Eigenvalues固有値こゆうち, diagonalization対角化たいかくか, and further topics - Basic Exercises

1Exercise plan

An eigenvector固有こゆうベクトル is a direction whose orientation does not change under a linear transformation線型変換せんけいへんかん. Diagonalization対角化たいかくか decomposes the space into such directions and simplifies the transformation to scaling along each direction.

Quadratic forms二次形式にじけいしき, minimal polynomials最小多項式さいしょうたこうしき, Jordan form, SVD, and the pseudoinverse擬似逆行列ぎじぎゃくぎょうれつ are separated into related exercises.

For the eigenvalue problem固有値問題こゆうちもんだい, first solve $\det (A-\lambda I)=0$\det(A-\lambda I)=0, then compute $\ker (A-\lambda I)$\ker(A-\lambda I). For companion matrices, check that polynomial roots become eigenvalues.

2Problem 1

Find the eigenvalues固有値こゆうち and eigenvectors固有こゆうベクトル of

and state the geometric meaning.

2.1Answer example

The eigenvalues are $2$2 and $3$3. Eigenvectors for $\lambda =2$\lambda=2 are nonzero multiples of $e_1$e_1, and eigenvectors for $\lambda =3$\lambda=3 are nonzero multiples of $e_2$e_2.

2.2Explanation

$A$A stretches the $x$x-axis direction by 2 and the $y$y-axis direction by 3. The axis directions do not change, so the standard basis標準基底ひょうじゅんきてい vectors are eigenvectors. The zero vector零れいベクトル is not an eigenvector because it has no direction.

3Problem 2

Determine whether

is diagonalizable対角化可能たいかくかかのう.

3.1Answer example

The only eigenvalue is $\lambda =1$\lambda=1. Since

its eigenspace固有空間こゆうくうかん is $y=0$y=0, the 1-dimensional space spanned by $e_1$e_1. A $2\times 2$2\times2 matrix needs two linearly independent一次独立いちじどくりつ eigenvectors to be diagonalized, so $B$B is not diagonalizable.

3.2Explanation

Finding eigenvalues is not enough. Diagonalization対角化たいかくか requires a basis of eigenvectors filling the whole space.

4Problem 3

Orthogonally diagonalize直交対角化ちょっこうたいかくか

4.1Answer example

The eigenvalues are $3$3 and $1$1. Corresponding unit eigenvectors are

Thus

and

4.2Explanation

Because $C$C is a symmetric matrix対称行列たいしょうぎょうれつ, eigenvectors belonging to distinct eigenvalues can be chosen orthogonal直交ちょっこう. A change of basis基底変換きていへんかん by an orthogonal matrix直交行列ちょっこうぎょうれつ preserves lengths and angles.

5Problem 4

Diagonalize対角化たいかくか

and find $A^n$A^n.

5.1Answer example

For $\lambda =2$\lambda=2, an eigenvector is $(1,0{)}^T$(1,0)^T. For $\lambda =3$\lambda=3, one may take $(1,1{)}^T$(1,1)^T. Therefore

and $A=PD{P}^{-1}$A=PDP^{-1}. Hence

5.2Explanation

Diagonalization converts powers into powers of scale factors along eigendirections固有方向こゆうほうこう.

6Problem 5

Solve

6.1Answer example

The components satisfy $x_{1^{\prime }}\left(t\right)=x_1\left(t\right)$x_1'(t)=x_1(t) and $x_{2^{\prime }}\left(t\right)=-2x_2\left(t\right)$x_2'(t)=-2x_2(t). Therefore

6.2Explanation

Eigenvalues determine growth and decay in time evolution: the first direction grows by $e^t$e^t and the second decays by $e^{-2t}$e^{-2t}.

7Supplementary theorem and worked-example problems

7.1Problem 6

Let $A$A be a real symmetric matrix対称行列たいしょうぎょうれつ. Prove that eigenvectors corresponding to distinct eigenvalues are orthogonal直交ちょっこう.

7.2Answer example

Let $Av=\lambda v$Av=\lambda v and $Aw=\mu w$Aw=\mu w with $\lambda \ne \mu$\lambda\ne\mu. Since $A^T=A$A^T=A,

Thus $(\lambda -\mu )⟨v,w⟩=0$(\lambda-\mu)\langle v,w\rangle=0. Because $\lambda \ne \mu$\lambda\ne\mu, we get $⟨v,w⟩=0$\langle v,w\rangle=0.

7.3Explanation

This is the basic orthogonality mechanism behind the spectral theorem. Symmetry lets the matrix move from one side of the inner product to the other.

7.4Problem 7

Use the spectral theoremスペクトル定理 to write

as a sum of eigenvalue times orthogonal projections.

7.5Answer example

From Problem 3,

with eigenvalues $3$3 and $1$1. Therefore

Explicitly,

Adding gives

7.6Explanation

The spectral theorem can be read as a projection decomposition. The matrix stretches the $q_i$q_i direction by ${\lambda }_i$\lambda_i and adds the orthogonal components back together.

7.7Problem 8

For

compute $A^n{x}_0$A^n x_0 and explain the long-term direction after normalization.

