3.7.1 Quadratic forms and positive definiteness

What a quadratic form is

Lesson 3.2.3 multiplied a row, a square matrix and a column to get one number. Use the same vector on both sides and the result is a quadratic form. For a symmetric $n \times n$ matrix $A$ and a vector $\mathbf{x}$ in $\mathbb{R}^n$ , it is

q(\mathbf{x}) = \mathbf{x}^\top A\,\mathbf{x}

The shapes are $(1 \times n)(n \times n)(n \times 1)$ , so $q(\mathbf{x})$ is a single number. It is called quadratic because every term multiplies two components of $\mathbf{x}$ .

Write it out for $A = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$ and $\mathbf{x} = (x_1, x_2)$ . First $A\mathbf{x} = (ax_1 + bx_2,\ bx_1 + cx_2)$ , then the dot product with $\mathbf{x}$ :

\begin{aligned} q(\mathbf{x}) &= x_1(ax_1 + bx_2) + x_2(bx_1 + cx_2) \\ &= ax_1^2 + 2bx_1x_2 + cx_2^2 \end{aligned}

The diagonal entries multiply the squares. The off-diagonal entry $b$ appears twice, once in each corner, so the cross term is $2bx_1x_2$ .

Rule: Quadratic form

\begin{gathered} q(\mathbf{x}) = \mathbf{x}^\top A\,\mathbf{x} = \textstyle\sum_i\sum_j a_{ij}\,x_i\,x_j \\[4pt] n = 2:\ ax_1^2 + 2bx_1x_2 + cx_2^2 \end{gathered}

A symmetric matrix $A$ turns each vector $\mathbf{x}$ into one number. The diagonal of $A$ gives the squared terms; each off-diagonal pair $a_{ij} = a_{ji}$ gives the cross term $2a_{ij}x_ix_j$ .

Take lesson 3.4.1's matrix $A = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}$ , so $q(\mathbf{x}) = 3x_1^2 + 2x_1x_2 + 2x_2^2$ . At $\mathbf{x} = (1, -1)$ , $A\mathbf{x} = (2, -1)$ and $\mathbf{x}^\top A\mathbf{x} = 2 + 1 = 3$ ; the formula gives $3 - 2 + 2 = 3$ too.

$\mathbf{x}$	$(1, 0)$	$(0, 1)$	$(1, 1)$	$(1, -1)$	$(2, -1)$	$(-1, 3)$
$q(\mathbf{x})$	3	2	7	3	10	15

Three facts follow from the formula:

Only the symmetric part counts. $B = \begin{pmatrix} 3 & 2 \\ 0 & 2 \end{pmatrix}$ is not symmetric, but its cross term is $(2 + 0)x_1x_2$ , the same as $A$ 's. Any square $B$ gives the form of $\frac{1}{2}(B + B^\top)$ , which is why quadratic forms use symmetric matrices.
Scaling $\mathbf{x}$ by $c$ scales $q$ by $c^2$ . $q(2, -2) = 12 = 4 \times q(1, -1)$ , and $q(-\mathbf{x}) = q(\mathbf{x})$ . So the sign of $q$ depends only on the direction of $\mathbf{x}$ .
$q(\mathbf{0}) = 0$ for every $A$ .

With three or more components the double sum has a term for every pair $(i, j)$ : case study 3.2.5 met it as the portfolio variance $\mathbf{w}^\top\Sigma\mathbf{w}$ .

3x_1^2 + 2x_1x_2 + 2x_2^2

import numpy as np def q(A, x): return x @ A @ x A = np.array([[3, 1], [1, 2]]) for x in [(1, 0), (0, 1), (1, 1), (1, -1), (2, -1), (-1, 3)]: x = np.array(x) print(x, q(A, x), 3 * x[0]**2 + 2 * x[0] * x[1] + 2 * x[1]**2) B = np.array([[3, 2], [0, 2]]) print(q(B, np.array([2, -1])), q(A, np.array([2, -1]))) print((B + B.T) / 2) print(q(A, np.array([2, -2])), 4 * q(A, np.array([1, -1])), q(A, np.array([-1, 1])))

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Four kinds of symmetric matrix

The question that matters is the sign of $q(\mathbf{x})$ over all $\mathbf{x} \neq \mathbf{0}$ . Four matrices show the possibilities:

\begin{gathered} A = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}, \quad J = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \\[6pt] N = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}, \quad -S = \begin{pmatrix} -2 & -1 \\ -1 & -2 \end{pmatrix} \end{gathered}

Here $S = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ is lesson 3.2.3's symmetric matrix, and $J$ has every entry 1.

