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Properties of Eigenvalues and Eigenvectors

The previous section introduced eigenvalues and eigenvectors, and concentrated on their existence and determination.

Properties of Eigenvalues and Eigenvectors

The previous section introduced eigenvalues and eigenvectors, and concentrated on their existence and determination. This section will be more about theorems, and the various properties eigenvalues and eigenvectors enjoy. Like a good $4\times 100\text{ meter}$ relay, we will lead-off with one of our better theorems and save the very best for the anchor leg.

How can you identify a nonsingular matrix just by looking at its eigenvalues?
How many different eigenvalues may a square matrix of size $n$ have?
What is amazing about the eigenvalues of a Hermitian matrix and why is it amazing?

Basic Properties of Eigenvalues

Theorem: Eigenvectors with Distinct Eigenvalues are Linearly Independent

Suppose that $A$ is an $n\times n$ square matrix and $S=\set{\vectorlist{x}{p}}$ is a set of eigenvectors with eigenvalues $\scalarlist{\lambda}{p}$ such that $\lambda_i\neq\lambda_j$ whenever $i\neq j$. Then $S$ is a linearly independent set.

Proof

If $p=1$, then the set $S=\set{\vect{x}_1}$ is linearly independent since eigenvectors are nonzero (), so assume for the remainder that $p\geq 2$.

We will prove this result by contradiction (). Suppose to the contrary that $S$ is a linearly dependent set. Define $S_i=\set{\vectorlist{x}{i}}$ and let $k$ be an integer such that $S_{k-1}=\set{\vectorlist{x}{k-1}}$ is linearly independent and $S_k=\set{\vectorlist{x}{k}}$ is linearly dependent. We have to ask if there is even such an integer $k$? First, since eigenvectors are nonzero, the set $\set{\vect{x}_1}$ is linearly independent. Since we are assuming that $S=S_p$ is linearly dependent, there must be an integer $k$, $2\leq k\leq p$, where the sets $S_i$ transition from linear independence to linear dependence (and stay that way). In other words, $\vect{x}_k$ is the vector with the smallest index that is a linear combination of just vectors with smaller indices.

Since $\set{\vectorlist{x}{k}}$ is a linearly dependent set there must be scalars, $\scalarlist{a}{k}$, not all zero (), so that \zerovector=\lincombo{a}{x}{k}

Then, \zerovector &=\left(A-\lambda_kI_n\right)\zerovector &&\\ &=\left(A-\lambda_kI_n\right)\left(\lincombo{a}{x}{k}\right) &&\\ &=\left(A-\lambda_kI_n\right)a_1\vect{x}_1+ \cdots+ \left(A-\lambda_kI_n\right)a_k\vect{x}_k &&\\ &=a_1\left(A-\lambda_kI_n\right)\vect{x}_1+ \cdots+ a_k\left(A-\lambda_kI_n\right)\vect{x}_k &&\\ &=a_1\left(A\vect{x}_1-\lambda_kI_n\vect{x}_1\right)+ \cdots+ a_k\left(A\vect{x}_k-\lambda_kI_n\vect{x}_k\right) &&\\ &=a_1\left(A\vect{x}_1-\lambda_k\vect{x}_1\right)+ \cdots+ a_k\left(A\vect{x}_k-\lambda_k\vect{x}_k\right) &&\\ &=a_1\left(\lambda_1\vect{x}_1-\lambda_k\vect{x}_1\right)+ \cdots+ a_k\left(\lambda_k\vect{x}_k-\lambda_k\vect{x}_k\right) &&\\ &=a_1\left(\lambda_1-\lambda_k\right)\vect{x}_1+ \cdots+ a_k\left(\lambda_k-\lambda_k\right)\vect{x}_k &&\\ &=a_1\left(\lambda_1-\lambda_k\right)\vect{x}_1+ \cdots+ a_{k-1}\left(\lambda_{k-1}-\lambda_k\right)\vect{x}_{k-1}+ a_k\left(0\right)\vect{x}_k &&\\ &=a_1\left(\lambda_1-\lambda_k\right)\vect{x}_1+ \cdots+ a_{k-1}\left(\lambda_{k-1}-\lambda_k\right)\vect{x}_{k-1}+ \zerovector &&\\ &=a_1\left(\lambda_1-\lambda_k\right)\vect{x}_1+ \cdots+ a_{k-1}\left(\lambda_{k-1}-\lambda_k\right)\vect{x}_{k-1} &&

This equation is a relation of linear dependence on the linearly independent set $\set{\vectorlist{x}{k-1}}$, so the scalars must all be zero. That is, $a_i\left(\lambda_i-\lambda_k\right)=0$ for $1\leq i\leq k-1$. However, we have the hypothesis that the eigenvalues are distinct, so $\lambda_i\neq\lambda_k$ for $1\leq i\leq k-1$. Thus $a_i=0$ for $1\leq i\leq k-1$.

