.

The calculation is 3Γ—2+0Γ—2=6, meaning that we can now write 𝐴v as a scaling of the vector v: 𝐴v=213022=66=322.

Since it is possible to write 𝐴v=𝑑v when 𝑑=3, we conclude from the definition that v is an eigenvector of 𝐴 with an eigenvalue of 3.

In contrast, let us consider the same matrix 𝐴 and a new vector Μƒv=[4βˆ’3].

Replicating the steps from the above working, we deduce that 𝐴̃v=21304βˆ’3=512.

It is evidently not possible to take the resultant vector 512 and write this as a constant multiple of the original vector Μƒv. This means that Μƒv is not an eigenvector of the matrix 𝐴.

Although it is always possible to check whether or not a given vector is an eigenvector of a particular square matrix, in reality this is of little help when working with a square matrix without any prior knowledge. In this situation, it is necessary to first calculate the eigenvalues and then the eigenvectors. In this explainer, we will only show how to calculate the eigenvalues, with the calculation of eigenvectors being reserved for another explainer.

Definition: Characteristic Polynomial

For a square matrix 𝐴 of order 𝑛×𝑛, consider the β€œcharacteristic matrix,” which is defined as 𝐴=π‘‘πΌπ‘›βˆ’π΄, where 𝐼𝑛 is the 𝑛×𝑛 identity matrix. Then, the β€œcharacteristic polynomial” is the determinant of this matrix; that is, 𝑃(𝑑)=|𝐴|.

As we will soon see, the characteristic polynomial of an 𝑛×𝑛 matrix is a polynomial, where the largest order term is π‘₯𝑛. By the fundamental theorem of arithmetic, this means that there will be at most 𝑛 unique solutions over the real numbers. To demonstrate this, we will consider the 2Γ—2 matrix 𝐴=ο”βˆ’124βˆ’3.

To calculate the characteristic polynomial, we first calculate the characteristic matrix: 𝐴=𝑑𝐼2βˆ’π΄=𝑑1001ο βˆ’ο”βˆ’124βˆ’3=𝑑00π‘‘ο βˆ’ο”βˆ’124βˆ’3=𝑑+1βˆ’2βˆ’4𝑑+3.

We now evaluate the determinant of this 2Γ—2 matrix by using the standard approach to obtain the characteristic polynomial: 𝑃(𝑑)=|𝐴|=||𝑑+1βˆ’2βˆ’4𝑑+3||=(𝑑+1)Γ—(𝑑+3)βˆ’(βˆ’2)Γ—(βˆ’4)=𝑑2+4π‘‘βˆ’5=(𝑑+5)(π‘‘βˆ’1).

The characteristic polynomial of 𝐴 is quadratic and therefore has at most 2 real roots. It is these roots that are of interest in relation to the eigenvalues of the matrix 𝐴.

Theorem: Eigenvalues of a Matrix

The eigenvalues of a square matrix 𝐴 with order 𝑛×𝑛 are roots of the characteristic polynomial 𝑃(𝑑).

We will demonstrate this theorem in relation to the above matrix: 𝐴=ο”βˆ’124βˆ’3.

The characteristic polynomial was 𝑃(𝑑)=(π‘‘βˆ’1)(𝑑+5), and the roots of this (which are the eigenvalues of 𝐴) are 𝑑=βˆ’5 and 𝑑=1. We can check this from the definition of the eigenvalues and eigenvectors, which states that v is an eigenvector of the matrix 𝐴 if it satisfies the equation 𝐴v=𝑑v, where 𝑑 is the eigenvalue corresponding to the eigenvector v. Suppose we knew a priori that two eigenvectors of 𝐴 are v=1βˆ’2,w=11.

We can then use the definition of an eigenvector to calculate the eigenvalue. First, we substitute v, as shown: 𝐴v=ο”βˆ’124βˆ’31βˆ’2=ο”βˆ’510=βˆ’51βˆ’2=𝑑v.