$\mathbf{x}$	$(1, 0)$	$(0, 1)$	$(1, 1)$	$(1, -1)$	$(2, -1)$
$A$ : $3x_1^2 + 2x_1x_2 + 2x_2^2$	3	2	7	3	10
$J$ : $(x_1 + x_2)^2$	1	1	4	0	1
$N$ : $x_1^2 + 4x_1x_2 + x_2^2$	1	1	6	$-2$	$-3$
$-S$ : $-2x_1^2 - 2x_1x_2 - 2x_2^2$	$-2$	$-2$	$-6$	$-2$	$-6$

$J$ 's form is $x_1^2 + 2x_1x_2 + x_2^2 = (x_1 + x_2)^2$ , a square: never negative, and 0 along the whole line $x_2 = -x_1$ . $N$ is positive at $(1, 1)$ and negative at $(1, -1)$ .

Rule: Definite, semidefinite, indefinite

\begin{gathered} \text{positive definite:} \\ \mathbf{x}^\top A\mathbf{x} > 0 \text{ for every } \mathbf{x} \neq \mathbf{0} \\[6pt] \text{positive semidefinite:} \\ \mathbf{x}^\top A\mathbf{x} \ge 0 \text{ for every } \mathbf{x} \end{gathered}

Negative definite and negative semidefinite are the same with $< 0$ and $\le 0$ . A matrix whose form takes both signs is indefinite.

Every positive definite matrix is also positive semidefinite. When this module calls a matrix semidefinite, it means that $q$ is also 0 at some $\mathbf{x} \neq \mathbf{0}$ , as for $J$ .

So $A$ looks positive definite, $J$ positive semidefinite, $N$ indefinite and $-S$ negative definite. A table can only rule things out: one negative value proves that $N$ is not semidefinite. No finite table proves that $A$ is positive everywhere; the next two sections give tests that do.

One quick check comes free. $q(\mathbf{e}_i) = a_{ii}$ , the $i$ -th diagonal entry, so a positive definite matrix has every diagonal entry positive. That is needed but not enough: $N$ has 1 and 1 on its diagonal and is indefinite.

q

import numpy as np examples = { "A": [[3, 1], [1, 2]], "J": [[1, 1], [1, 1]], "N": [[1, 2], [2, 1]], "-S": [[-2, -1], [-1, -2]], } points = [(1, 0), (0, 1), (1, 1), (1, -1), (2, -1)] for name, M in examples.items(): M = np.array(M) print(name, [int(np.array(x) @ M @ np.array(x)) for x in points])

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The eigenvalue test

Lesson 3.6.3 showed that a symmetric matrix has real eigenvalues $\lambda_1, \ldots, \lambda_n$ and perpendicular unit eigenvectors $\mathbf{q}_1, \ldots, \mathbf{q}_n$ , with $A = Q\Lambda Q^\top$ . Here $Q$ has the eigenvectors as its columns and $\Lambda$ is the diagonal matrix of eigenvalues.

Measure $\mathbf{x}$ along the eigenvectors instead of the axes: $y_i = \mathbf{q}_i \cdot \mathbf{x}$ , or $\mathbf{y} = Q^\top\mathbf{x}$ . Then $\mathbf{x}^\top A\mathbf{x} = \mathbf{x}^\top Q\Lambda Q^\top\mathbf{x} = \mathbf{y}^\top\Lambda\mathbf{y}$ , and the cross terms disappear:

\mathbf{x}^\top A\,\mathbf{x} = \lambda_1 y_1^2 + \cdots + \lambda_n y_n^2

Take $S = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ . Its eigenvalues are 3, along $\mathbf{q}_1 = \frac{1}{\sqrt{2}}(1, 1)$ , and 1, along $\mathbf{q}_2 = \frac{1}{\sqrt{2}}(1, -1)$ . So $y_1 = \frac{x_1 + x_2}{\sqrt{2}}$ , $y_2 = \frac{x_1 - x_2}{\sqrt{2}}$ , and

\begin{aligned} q(\mathbf{x}) &= 2x_1^2 + 2x_1x_2 + 2x_2^2 \\ &= 3y_1^2 + y_2^2 \\ &= 1.5(x_1 + x_2)^2 \\ &\quad + 0.5(x_1 - x_2)^2 \end{aligned}

Expand the right side to check: $1.5 + 0.5 = 2$ for each square, and $3 - 1 = 2$ for the cross term. At $(2, -1)$ both sides give 6.

