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Eigenvalues and Eigenvectors

The Heart of Transformations

Linear transformations can stretch, rotate, and shear vectors, changing both their length and direction. But for any given transformation, there are often special vectors that have a simpler relationship with it. These are the eigenvectors.

An eigenvector is a non-zero vector whose direction doesn't change when the linear transformation is applied. It might get stretched or squished, but it still points along the same line. The factor by which it's stretched or squished is its corresponding eigenvalue, represented by the Greek letter lambda (λλ).

Av=λvA\vec{v} = \lambda\vec{v}

Here, AA is the matrix representing the linear transformation, v\vec{v} is the eigenvector, and λλ is the scalar eigenvalue. This equation states that applying the transformation AA to the eigenvector v\vec{v} gives the same result as simply scaling v\vec{v} by the amount λλ.

Eigenvectors form the 'axes' of a transformation. They reveal the fundamental directions along which the transformation acts purely by stretching or compressing. This property makes them incredibly useful for understanding the behavior of complex systems.

Finding Eigenvalues

To find these special vectors, we start by rearranging the main equation:

Av=λvA\vec{v} = \lambda\vec{v} Avλv=0A\vec{v} - \lambda\vec{v} = 0

To turn the scalar λλ into a matrix so we can subtract it from AA, we multiply it by the identity matrix II. The identity matrix is the matrix equivalent of the number 1; multiplying by it doesn't change a vector.

(AλI)v=0(A - \lambda I)\vec{v} = 0

We are looking for a non-zero vector v\vec{v} that solves this equation. A system of equations like this only has a non-zero solution if the matrix (AλI)(A - \lambda I) is singular, meaning its determinant is zero. This gives us a way to find the eigenvalues.

The eigenvalues of a matrix AA are the values of λλ that solve the characteristic equation: det(AλI)=0\det(A - \lambda I) = 0.

The expression det(AλI)\det(A - \lambda I) is a polynomial in λλ, called the characteristic polynomial. Let's find the eigenvalues for a simple 2x2 matrix.

A=(4123)A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}

First, we set up (AλI)(A - \lambda I):

AλI=(4123)λ(1001)=(4λ123λ)A - \lambda I = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{pmatrix}

Next, we find the determinant and set it to zero:

det(AλI)=(4λ)(3λ)(1)(2)=0\det(A - \lambda I) = (4-\lambda)(3-\lambda) - (1)(2) = 0

Expanding this gives us the characteristic polynomial:

124λ3λ+λ22=012 - 4\lambda - 3\lambda + \lambda^2 - 2 = 0 λ27λ+10=0\lambda^2 - 7\lambda + 10 = 0

Factoring the polynomial, we get (λ5)(λ2)=0(\lambda - 5)(\lambda - 2) = 0. The solutions are our eigenvalues: λ1=5λ_1 = 5 and λ2=2λ_2 = 2. Now that we have the eigenvalues, we can find their corresponding eigenvectors by plugging them back into the equation (AλI)v=0(A - \lambda I)\vec{v} = 0.

Diagonalization

One of the most powerful applications of eigenvectors is diagonalization. If an n×nn \times n matrix AA has nn linearly independent eigenvectors, we can break it down into a simpler form.

Specifically, we can write AA as a product of three matrices:

A=PDP1A = PDP^{-1}

Here, PP is a matrix whose columns are the eigenvectors of AA. And DD is a diagonal matrix, which means it has zeros everywhere except for the main diagonal. The entries on its diagonal are the eigenvalues of AA, in the same order as their corresponding eigenvectors in PP.

Why is this useful? Diagonal matrices are very easy to work with. For instance, calculating powers of a diagonal matrix is simple: you just raise its diagonal elements to that power. This makes calculating powers of the original matrix AA much easier, too.

For example, A2=(PDP1)(PDP1)=PD(P1P)DP1=PDIDP1=PD2P1A^2 = (PDP^{-1})(PDP^{-1}) = PD(P^{-1}P)DP^{-1} = PDIDP^{-1} = PD^2P^{-1}.

In general, calculating AkA^k is as simple as finding DkD^k and multiplying by PP and P1P^{-1}.

Ak=PDkP1A^k = PD^kP^{-1}

This trick is a huge computational shortcut and is fundamental to understanding how systems evolve over time, especially in quantum mechanics.

Eigenvalues in Quantum Mechanics

In quantum mechanics, the state of a system is described by a vector in a vector space. Physical properties you can measure, like position, momentum, or energy, are represented by linear operators.

When you measure a physical property, the system collapses into one of the operator's eigenstates (another name for eigenvectors). The value you measure is the corresponding eigenvalue. This is a core principle of quantum mechanics.

The eigenvalues of a quantum mechanical operator represent the possible measurement outcomes of a physical observable.

For example, the energy of a quantum system is described by an operator called the Hamiltonian, often written as HH. The possible energy levels of the system are the eigenvalues of HH. When you measure the system's energy, you will always get one of these eigenvalues. The state of the system after the measurement will be the corresponding energy eigenstate.

This is described by the time-independent Schrödinger equation, which is an eigenvalue equation:

H^ψ=Eψ\hat{H}|\psi\rangle = E|\psi\rangle

Here, H^\hat{H} is the Hamiltonian operator, ψ|\psi\rangle is the energy eigenstate (the eigenvector), and EE is the energy value (the eigenvalue).

Diagonalization is also crucial for understanding how a quantum state evolves. If you know the energy eigenstates of a system, you can express any state as a linear combination of them. Since the evolution of each eigenstate is simple (it just picks up a phase factor related to its energy eigenvalue), you can calculate the evolution of the entire system by adding up the evolutions of its component parts.

Eigenvalues and eigenvectors are not just abstract mathematical tools. They are the language used to describe the fundamental, quantized nature of reality.

Quiz Questions 1/5

What is the defining characteristic of an eigenvector when a linear transformation is applied to it?

Quiz Questions 2/5

To find the eigenvalues of a matrix A, we solve the characteristic equation. Which of the following represents that equation?