As can be seen, the eigenvalue is βˆ’5, which is one of the two values that we calculated above. If we take the second given eigenvector w, then the result is even simpler: 𝐴w=ο”βˆ’124βˆ’311=11=111=𝑑w.

We have found that the scaling constant is 1, which is the other eigenvalue that we expected.

At this stage, it should be clarified that we are not stating that v and w are the only eigenvectors of 𝐴. In fact, we could have taken any nonzero scalar multiple of v or w, and these would have been eigenvectors with exactly the same eigenvalues. It is not quite the correct approach to ask how many eigenvectors there are of a matrix, as the matter is more delicate than this and requires a full consideration of β€œeigenspaces.” This is a topic that should be covered separately, and for the rest of this explainer we will focus only on correctly calculating the eigenvalues of a matrix.

Example 1: Eigenvalues of a 2 Γ— 2 Matrix

Calculate the eigenvalues of the matrix 𝐴=4βˆ’1βˆ’25.

Answer

We begin with the characteristic matrix corresponding to 𝐴: 𝐴=𝑑𝐼2βˆ’π΄=𝑑1001ο βˆ’ο”4βˆ’1βˆ’25=𝑑00π‘‘ο βˆ’ο”4βˆ’1βˆ’25=ο”π‘‘βˆ’412π‘‘βˆ’5.

Now, evaluating the determinant of 𝐴 gives the characteristic polynomial 𝑃(𝑑)=|𝐴|=(π‘‘βˆ’4)Γ—(π‘‘βˆ’5)βˆ’1Γ—2=𝑑2βˆ’9𝑑+18=(π‘‘βˆ’3)(π‘‘βˆ’6).

The eigenvalues are the roots of the characteristic polynomial, which in this case implies that the eigenvalues are 𝑑=3 and 𝑑=6.

In the question above, we can confirm that 𝑑=3 and 𝑑=6 are indeed two eigenvalues of 𝐴, as can be checked with the example eigenvectors v=33,w=ο”βˆ’24.

In the two examples above, we were given a matrix and were able to calculate two unique, real eigenvalues that were the roots of a quadratic equation. We already know that a general quadratic equation does not always have two unique, real roots, but it could also have one repeated real root or even no real roots.

Example 2: Eigenvalues of a 2 Γ— 2 Matrix

Calculate the eigenvalues of the matrix 𝐴=1βˆ’337.

Answer

Before calculating the characteristic polynomial, we construct the characteristic matrix 𝐴=𝑑𝐼2βˆ’π΄=𝑑1001ο βˆ’ο”1βˆ’337=𝑑00π‘‘ο βˆ’ο”1βˆ’337=ο”π‘‘βˆ’13βˆ’3π‘‘βˆ’7.

Then, the determinant is evaluated to give the characteristic polynomial 𝑃(𝑑)=(π‘‘βˆ’1)Γ—(π‘‘βˆ’7)βˆ’3Γ—(βˆ’3)=𝑑2βˆ’8𝑑+16=(π‘‘βˆ’4)2.

There is one repeated root of the characteristic polynomial 𝑃(𝑑)=0, which implies that the only eigenvalue of 𝐴 is 𝑑=4.

We can check that one of the eigenvectors is v=ο”βˆ’11.

The matrix in the above example having only one, repeated eigenvalue is clearly different to the previous examples where there were two, unique eigenvalues. We can reasonably expect this difference to influence the properties of any corresponding eigenvectors. To describe this relationship in full, it is necessary to study eigenspaces and how these characteristics can be used to classify linear maps described by matrices.

We will give one further example as to how we can calculate the eigenvalues of a square matrix. Rather than work with a 2Γ—2 matrix, we will show how the technique above can extend to a 3Γ—3 matrix. Conceptually, there is little extra difference in finding the eigenvalues of a square matrix with a larger order, although evaluating the determinant of a larger matrix is more complicated.