A sum $\lambda_1y_1^2 + \cdots + \lambda_ny_n^2$ is positive for every $\mathbf{y} \neq \mathbf{0}$ exactly when every $\lambda_i > 0$ . And $\mathbf{y} = \mathbf{0}$ only when $\mathbf{x} = \mathbf{0}$ , because $Q$ only turns or reflects the plane.

Rule: The eigenvalue test

\begin{gathered} \text{every } \lambda_i > 0 \\ \iff \text{positive definite} \\[6pt] \text{every } \lambda_i \ge 0 \\ \iff \text{positive semidefinite} \end{gathered}

All $\lambda_i < 0$ means negative definite. Eigenvalues of both signs mean indefinite: $q$ is positive along an eigenvector with $\lambda_i > 0$ and negative along one with $\lambda_i < 0$ .

matrix	eigenvalues	kind
$A$	1.382, 3.618	positive definite
$J$	0, 2	positive semidefinite
$N$	$-1$ , 3	indefinite
$-S$	$-3$ , $-1$	negative definite

The eigenvectors show where the signs are. $N$ has $\lambda = -1$ along $\frac{1}{\sqrt{2}}(1, -1)$ , and there $q = -1$ . $J$ has $\lambda = 0$ along $(1, -1)$ , the line where $(x_1 + x_2)^2 = 0$ .

For a unit vector, $y_1^2 + \cdots + y_n^2 = 1$ , so $q$ is an average of the eigenvalues with weights $y_i^2$ . It therefore runs from the smallest eigenvalue to the largest: from 1 to 3 for $S$ , from $-1$ to 3 for $N$ .

For $2 \times 2$ matrices lesson 3.6.1 gives a shortcut: $\lambda_1\lambda_2 = \det A$ and $\lambda_1 + \lambda_2$ is the trace, the sum of the diagonal. $\det A > 0$ means the two eigenvalues have the same sign, and the trace says which. $\det N = -3 < 0$ means opposite signs: indefinite.

q(\mathbf{x})

import numpy as np S = np.array([[2, 1], [1, 2]]) lam, Q = np.linalg.eigh(S) Q = Q * np.sign(Q[0]) # make each eigenvector's first entry positive print(np.round(lam, 4), np.round(Q, 4)) for x in [(1, 0), (1, 1), (2, -1), (0.3, 0.7)]: x = np.array(x) y = Q.T @ x # coordinates along the eigenvectors print(x, round(x @ S @ x, 6), round(np.sum(lam * y**2), 6), round(1.5 * (x[0] + x[1])**2 + 0.5 * (x[0] - x[1])**2, 6))

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Testing on a computer

np.linalg.eigvalsh returns just the eigenvalues of a symmetric matrix, smallest first: it is eigh without the eigenvectors. Testing "every $\lambda \ge 0$ " needs care. A zero eigenvalue often comes out as a tiny number such as $10^{-16}$ , positive or negative depending on the computer.

So compare with a small tolerance, scaled to the size of the matrix, as lesson 3.4.1 did for determinants. classify below treats any eigenvalue within $10^{-10}$ times the largest one as zero.

N

import numpy as np def classify(M, tol=1e-10): lam = np.linalg.eigvalsh(M) small = tol * np.abs(lam).max() if lam.min() > small: return "positive definite" if lam.min() >= -small: return "positive semidefinite" if lam.max() < -small: return "negative definite" if lam.max() <= small: return "negative semidefinite" return "indefinite" for name, M in {"A": [[3, 1], [1, 2]], "J": [[1, 1], [1, 1]], "N": [[1, 2], [2, 1]], "-S": [[-2, -1], [-1, -2]]}.items(): M = np.array(M, dtype=float) print(name, np.round(np.linalg.eigvalsh(M), 4) + 0.0, round(np.trace(M), 6), round(np.linalg.det(M), 6) + 0.0, classify(M)) N = np.array([[1, 2], [2, 1]]) u = np.array([1, -1]) / np.sqrt(2) print(round(u @ N @ u, 6))

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The test by leading minors

Eigenvalues are hard to find by hand beyond $2 \times 2$ . Determinants give a test you can do on paper.

