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Linear Algebra
by Samuel S. Watson
Introduction
Vectors
Span
Linear Independence
Vector Spaces
Linear Transformations
Matrices
Dot Products
Transposes
Orthogonality
Eigenanalysis
Singular Value Decomposition
Determinants

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อภิธานศัพท์

Linear dependence lemma
algebraically
Basis equality theorem
Dimension
Dot Product Cosine Formula
Full rank
geometrically
Gram matrix
Linear Combination
List
Matrix product dot formula
Multiple extension principle
Norm
Orthogonal Complement
Positive definite
Product Column Rule
Rank deficient
Span
Spectral Theorem
SVD Figure
Zero Vector

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Linear AlgebraDeterminants

เวลาอ่านหนังสือ: ~25 min

The determinant of a square matrix A is a single number which captures some important information about how the transformation \mathbf{x}\mapsto A\mathbf{x} behaves. In this section, we will develop a geometrically-motivated definition of the determinant.

Exercise
Suppose that R is a region in \mathbb{R}^n and that A is an n \times n matrix. Consider the singular value decomposition A = U \Sigma V'.

  • Let L_1(\mathbf{x}) = V'\mathbf{x}. By what factor does L_1 transform volumes?
  • Let L_2(\mathbf{x}) = \Sigma\mathbf{x}. In terms of the entries of \Sigma, by what factor does L_1 transform volumes?
  • Let L_3(\mathbf{x}) = U\mathbf{x}. By what factor does L_3 transform volumes?


Solution. Since U and V are orthogonal, L_1 and L_3 both preserve volumes. So they multiply volumes by a factor of 1. Since L_2 scales volumes by a factor of \sigma_1 along the first axis, \sigma_2 along the second, and so on, it scales volumes by a factor of \sigma_1 \sigma_2 \ldots \sigma_n.

Volume scale factor

We see from this exercise that a linear transformation T from \mathbb{R}^n to \mathbb{R}^n scales the volume of any n-dimensional region by the same factor: the volume scale factor of T.

Exercise
Find the volume scale factor of the matrix A = \begin{bmatrix} 1 & 0 & 0 \\\ 0 & 0 & 1 \\\ 0 & k & 0 \end{bmatrix} by describing how the matrix transforms a region in \mathbb{R}^3.

Solution. Since A[x,y,z] = [x,z,ky], we see that A stretches (or compresses) regions in \mathbb{R}^3 by a factor k along the y-axis and then reflects across the plane y = z. For example, the unit cube is mapped to a 1 \times 1 \times k box Since such a box has volume k, the volume scale factor of S is k.

Orientation factor

Another geometrically relevant piece of information about T is whether it preserves or reverses orientations. For example, rotations in \mathbb{R}^2 are orientation preserving, while reflections are orientation reversing. Let's define the orientation factor of T to be +1 if T is orientation preserving and -1 if T is orientation reversing.

Definition
We define the determinant of a transformation T to be the product of its orientation factor and its volume scale factor.

We define the determinant of a matrix A to be the determinant of the corresponding linear transformation \mathbf{x}\mapsto A\mathbf{x}.

Exercise
Interpret A = \begin{bmatrix} 0 & -1 \\\ -1 & 0 \end{bmatrix} geometrically and use this interpretation to find \det A, the determinant of A.

Solution. Since A\begin{bmatrix} x \\\ y \end{bmatrix} = \begin{bmatrix} -y \\\ -x \end{bmatrix}, A reflects points in \mathbb{R}^2 across the line y = -x. Therefore, it preserves areas and reverses orientations. So its determinant is -1.

There is relatively simple formula for \det A in terms of the entries of A. For example,

\begin{align*}\left|\begin{array}{cc} a & b \\\ c & d \end{array}\right| = ad - bc\end{align*}

is the determinant of a 2 \times 2 matrix. However this formula is terribly inefficient if A has many entries (it has n! terms for an n\times n matrix), and all scientific computing environments have a det function which uses much faster methods.

Exercise
For various values of n, use the expression np.linalg.det(np.random.randint(-9,10,(n,n))) det(rand(-9:9, n, n)) to find the determinant of an n\times n matrix filled with random single-digit numbers. How large does n have to be for the determinant to be large enough to consistently overflow?

Solution. Trial and error reveals that this determinant starts to consistently return inf Inf at n = 187.