Example 3: Eigenvalues of a 3 Γ— 3 Matrix

Calculate the eigenvalues of the matrix 𝐴=11βˆ’120βˆ’1βˆ’223.

Answer

Calculating the characteristic polynomial will require calculating the determinant of the characteristic matrix 𝐴=𝑑𝐼3βˆ’π΄=π‘‘ο˜100010001ο€βˆ’ο˜11βˆ’120βˆ’1βˆ’223=ο˜π‘‘000𝑑000π‘‘ο€βˆ’ο˜11βˆ’120βˆ’1βˆ’223=ο˜π‘‘βˆ’1βˆ’11βˆ’2𝑑12βˆ’2π‘‘βˆ’3.

The determinant of a 3Γ—3 matrix can be evaluated using Sarrus’ rule or any viable method. We choose to expand the determinant along the first row, which is equivalent to Sarrus’ rule. We highlight the relevant entries in 𝐴: 𝐴=ο˜π‘‘βˆ’1βˆ’11βˆ’2𝑑12βˆ’2π‘‘βˆ’3, where the highlighted entries are Μƒπ‘Ž11=π‘‘βˆ’1, Μƒπ‘Ž12=βˆ’1, and Μƒπ‘Ž13=1. The corresponding matrix minors are 𝐴11=𝑑1βˆ’2π‘‘βˆ’3,𝐴12=ο”βˆ’212π‘‘βˆ’3,𝐴13=ο”βˆ’2𝑑2βˆ’2.

Calculating the determinants and simplifying the resulting expressions give ||𝐴11||=𝑑2βˆ’3𝑑+2,||𝐴12||=4βˆ’2𝑑,||𝐴13||=4βˆ’2𝑑.

The characteristic polynomial is the determinant of the characteristic matrix 𝐴, which can be calculated as |𝐴|=Μƒπ‘Ž11||𝐴11||βˆ’Μƒπ‘Ž12||𝐴12||+Μƒπ‘Ž13||𝐴13||=(π‘‘βˆ’1)×𝑑2βˆ’3𝑑+2ο…βˆ’(βˆ’1)Γ—(4βˆ’2𝑑)+(1)Γ—(4βˆ’2𝑑)=𝑑3βˆ’4𝑑2+5π‘‘βˆ’2+(4βˆ’2𝑑)+(4βˆ’2𝑑)=𝑑3βˆ’4𝑑2+𝑑+6=(𝑑+1)(π‘‘βˆ’2)(π‘‘βˆ’3).

The roots of 𝑃(𝑑) are the eigenvalues of 𝐴, therefore having the values 𝑑=βˆ’1, 𝑑=2, and 𝑑=3.

We could check that the three eigenvalues in the previous example will appear when we produce a suitable eigenvector. For example, if we had the vectors u=1βˆ’11,v=220,w=33βˆ’3, then we could check from the definition that these vectors are all eigenvectors of 𝐴 that correspond to the three eigenvalues of 𝑑=βˆ’1, 𝑑=2, and 𝑑=3.

This explainer has not discussed how to calculate the eigenvectors of a square matrix, which is not a particularly difficult method but is one that requires a separate and detailed explanation. In terms of calculating the eigenvalues, the method does not change depending on whether there are two real eigenvalues, one repeated real eigenvalue, or no real eigenvalues. What does change, however, is the method for calculating and classifying the eigenvectors, meaning that the topic should be explored independently.

Key Points

  • The eigenvalues of an 𝑛×𝑛 matrix 𝐴 are the roots of the characteristic polynomial 𝑝(𝑑)=|π‘‘πΌπ‘›βˆ’π΄|, where 𝐼𝑛 is the 𝑛×𝑛 identity matrix.
  • For a matrix of order 𝑛×𝑛, there will be at most 𝑛 distinct real solutions, with the possibility of the roots being repeated.
  • Calculating the eigenvalues is usually the first step towards calculating the eigenvectors, unless one or more of the eigenvectors are particularly obvious!

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