The leading minors of $A$ are the determinants of its top-left corners (lesson 3.4.1). They run from the $1 \times 1$ corner $a_{11}$ , through the $2 \times 2$ corner, up to $\det A$ itself.

Rule: The leading-minor test

\begin{gathered} 2 \times 2:\ a > 0 \text{ and } ac - b^2 > 0 \\[4pt] \text{every leading minor} > 0 \end{gathered}

A symmetric matrix is positive definite exactly when all its leading minors are positive. It is negative definite exactly when they alternate in sign, starting negative: $-, +, -, \ldots$

Why it works for $2 \times 2$ : complete the square. For $a \neq 0$ ,

\begin{aligned} & ax_1^2 + 2bx_1x_2 + cx_2^2 \\ &= a\Big(x_1 + \frac{b}{a}x_2\Big)^2 \\ &\quad + \frac{ac - b^2}{a}\,x_2^2 \end{aligned}

If $a > 0$ and $ac - b^2 > 0$ , both coefficients are positive. The sum of the two squares is then 0 only if $x_2 = 0$ and then $x_1 = 0$ .

For $A$ : $3\big(x_1 + \frac{1}{3}x_2\big)^2 + \frac{5}{3}x_2^2$ , which at $(1, -1)$ is $\frac{4}{3} + \frac{5}{3} = 3$ , as in the table. Lesson 3.7.2 turns this completed square into a factorisation of $A$ .

Two $3 \times 3$ examples. Lesson 3.2.3's

K = \begin{pmatrix} 4 & -1 & 0 \\ -1 & 3 & 2 \\ 0 & 2 & 5 \end{pmatrix}

has leading minors 4, $4 \cdot 3 - (-1)(-1) = 11$ and $\det K = 39$ , all positive: positive definite.

T = \begin{pmatrix} 2 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 1 \end{pmatrix}

passes the first two, 2 and 3, but $\det T = -3$ . So $T$ is not positive definite, and indeed $q(1, 0, -1) = 2 - 4 + 1 = -1$ . Its eigenvalues are $-0.5758$ , 1.1873 and 4.3885: indefinite.

Where the test breaks: semidefinite matrices. Leading minors that are $\ge 0$ do not prove semidefinite. $E = \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}$ has leading minors 0 and 0, yet $q(\mathbf{x}) = -x_2^2$ , which is negative at $(0, 1)$ . For semidefinite questions, use the eigenvalues.

q(1, 0, -1)

import numpy as np def leading_minors(M): M = np.array(M, dtype=float) return [round(float(np.linalg.det(M[:k, :k])), 6) + 0.0 for k in range(1, len(M) + 1)] examples = { "A": [[3, 1], [1, 2]], "K": [[4, -1, 0], [-1, 3, 2], [0, 2, 5]], "T": [[2, 1, 2], [1, 2, 1], [2, 1, 1]], "-S": [[-2, -1], [-1, -2]], "E": [[0, 0], [0, -1]], } for name, M in examples.items(): print(name, leading_minors(M), np.round(np.linalg.eigvalsh(np.array(M, dtype=float)), 4) + 0.0) T = np.array([[2, 1, 2], [1, 2, 1], [2, 1, 1]]) x = np.array([1, 0, -1]) print(x @ T @ x) x1, x2 = 1, -1 print(round(3 * (x1 + x2 / 3)**2 + (5 / 3) * x2**2, 10))

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Level curves: ellipses and hyperbolas

A level curve of $q$ is the set of points where $q(\mathbf{x})$ takes one fixed value, like a contour line on a map. Module 4.1 draws them for any function of two variables; for a quadratic form, the eigenvalues decide their shape.

Along the eigenvector $\mathbf{q}_i$ , $q = \lambda_i y_i^2$ , so $q = 1$ where $y_i = 1/\sqrt{\lambda_i}$ . A large eigenvalue makes $q$ climb fast, so the curve is close to the origin in that direction.

Level curves of $S$ , $J$ and $N$ . Positive definite gives ellipses, here with semi-axes $1/\sqrt{3} \approx 0.58$ and 1 for $q = 1$ . A zero eigenvalue gives parallel lines, and eigenvalues of both signs give hyperbolas, with $q < 0$ in the shaded sectors.