Exercise
Suppose that A and B are 3 \times 3 matrices, with determinant 5 and \frac{1}{2} respectively. Suppose that R \subset \mathbb{R}^3 is a 3D region modeling a fish whose volume is 14. What is the volume of the transformed fish BA(R)?

19.5
35
12
16.5

Solution. The volume of A(R) is 5 \cdot 14 = 70. The volume of BA(R) = B(A(R)) is \tfrac{1}{2} \cdot 70 = 35.

Exercise
Let R \subset \mathbb{R}^3 be 3D region modeling a fish, and suppose A an invertible 3 \times 3 matrix. If R has volume 15 and A^{-1}(R) has volume 5, then the determinant of A is equal to ???

?

Solution. We can see that the matrix A^{-1} scales volumes by \frac{1}{3}, and hence \det A^{-1} = \frac{1}{3}. This implies that \det A = 3.

Determinants can be used to check whether a matrix is invertible, since A is noninvertible if and only if it maps \mathbb{R}^n to a lower-dimensional subspace of \mathbb{R}^n, and in that case A squishes positive-volume regions down to zero-volume regions.

Exercise
Let A = \begin{bmatrix} 2 & -2 \\\ -4 & 0 \end{bmatrix}. Find the values of \lambda \in \mathbb{R} for which the equation A \mathbf{v} = \lambda \mathbf{v} has nonzero solutions for \mathbf{v}.

Solution. We can rewrite A\mathbf{v} = \lambda \mathbf{v} as A\mathbf{v} = (\lambda I) \mathbf{v}, where I is the identity matrix. We can rearrange this to give the equation (A - \lambda I)\mathbf{v} = 0. This has a nontrivial solution if (A - \lambda I) has a nonzero nullspace. Since A - \lambda I is a square matrix, this is equivalent to it having determinant zero.

\begin{align*}\det \left(A - \lambda I \right) = \det \left(\begin{bmatrix} 2-\lambda & -2\\ -4 &- \lambda \end{bmatrix} \right) = -\lambda (2-\lambda) - 8\end{align*}

Setting this equal to zero gives

\begin{align*}\lambda^2 - 2\lambda - 8 = 0\end{align*}

The left-hand side can be factored

\begin{align*}(\lambda - 4)(\lambda + 2) = 0\end{align*}

Thus our two solutions are \lambda = 4,-2.

Exercise
For an n \times n square matrix, which of the following is the relationship between \det A and \det (3A)?

XEQUATIONX3239XEQUATIONX.
XEQUATIONX3240XEQUATIONX.
XEQUATIONX3241XEQUATIONX.
\det(3A) = 3^{n} \det (A).

Solution. The answer is (4) \det(3A) = 3^{n} \det (A). There are two ways to see this, algebraically

Mathematical reasoning is algebraic if it involves the manipulation of symbols representing mathematical objects. Algebraic reasoning complements geometric reasoning, which involves figures or other non-symbolic visualizations.

For example, (a+b)^2 = a^2 + 2ab + b^2 may be proved algebraically using the distributive property or geometrically by representing the left side as the area of an (a+b) \times (a+b) square and the right side as the sum of the areas of four rectangles into which the square is divided.

and geometrically

Mathematical reasoning is geometric if it involves spatial representations of the mathematical objects under consideration. Geometric reasoning complements algebraic reasoning, which involves symbolic manipulations.

For example, (a+b)^2 = a^2 + 2ab + b^2 may be proved algebraically using the distributive property or geometrically by representing the left side as the area of an (a+b) \times (a+b) square and the right side as the sum of the areas of four rectangles into which the square is divided.

.

To check that this is the right answer using algebra, let A = I_{n} be the n \times n identity matrix, with determinant 1. The matrix 3A is diagonal, with threes on the diagonal. Its determinant is the product of the entries on its diagonal, 3 \times 3 \times \cdots \times 3 = 3^{n}.

Geometrically, we know that the determinant of A measures how much A scales volume. The matrix 3A scales by a factor of three more in each dimension. Since there are n dimensions, the total scaling of volume is multiplied by a factor 3^n.

Exercise
Is every matrix with positive determinant positive definite? ???

Solution. No. Consider the negation of the 2 \times 2 identity matrix. It has determinant 1, yet its eigenvalues are both negative.

Congratulations! You have completed the Data Gymnasia Linear Algebra course.

Bruno Bruno