7.8Answer example

Since $A$A is diagonal,

For large $n$n, the first component dominates. After normalization, the direction approaches $e_1$e_1.

7.9Explanation

Powers of a diagonalizable matrix amplify the eigendirection with the largest eigenvalue magnitude, provided the initial vector has a nonzero component in that direction.

8Related exercises

9Related lectures

10Proof exercise: basic theorem on eigenvalues and diagonalization

10.1Problem

Prove that eigenvectors corresponding to distinct eigenvalues固有値こゆうち are linearly independent線型独立せんけいどくりつ. Also explain why diagonalizability is equivalent to having a basis基底きてい of eigenvectors.

10.2Answer

Assume $\sum a_i{v}_i=0$\sum a_iv_i=0 and apply $A$A. Subtract the equation obtained by multiplying the original relation by one eigenvalue. This gives a relation among fewer eigenvectors. By induction, all coefficients are 0.

If $A=PD{P}^{-1}$A=PDP^{-1}, then $AP=PD$AP=PD. Comparing columns shows that the columns of $P$P are eigenvectors. Conversely, if a basis of eigenvectors is placed in the columns of $P$P, then $AP=PD$AP=PD, hence $A=PD{P}^{-1}$A=PDP^{-1}.

10.3Explanation

Diagonalization is exactly the operation of replacing the basis by eigenvectors.

11Related lectures

12Supplementary theorem and worked-example problems

12.1Problem 6

Let $A$A be a real symmetric matrix対称行列たいしょうぎょうれつ. Prove that eigenvectors corresponding to distinct eigenvalues are orthogonal直交ちょっこう.

12.2Answer example

Let $Av=\lambda v$Av=\lambda v and $Aw=\mu w$Aw=\mu w with $\lambda \ne \mu$\lambda\ne\mu. Since $A^T=A$A^T=A,

Thus $(\lambda -\mu )⟨v,w⟩=0$(\lambda-\mu)\langle v,w\rangle=0. Because $\lambda \ne \mu$\lambda\ne\mu, we get $⟨v,w⟩=0$\langle v,w\rangle=0.

12.3Explanation

This is the basic orthogonality mechanism behind the spectral theorem. Symmetry lets the matrix move from one side of the inner product to the other.

12.4Problem 7

Use the spectral theoremスペクトル定理 to write

as a sum of eigenvalue times orthogonal projections.

12.5Answer example

From Problem 3,

with eigenvalues $3$3 and $1$1. Therefore

Explicitly,

Adding gives

12.6Explanation

The spectral theorem can be read as a projection decomposition. The matrix stretches the $q_i$q_i direction by ${\lambda }_i$\lambda_i and adds the orthogonal components back together.

12.7Problem 8

For

compute $A^n{x}_0$A^n x_0 and explain the long-term direction after normalization.

12.8Answer example

Since $A$A is diagonal,

For large $n$n, the first component dominates. After normalization, the direction approaches $e_1$e_1.

12.9Explanation

Powers of a diagonalizable matrix amplify the eigendirection with the largest eigenvalue magnitude, provided the initial vector has a nonzero component in that direction.

12.10Problem 9

Solve the eigenvalue problem固有値問題こゆうちもんだい for

Find the eigenvalues and the corresponding eigenspaces.

12.11Answer example

First,

Thus the eigenvalues are $1$1 and $2$2. For $\lambda =1$\lambda=1,

so

For $\lambda =2$\lambda=2,

so

12.12Explanation

This problem checks the two-stage eigenvalue-problem workflow: find $\lambda$\lambda first, then compute the kernel.

12.13Problem 10

Construct the companion matrix同伴行列どうはんぎょうれつ of

and give one eigenvector for $\lambda =2$\lambda=2.

12.14Answer example

Here $c_2=-6,c_1=11,c_0=-6$c_2=-6,\ c_1=11,\ c_0=-6, so

Since $p\left(2\right)=0$p(2)=0,

is an eigenvector for $\lambda =2$\lambda=2. Indeed,

12.15Explanation

The signs in the last row are the common error point: the entries are $-c_0,-c_1,\dots ,-c_{n-1}$-c_0,-c_1,\ldots,-c_{n-1}.

13Proof exercise: basic theorem on eigenvalues and diagonalization

13.1Problem

Prove that eigenvectors corresponding to distinct eigenvalues固有値こゆうち are linearly independent線型独立せんけいどくりつ. Also explain why diagonalizability is equivalent to having a basis基底きてい of eigenvectors.

13.2Answer

Assume $\sum a_i{v}_i=0$\sum a_iv_i=0 and apply $A$A. Subtract the equation obtained by multiplying the original relation by one eigenvalue. This gives a relation among fewer eigenvectors. By induction, all coefficients are 0.

If $A=PD{P}^{-1}$A=PDP^{-1}, then $AP=PD$AP=PD. Comparing columns shows that the columns of $P$P are eigenvectors. Conversely, if a basis of eigenvectors is placed in the columns of $P$P, then $AP=PD$AP=PD, hence $A=PD{P}^{-1}$A=PDP^{-1}.

13.3Explanation

Diagonalization is exactly the operation of replacing the basis by eigenvectors.