Positive definite: each level curve $q = k > 0$ is an ellipse, with its axes along the eigenvectors and semi-axes $\sqrt{k/\lambda_i}$ . Seen as a height over the plane, $q$ is a bowl with its bottom at the origin.
Semidefinite, one zero eigenvalue: $q$ does not grow along that eigenvector, so the curves are parallel lines. The surface is a trough.
Indefinite: the curves are hyperbolas. $N$ has the same eigenvectors as $S$ , and $q = 3y_1^2 - y_2^2$ . It is 0 on the two lines $y_2 = \pm\sqrt{3}\,y_1$ , which separate the positive part from the negative part. The surface is a saddle.

Rule: Level curves of a quadratic form

\begin{gathered} q = 1 \text{ meets the axis of } \mathbf{q}_i \\[4pt] \text{at distance } 1/\sqrt{\lambda_i} \text{ if } \lambda_i > 0 \end{gathered}

All $\lambda_i > 0$ : ellipses. One $\lambda_i = 0$ , the other positive: parallel lines. Eigenvalues of both signs: hyperbolas.

S

import numpy as np S = np.array([[2, 1], [1, 2]]) N = np.array([[1, 2], [2, 1]]) q1 = np.array([1, 1]) / np.sqrt(2) q2 = np.array([1, -1]) / np.sqrt(2) for p in [q1 / np.sqrt(3), q2, 2 * q1 / np.sqrt(3), 2 * q2]: print("S", np.round(p, 4), round(p @ S @ p, 6)) for p in [q1 / np.sqrt(3), q2, q1 + np.sqrt(3) * q2]: print("N", np.round(p, 4), round(p @ N @ p, 6) + 0.0)

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Matrices that are always semidefinite

For any $m \times n$ matrix $M$ , lesson 3.2.3 showed that $M^\top M$ is symmetric. Its form is a squared length:

\begin{aligned} \mathbf{x}^\top M^\top M\,\mathbf{x} &= (M\mathbf{x})^\top(M\mathbf{x}) \\ &= \lVert M\mathbf{x} \rVert^2 \ge 0 \end{aligned}

So $M^\top M$ is always positive semidefinite. It is positive definite exactly when $M\mathbf{x} = \mathbf{0}$ only for $\mathbf{x} = \mathbf{0}$ : when the columns of $M$ are independent, with rank $n$ (lesson 3.3.2).

Lesson 3.2.3's $M = \begin{pmatrix} 1 & 2 & 0 \\ 3 & -1 & 4 \end{pmatrix}$ is $2 \times 3$ , so its rank is at most 2 and $M^\top M$ is only semidefinite:

M^\top M = \begin{pmatrix} 10 & -1 & 12 \\ -1 & 5 & -4 \\ 12 & -4 & 16 \end{pmatrix}

At $\mathbf{x} = (1, 1, -1)$ , $M\mathbf{x} = (3, -2)$ and the form is $9 + 4 = 13$ . At $\mathbf{x} = (-8, 4, 7)$ , $M\mathbf{x} = \mathbf{0}$ and the form is 0. The eigenvalues are 0, 4.9525 and 26.0475; the smaller matrix $MM^\top$ has eigenvalues 4.9525 and 26.0475, both positive.

Covariance matrices. In case study 3.2.5, $\mathbf{w}^\top\Sigma\mathbf{w}$ is the variance of a portfolio, and a variance is never negative. So a covariance matrix must be positive semidefinite.

A sample covariance matrix is built as $M^\top M$ divided by $n - 1$ , from the centred returns, so it always is. The three assets' $\Sigma$ has eigenvalues 0.011741, 0.022502 and 0.075058: positive definite. Case study 3.7.4 uses these tests to catch correlation matrices that cannot be real.

M^\top M

import numpy as np M = np.array([[1, 2, 0], [3, -1, 4]]) G = M.T @ M print(G) x = np.array([1, 1, -1]) print(M @ x, x @ G @ x) z = np.array([-8, 4, 7]) print(M @ z, z @ G @ z) lam = np.linalg.eigvalsh(G) print(np.round(lam, 4) + 0.0, lam.min() >= -1e-12, np.linalg.matrix_rank(G)) print(np.round(np.linalg.eigvalsh(M @ M.T), 4)) sigma = np.array([0.12, 0.18, 0.25]) R = np.array([[1.0, 0.3, 0.1], [0.3, 1.0, 0.5], [0.1, 0.5, 1.0]]) Sigma = np.diag(sigma) @ R @ np.diag(sigma) print(np.round(np.linalg.eigvalsh(Sigma), 6))

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Lesson 4.3.2 applies the same tests to the matrix of second derivatives of a function, to tell a minimum from a maximum or a saddle.

