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, and Jordan form - Basic Exercises

date2026-06-06document_iddoc_c93a95d880e00a983edaf7a88b281565description二次形式、正定値性、最小多項式、ジョルダン標準形を、対角化できる場合とできない場合の違いから確認する演習である。prerequisites二次形式と正定値行列 / 最小多項式の基本 / ジョルダン標準形の入口type問題演習content_typeexercisestatusactive

mathlinear-algebraexercisequadratic-formminimal-polynomialjordan-form

data/lecture/math/linear-algebra/quadratic-forms-and-positive-definite-matrices.lecture.n.md data/lecture/math/linear-algebra/minimal-polynomial-basics.lecture.n.md data/lecture/math/linear-algebra/introduction-to-jordan-canonical-form.lecture.n.md

1Exercise plan

A quadratic form二次形式にじけいしき studies signs and stretching along directions. A minimal polynomial最小多項式さいしょうたこうしき and Jordan normal formジョルダン標準形ひょうじゅんけい record non-diagonalizable information that eigenvalues固有値こゆうち alone do not show.

2Problem 1

Determine whether

q (x, y) = 2 x^{2} + 2 x y + 2 y^{2}

is positive definite正定値せいていち.

2.1Answer example

The corresponding symmetric matrix対称行列たいしょうぎょうれつ is

C = (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}) .

Its eigenvalues固有値こゆうち are $3$ and $1$ , both positive. Therefore $q$ is positive definite.

2.2Explanation

A quadratic form can be read along eigendirections固有方向こゆうほうこう. If every eigenvalue coefficient is positive, every nonzero input gives a positive value.

3Problem 2

Write

q (x, y, z) = 2 x^{2} + 2 x y + 3 y^{2} - z^{2}

as a symmetric matrix対称行列たいしょうぎょうれつ, and classify it as positive definite, positive semidefinite非負定値ひふていち, or indefinite不定ふてい.

3.1Answer example

The corresponding matrix is

S = (\begin{matrix} 2 & 1 & 0 \\ 1 & 3 & 0 \\ 0 & 0 & - 1 \end{matrix}) .

Since $q (1, 0, 0) = 2 > 0$ and $q (0, 0, 1) = - 1 < 0$ , the form is indefinite.

3.2Explanation

Positive definiteness requires positivity for all nonzero vectors. If both positive and negative directions exist, the form is indefinite.

4Problem 3

For

J = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}),

explain why the minimal polynomial最小多項式さいしょうたこうしき is $(t - 1)^{2}$ .

4.1Answer example

J - I = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) \neq 0,

but

(J - I)^{2} = (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}) .

Thus the monic polynomial of lowest degree sending $J$ to the zero matrix is $(t - 1)^{2}$ .

4.2Explanation

The eigenvalue alone suggests $(t - 1)$ , but $J - I \neq 0$ shows that a nilpotent off-diagonal part remains. Jordan form records exactly this information.

5Problem 4

Find the minimal polynomials最小多項式さいしょうたこうしき of

D = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), E = (\begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix}), J = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}),

and explain what the characteristic polynomial特性多項式とくせいたこうしき alone cannot distinguish.

5.1Answer example

Since $D - I = 0$ ,

m_{D} (t) = t - 1.

For $E$ , $(E - I) (E - 2 I) = 0$ and neither single factor suffices, so

m_{E} (t) = (t - 1) (t - 2) .

For $J$ , $J - I \neq 0$ but $(J - I)^{2} = 0$ , so

m_{J} (t) = (t - 1)^{2.}

5.2Explanation

Both $D$ and $J$ have characteristic polynomial $(t - 1)^{2}$ . But $D$ is diagonal and $J$ is not diagonalizable対角化可能たいかくかかのう. The minimal polynomial detects this through whether a repeated factor is needed.

6Problem 5

Suppose $N = A - λ I$ acts nilpotently冪零べきれい on a 4-dimensional space and

\dim \ker N = 2, \dim \ker N^{2} = 3, \dim \ker N^{3} = 4.

Find the sizes of the Jordan blocks for $λ$ .

6.1Answer example

$\dim \ker N$ gives the number of Jordan blocks, so there are 2 blocks. The differences are

2, 1, 1.

This means there are 2 blocks of size at least 1, 1 block of size at least 2, and 1 block of size at least 3. Hence the block sizes are 3 and 1.

6.2Explanation

The growth of $\dim \ker N^{k}$ records how long the Jordan chainsジョルダン鎖さ are.

7Problem 6

For

A = (\begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{matrix}),

let $N = A - 2 I$ . Find $\ker N$ , $\ker N^{2}$ , and $\ker N^{3}$ , construct one Jordan chainジョルダン鎖さ, and compute $e^{t A}$ .