This reduces the original relation of linear dependence on $\set{\vectorlist{x}{k}}$ to the simpler equation $a_k\vect{x}_k=\zerovector$. By we conclude that $a_k=0$ or $\vect{x}_k=\zerovector$. Eigenvectors are never the zero vector (), so $a_k=0$. So all of the scalars $a_i$, $1\leq i\leq k$ are zero, contradicting their introduction as the scalars creating a nontrivial relation of linear dependence on the set $\set{\vectorlist{x}{k}}$. With a contradiction in hand, we conclude that $S$ must be linearly independent.

Condensed: the full section is in Beezer, A First Course in Linear Algebra.

Multiplicities of Eigenvalues

A polynomial of degree $n$ will have exactly $n$ roots. From this fact about polynomial equations we can say more about the algebraic multiplicities of eigenvalues.

Theorem: Number of Eigenvalues of a Matrix

Suppose that $\scalarlist{\lambda}{k}$ are the distinct eigenvalues of a square matrix $A$ of size $n$. Then \sum_{i=1}^{k}\algmult{A}{\lambda_i}=n

Proof

By the definition of the algebraic multiplicity (), we can factor the characteristic polynomial as \charpoly{A}{x}=c(x-\lambda_1)^{\algmult{A}{\lambda_1}}(x-\lambda_2)^{\algmult{A}{\lambda_2}}(x-\lambda_3)^{\algmult{A}{\lambda_3}}\cdots(x-\lambda_k)^{\algmult{A}{\lambda_k}} where $c$ is a nonzero constant. (We could prove that $c=(-1)^{n}$, but we do not need that specificity right now. See ) The left-hand side is a polynomial of degree $n$ by and the right-hand side is a polynomial of degree $\sum_{i=1}^{k}\algmult{A}{\lambda_i}$. So the equality of the polynomials' degrees gives the equality $\sum_{i=1}^{k}\algmult{A}{\lambda_i}=n$.

Condensed: the full section is in Beezer, A First Course in Linear Algebra.

Eigenvalues of Hermitian Matrices

Recall that a matrix is Hermitian (or self-adjoint) if $A=\adjoint{A}$ (). In the case where $A$ is a matrix whose entries are all real numbers, being Hermitian is identical to being symmetric (). Keep this in mind as you read the next two theorems. Their hypotheses could be changed to suppose $A$ is a real symmetric matrix.

Theorem: Hermitian Matrices have Real Eigenvalues

Suppose that $A$ is a Hermitian matrix and $\lambda$ is an eigenvalue of $A$. Then $\lambda\in{\mathbb R}$.

Proof

Let $\vect{x}\neq\zerovector$ be one eigenvector of $A$ for the eigenvalue $\lambda$. Then by we know $\innerproduct{\vect{x}}{\vect{x}}\neq 0$. So \lambda &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\lambda\innerproduct{\vect{x}}{\vect{x}} &&\\ &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\innerproduct{\vect{x}}{\lambda\vect{x}} &&\\ &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\innerproduct{\vect{x}}{A\vect{x}} &&\\ &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\innerproduct{A\vect{x}}{\vect{x}} &&\\ &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\innerproduct{\lambda\vect{x}}{\vect{x}} &&\\ &=\frac{1}{\innerproduct{\vect{x}}{\vect{x}}}\conjugate{\lambda}\innerproduct{\vect{x}}{\vect{x}} &&\\ &=\conjugate{\lambda} &&

If a complex number is equal to its conjugate, then it has a complex part equal to zero, and therefore is a real number.

Notice the appealing symmetry to the justifications given for the steps of this proof. In the center is the ability to pitch a Hermitian matrix from one side of the inner product to the other.

Look back and compare and . In the matrix has only real entries, yet the characteristic polynomial has roots that are complex numbers, and so the matrix has complex eigenvalues. However, in , the matrix has only real entries, but is also symmetric, and hence Hermitian. So by , we were guaranteed eigenvalues that are real numbers.

In many physical problems, a matrix of interest will be real and symmetric, or Hermitian. Then if the eigenvalues are to represent physical quantities of interest, guarantees that these values will not be complex numbers.

The eigenvectors of a Hermitian matrix also enjoy a pleasing property that we will exploit later.

Notice again how the key step in this proof is the fundamental property of a Hermitian matrix () the ability to swap $A$ across the two arguments of the inner product. We will build on these results and continue to see some more interesting properties in .

Condensed: the full section is in Beezer, A First Course in Linear Algebra.

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Запитання людей

What does a determinant mean geometrically?

It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.

What is an eigenvector?

A direction the matrix does not turn: it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.

Why is matrix multiplication not commutative?

Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.

Частини цієї сторінки адаптуються від Beezer, A First Course in Linear Algebra (GFDL 1.2). Засуджені і пояснені тут; помилки є нашими.

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