Common mistakes

The cross term is $2b$ . In $3x_1^2 + 4x_1x_2 + 2x_2^2$ the off-diagonal entry is 2, not 4: $A = \begin{pmatrix} 3 & 2 \\ 2 & 2 \end{pmatrix}$ .
Positive entries do not make a matrix positive definite. $N = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ is indefinite; $\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$ , with negative entries, is positive definite (eigenvalues 1 and 3).
A positive diagonal is needed, not enough. It rules out matrices with a diagonal entry $\le 0$ , nothing more.
A positive determinant is not enough. $-S$ has $\det = 3$ and is negative definite. Check every leading minor, or the trace too for $2 \times 2$ .
Leading minors $\ge 0$ do not prove semidefinite. $E = \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}$ passes and is not. Use the eigenvalues.
Use the symmetric matrix. $B = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}$ has eigenvalues 1 and 1, yet its form $x_1^2 + 4x_1x_2 + x_2^2$ is $-2$ at $(1, -1)$ . Test $\frac{1}{2}(B + B^\top)$ , whose eigenvalues are $-1$ and 3.
Allow for rounding. A zero eigenvalue prints as a tiny number of either sign. Compare with a tolerance, never with exactly 0.

Quadratic forms and positive definiteness

0 of 12 correct

1.
For $A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}$ , find $\mathbf{x}^\top A\mathbf{x}$ at $\mathbf{x} = (1, 2)$
2.
For $A = \begin{pmatrix} 4 & -3 \\ -3 & 1 \end{pmatrix}$ , what is the coefficient of $x_1x_2$ in $\mathbf{x}^\top A\mathbf{x}$ ?
3.
Write $5x_1^2 + 8x_1x_2 + 2x_2^2$ as $\mathbf{x}^\top A\mathbf{x}$ with $A$ symmetric. What is $a_{12}$ ?
4.
A symmetric matrix has eigenvalues 4 and $-0.5$ . Is it positive semidefinite? Type 1 for yes, 0 for no
5.
In eigenvector coordinates a form is $3y_1^2 + y_2^2$ . Find its value at $\mathbf{y} = (1, 2)$
6.
A symmetric matrix has eigenvalues 2 and 7. What is the smallest value of $\mathbf{x}^\top A\mathbf{x}$ over unit vectors $\mathbf{x}$ ?
7.
Find the second leading minor of $\begin{pmatrix} 1 & 3 \\ 3 & 4 \end{pmatrix}$
8.
Find the third leading minor, $\det A$ , of $\begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}$
9.
For which positive $k$ is $\begin{pmatrix} 1 & k \\ k & 9 \end{pmatrix}$ positive semidefinite but not positive definite?
10.
A positive definite matrix has eigenvalue 9 along $\mathbf{q}_1$ . How far from the origin does the ellipse $q = 1$ cross the line of $\mathbf{q}_1$ ?
11.
For $M = \begin{pmatrix} 1 & 2 \\ 0 & 1 \\ 1 & 0 \end{pmatrix}$ and $\mathbf{x} = (1, -1)$ , find $\mathbf{x}^\top M^\top M\mathbf{x}$
12.
$E = \begin{pmatrix} 0 & 0 \\ 0 & -2 \end{pmatrix}$ has leading minors 0 and 0. Find $\mathbf{x}^\top E\mathbf{x}$ at $\mathbf{x} = (0, 1)$

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

q(\mathbf{x}) = \mathbf{x}^\top A\mathbf{x} = \sum_i\sum_j a_{ij}x_ix_j

Quadratic forms and positive definiteness

What a quadratic form is

Rule: Quadratic form

Four kinds of symmetric matrix

Rule: Definite, semidefinite, indefinite

The eigenvalue test

Rule: The eigenvalue test

Testing on a computer

The test by leading minors

Rule: The leading-minor test

Level curves: ellipses and hyperbolas

Rule: Level curves of a quadratic form

Matrices that are always semidefinite

Common mistakes

Quadratic forms and positive definiteness

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