7.1Answer example

N = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}),

so $N (x, y, z)^{T} = (y, z, 0)^{T}$ . Therefore

\ker N = span (e_{1}), \ker N^{2} = span (e_{1}, e_{2}), \ker N^{3} = R^{3.}

Since

N e_{1} = 0, N e_{2} = e_{1}, N e_{3} = e_{2},

the vectors $e_{1}, e_{2}, e_{3}$ form a Jordan chain. Also $A = 2 I + N$ and $N^{3} = 0$ , so

e^{t A} = e^{2 t} (I + t N + \frac{t^{2}}{2} N^{2}) = e^{2 t} (\begin{matrix} 1 & t & \frac{t^{2}}{2} \\ 0 & 1 & t \\ 0 & 0 & 1 \end{matrix}) .

7.2Explanation

When eigenvectors are not enough, generalized eigenvectors一般化いっぱんか固有こゆうベクトル fill the missing directions. The nilpotent part冪零成分べきれいせいぶん produces polynomial factors in $e^{t A}$ .

8Related exercises

data/exercise/math/linear-algebra/eigenvalues-diagonalization-and-extensions.exercise.n.md data/exercise/math/linear-algebra/svd-and-pseudoinverses.exercise.n.md

9Related lectures

data/lecture/math/linear-algebra/quadratic-forms-and-positive-definite-matrices.lecture.n.md data/lecture/math/linear-algebra/minimal-polynomial-basics.lecture.n.md data/lecture/math/linear-algebra/introduction-to-jordan-canonical-form.lecture.n.md

10Supplementary theorem-check problems

10.1Problem 7

For

H = (\begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix}),

check positive definiteness正定値性せいていちせい by three methods: eigenvalues, leading principal minors, and completing the square.

10.2Answer example

The eigenvalues are roots of

\det (H - λ I) = (2 - λ) (3 - λ) - 1 = λ^{2} - 5 λ + 5,

so

λ = \frac{5 \pm \sqrt{5}}{2},

both positive. The leading principal minors are $2 > 0$ and $\det H = 5 > 0$ , also showing positive definiteness. Finally,

2 x^{2} + 2 x y + 3 y^{2} = 2 {(x + \frac{y}{2})}^{2} + \frac{5}{2} y^{2},

which is positive for every nonzero $(x, y)$ .

10.3Explanation

For a real symmetric $2 \times 2$ matrix, these three readings agree: positive eigenvalues, positive leading principal minors, and a sum of positive squares.

10.4Problem 8

State and prove, in the real symmetric case, why a quadratic form二次形式にじけいしき $q (x) = x^{T} A x$ is positive definite if and only if all eigenvalues of $A$ are positive.

10.5Answer example

By the spectral theorem, there is an orthogonal matrix $Q$ such that

Q^{T} A Q = diag (λ_{1}, \dots, λ_{n}) .

Put $y = Q^{T} x$ . Since $Q$ is invertible and preserves length, $x \neq 0$ if and only if $y \neq 0$ . Then

q (x) = x^{T} A x = y^{T} diag (λ_{1}, \dots, λ_{n}) y = \sum_{i} λ_{i} y_{i^{2.}}

If every $λ_{i} > 0$ , this sum is positive for every nonzero $y$ . Conversely, if some $λ_{j} [PARSE ERROR: Undefined("Command(\"le\")")] 0$ , taking $y = e_{j}$ gives $q (x) = λ_{j} [PARSE ERROR: Undefined("Command(\"le\")")] 0$ in the corresponding eigen-direction, so the form is not positive definite.

10.6Explanation

The spectral theorem turns a quadratic form into independent squared coordinates. Positive definiteness is therefore exactly positivity of every coefficient in those coordinates.

10.7Problem 9

For a complex matrix whose minimal polynomial splits, explain why being diagonalizable対角化可能たいかくかかのう is equivalent to the minimal polynomial最小多項式さいしょうたこうしき having no repeated linear factor. Check this on $D$ and $J$ from Problem 4.

10.8Answer example

If a matrix is diagonalizable with diagonal entries among $λ_{1}, \dots, λ_{s}$ , then

\prod_{i = 1}^{s} (t - λ_{i})

annihilates the matrix, so the minimal polynomial has only distinct linear factors. Conversely, if the minimal polynomial has no repeated factor, the primary decomposition has no nontrivial nilpotent part on any eigenspace, so the matrix is a direct sum of eigenspaces and is diagonalizable. For the diagonal matrix $D$ , the minimal polynomial has distinct factors. For the Jordan block $J$ , a squared factor is needed, so it is not diagonalizable.

10.9Explanation

Repeated factors in the minimal polynomial detect the nilpotent part of Jordan blocks. No repeated factor means every Jordan block has size 1.