Eigenvalues固有値こゆうち and eigenvectors固有こゆうベクトル

1Introduction

The central idea of this lecture is that an eigenvalue固有値こゆうち and an eigenvector固有こゆうベクトル describe a direction that a linear transformation線型変換せんけいへんかん does not turn, together with the scale factor on that direction. This is the starting point for expressing the essential structure構造こうぞう of a transformation in the simplest possible basis基底きてい.

If eigenvalues are treated only as roots of a characteristic equation, the reasons behind $\det (A-\lambda I)=0$\det(A-\lambda I)=0 and the distinction between algebraic and geometric multiplicity become unclear. Starting from the condition that a transformation preserves a direction makes the determinant condition, multiplicities, diagonalization, and applications follow naturally.

2Terms and definitions

2.1Eigenvalue固有値こゆうち and eigenvector固有こゆうベクトル

Let $A$A be an $n$n by $n$n square matrix正方行列せいほうぎょうれつ. A scalar $\lambda$\lambda is an eigenvalue固有値こゆうち of $A$A if there exists a nonzero vector $v$v such that

The nonzero vector $v$v is an eigenvector固有こゆうベクトル corresponding to $\lambda$\lambda.

2.2Characteristic polynomial特性多項式とくせいたこうしき

The polynomial

is called the characteristic polynomial特性多項式とくせいたこうしき of $A$A. It has degree $n$n, and the eigenvalues固有値こゆうち are precisely the roots of $p\left(\lambda \right)=0$p(\lambda)=0 over the field being considered.

3Plan

1. Translate $Av=\lambda v$Av=\lambda v into $\det (A-\lambda I)=0$\det(A-\lambda I)=0.
2. Explain how algebraic multiplicity and geometric multiplicity control diagonalizability.
3. Connect the Cayley-Hamilton theorem with powers of matrices and systems of differential equations.

The detailed solving workflow for the eigenvalue problem固有値問題こゆうちもんだい is treated in the next lecture.

4Intuitive explanation

A general linear transformation線型変換せんけいへんかん changes the directions of most vectors. A shear, rotation, or non-uniform stretch usually sends an arrow to a different direction.

An eigenvector固有こゆうベクトル is a direction that remains on the same line after the transformation. On that line the transformation looks like pure stretching or shrinking, not rotation or shearing. The scale factor is the eigenvalue固有値こゆうち.

Thus eigenvalues and eigenvectors are tools for finding the places where a complicated transformation can be read as "one scale factor per direction." This viewpoint leads directly to diagonalization対角化たいかくか.

5Precise explanation

5.11. From the definition to an equation

The definition of an eigenvector固有こゆうベクトル includes the nonzero condition:

This is equivalent to

At this point no determinant has been used; we have only moved all terms to one side. The determinant appears when we ask when the homogeneous system

has a nontrivial solution. If $A-\lambda I$A-\lambda I is invertible, the only solution is $x=0$x=0. Therefore a nontrivial solution exists exactly when

The two logical steps are

and

5.22. Properties of the characteristic polynomial特性多項式とくせいたこうしき

For an $n\times n$n\times n matrix行列ぎょうれつ $A$A,

When the characteristic polynomial splits into linear factors and the roots are counted with algebraic multiplicity,

For real matrices, real eigenvalues may not be enough. A rotation matrix can have complex eigenvalues, so one often extends scalars to the complex numbers when discussing all roots of the characteristic polynomial.

An important consequence is

Equivalently, $A$A is singular exactly when $0$0 belongs to its spectrum.

5.33. Algebraic and geometric multiplicity

The algebraic multiplicity代数的重複度だいすうてきちょうふくど of an eigenvalue固有値こゆうち ${\lambda }_0$\lambda_0 is its multiplicity as a root of the characteristic polynomial特性多項式とくせいたこうしき.

The geometric multiplicity幾何的重複度きかてきちょうふくど is the dimension of the eigenspace固有空間こゆうくうかん

For an $n\times n$n\times n matrix,

For every eigenvalue,

If the characteristic polynomial splits over the field under consideration, then $A$A is diagonalizable exactly when every eigenvalue has equal algebraic and geometric multiplicity:

Here $$\mathrm{Spec}\left(A\right)$$\operatorname{Spec}(A) is the set of all eigenvalues固有値こゆうち of $$A$$A. A matrix $$A$$A is diagonalizable対角化可能たいかくかかのう when there is a basis of eigenvectors, equivalently when $$A=PD{P}^{-1}$$A=PDP^{-1} for some invertible matrix $$P$$P and diagonal matrix $$D$$D. The proof of this criterion is treated in the next diagonalization lecture.

Example of failure: for

$\lambda =1$\lambda=1 has AM $=2$=2 but GM $=1$=1, so $A$A is not diagonalizable.

5.44. Concrete example

Let

The characteristic polynomial is

so the eigenvalues are $\lambda =2,3$\lambda=2,3. For $\lambda =3$\lambda=3,

gives $v\propto \left(\begin{array}{c}1\\ 0\end{array}\right)$v\propto \begin{pmatrix}1\\0\end{pmatrix}. For $\lambda =2$\lambda=2,

gives $v\propto \left(\begin{array}{c}1\\ -1\end{array}\right)$v\propto \begin{pmatrix}1\\-1\end{pmatrix}.

5.55. The Cayley-Hamilton theorem

A matrix satisfies its own characteristic polynomial. If

then

where $0$0 denotes the zero matrix. If one uses $p\left(\lambda \right)=\det (A-\lambda I)$p(\lambda)=\det(A-\lambda I) instead, the leading coefficient differs by a sign, but the theorem still says that the matrix annihilates itself through its characteristic polynomial.

For

the characteristic polynomial is ${\lambda }^2-(a+d)\lambda +(ad-bc)$\lambda^2-(a+d)\lambda+(ad-bc). Hence

When $\det A\ne 0$\det A\ne0, this relation can be rearranged to express $A^{-1}$A^{-1} using powers of $A$A.

5.66. Application to systems of differential equations

For

the solution is

If $A=PD{P}^{-1}$A=PDP^{-1} is diagonalizable, then

The real parts of the eigenvalues固有値こゆうち determine stability: negative real parts indicate decay, while positive real parts indicate growth.

6Criteria

• To find eigenvalues, solve $\det (A-\lambda I)=0$\det(A-\lambda I)=0 and then solve $(A-\lambda I)v=0$(A-\lambda I)v=0 for nonzero $v$v.
• If all eigenvalues are distinct, $A$A is diagonalizable; this is sufficient but not necessary.
• Use $\mathrm{t}\mathrm{r}A$\mathrm{tr}\,A and $\det A$\det A as checks: they match the sum and product of eigenvalues counted with multiplicity.
• For $x^{\prime }=A\mathit{x}$\boldsymbol{x}'=A\boldsymbol{x}, stability is controlled by the signs of the real parts of eigenvalues.

7Scope and limitations

Even real matrices can have complex eigenvalues固有値こゆうち. The diagonalization problem extends to the spectral theorem for symmetric or normal matrices and to Jordan normal form in the general case. In operator theory, eigenfunctions of differential operators, such as $e^{\lambda x}$e^{\lambda x}, are infinite-dimensional versions of the same idea.

8Final forms

9In one sentence

An eigenvalue固有値こゆうち is the scale factor along a direction that the transformation does not turn, the characteristic polynomial特性多項式とくせいたこうしき compresses that information into a degree-$n$n polynomial, and diagonalization対角化たいかくか uses eigenvectors as the coordinate axes of the transformation.

10Theorem and proof: eigenvectors for distinct eigenvalues are linearly independent

Let ${\lambda }_1,\dots ,{\lambda }_k$\lambda_1,\ldots,\lambda_k be distinct eigenvalues of $A$A, and let $v_i\ne 0$v_i\ne0 be an eigenvector for ${\lambda }_i$\lambda_i. Then $v_1,\dots ,v_k$v_1,\ldots,v_k are linearly independent線型独立せんけいどくりつ.

We prove this by induction on $k$k. The case $k=1$k=1 is immediate because one nonzero vector is linearly independent.

Assume

Applying $A$A gives

Subtract ${\lambda }_k$\lambda_k times the first equation:

By the induction hypothesis, $v_1,\dots ,v_{k-1}$v_1,\ldots,v_{k-1} are linearly independent, so $a_i({\lambda }_i-{\lambda }_k)=0$a_i(\lambda_i-\lambda_k)=0 for $i<k$i<k. Since ${\lambda }_i\ne {\lambda }_k$\lambda_i\ne\lambda_k, we get $a_i=0$a_i=0 for $i<k$i<k. The original equation then gives $a_k{v}_k=0$a_kv_k=0, and $v_k\ne 0$v_k\ne0 implies $a_k=0$a_k=0.

All coefficients are zero, so the vectors are linearly independent.

11Exercise link

12Related links