[Finite-Dimensional Linear Algebra] 4.1 + 4.2 + 4.3 The Determinant Function

 

 

Analysis of matrices is central in the study of finite-dimensional linear algebra, because any linear operator mapping one finite-dimensional space into another can be represented by a matrix. The simplest kind of matrix ia a diagonal matrix. The matrix `A \in F^{m \times n}` is diagonal if `A_{ij} = 0` for all `i \neq j`

 

To simplify matrix-based calculations, we want to transform matrices into diagonal matrices, and the key to diagonalization is identifying the eigenvalues and eigenvectors of a matrix. 

Definition of Eigenvalues and Eigenvectors 

Let `X` be a vector space over a field `F`, and let `T: X \rightarrow X`. If there exists a nonzero vector `x \in X` and a scalar `\lambda \in F` such that ``T(x) = \lambda x,``

then we call 

• `\lambda` an eigenvalue of `T` 
• `x` an eigenvector of `T` corresponding to `\lambda`
• `\lambda, x` an eigenpair of `T`

Figure 4.1.1. Eigenvectors `v_1, v_2` of matrix transformation `A` with a positive eigenvalue `\lambda _1` and a negative eigenvalue `\lambda _2`

In simpler terms, an eigenvector `x` is any vector for which the effect of linear transformation `T` is equivalent to scalar multiplication of the vector, and an eigenvalue `\lambda` is the multiplying factor. This means that `T(x)` and `\lambda x` are colinear with the origin, i.e. on the same line through the origin, as shown in Figure 4.1.1.

 

For the matrix form `Ax = \lambda x,` when matrix `A \in F^{n \times n}` is given, `\lambda` is an eigenvalue of `A` iff ``\lambda x - Ax = 0, x \neq 0,`` which is equivalent to ``(\lambda I - A)x = 0, x \neq 0.``

Therefore, we can say that `\lambda` is an eigenvalue of `A` iff the matrix `(\lambda I - A)` is singular, and we can find the eigenvalues of `A` by solving the equation ``det(\lambda I - A) = 0.`` The determinant function det: `F^{n \times n} \rightarrow F` is a special function with the property such that `det(B) = 0` iff `B` is singular. 

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Definition of the Determinant Function

Let `F` be a field and `n` a positive integer. The determinant function `det: (F^n)^n \rightarrow F` is defined by the following properties: 

1. If `\{ e_1, e_2, \dots ,e_n \}` is the standard basis for `F^n`, then `` det(e_1, e_2, \dots , e_n) = 1 ``
2. For any `A_1, A_2, \dots , A_n \in F^n`, any `\lambda \in F`, and any `j`, `1 \leq j \leq n,`  `` det( A_1, \dots , \lambda A_j, \dots , A_n ) = \lambda det(A_1, \dots , A_n). ``
3. For any `A_1, A_2, \dots , A_n \in F^n`, any `\lambda \in F`, and any `i \neq j`, `1 \leq i, j \leq n,` `` det( A_1, \dots , A_j + \lambda A_i, \dots , A_n ) = det(A_1, \dots , A_n). ``

The determinant function can be regarded as a function of the columns of a matrix. Thus, `det(A)` and `det(A_1, \dots , A_n)` can be used interchangeably.

The properties of a determinant function allow it to be characterised as a volume function. Geometrically, `det(A)` represents the signed volume of the parallelopiped in `F^n` determined by the vectors ` A_1, \dots , A_n.`

 

Figure 4.1.2. Geometric interpretation of determinants

For example, the defining property (3) expresses the geometric fact that the area of a parallelogram is determined by the length of a base and the corresponding perpendicular height. 

 

Figure 4.1.3. Parallelogram determined by `A_1, A_2` and by `A_1, A_2+\lambda A_1`

In the above figure, the two parallelograms have the same area: `` det(A_1, A_2 + \lambda A_1) = det(A_1, A_2),`` hence demonstrating how the properties of a determinant function can be interpreted geometrically. 

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Properties of the Determinant Function

Theorem Suppose `det: (F^n)^n \rightarrow F` satisfies the above definition. Suppose `A_1, A_2, \dots , A_n \in F^n`. Then: 

1. If ` 1 \leq j \leq n ` and `\lambda _i \in F` for `i \neq j`, then ``det(A_1, \dots , A_j + \sum_{i \neq j}^{ } \lambda_i A_i, \dots , A_n ) = det(A_1, \dots , A_j , \dots , A_n).``
2. If one of the vectors ` A_1, A_2, \dots , A_n` is the zero vector, then ``det( A_1, A_2, \dots , A_n) = 0.``
3. If `\{ A_1, A_2, \dots , A_n \}` is linearly dependent, then ``det( A_1, A_2, \dots , A_n) = 0.``
4. Interchanging two arguments changes the sign of the output: ``det( A_1, \dots , A_j, \dots , A_i, \dots , A_n) = -det(A_1, \dots , A_i, \dots , A_j, \dots , A_n).``
5. If `B_j` is any vector in `F_n`, then ``det( A_1, \dots , A_j + B_j, \dots , A_n) = det(A_1, \dots , A_j, \dots , A_n) + det(A_1, \dots , B_j, \dots , A_n).``

These properties show that det is multilinear, that is, linear in any one argument (column) when the other arguments are kept unchanged. 

Expansion of the Determinant Function

If `det: (F^n)^n \rightarrow F` satisfies the definition, then `det(A_1, A_2, \dots , A_n) = \sum_{\tau \in S_n}^{} \sigma (\tau )A_{\tau (1),1} A_{\tau (2), 2} \cdots A_{\tau (n), n}.`

This formula is referred to as the complete expansion of the determinant, and it shows that the determinant is a unique function on `(F^n)^n`. Using the properties of the determinant, particularly multilinearity, we can derive this formula.

First, replace each column vector `A` to a linear combination of basis vectors: ``det(A_1, A_2, \dots, A_n) = det(\sum_{i_1 = 1}^{n} A_{i_1, 1} e_{i_1}, \sum_{i_2 = 1}^{n} A_{i_2, 1} e_{i_2}, \dots , \sum_{i_n=1}^{n} A_{i_n, 1} e_{i_n}).`` where `A_{i, j}` represents components of `A_j.`  This can be simplified to ``\sum_{i_1=1}^{n} \sum_{i_2 = 1}^{n} \cdots \sum_{i_n = 1}^{n} det(e_{i_1}, e_{i_2}, \dots , e_{i_n})A_{i_1, 1}A_{i_2, 2} \cdots A_{i_n, n}.`` If `(i_1, i_2, \dots , i_n)` is not a permutation of `(1, 2, \dots , n)`, then ``det(e_{i_1}, e_{i_2}, \dots , e_{i_n}) = 0.`` On the other hand, if it is a permutation, then, since `det(e_1, e_2, \dots , e_n) = 1`, ``det(e_{i_1}, e_{i_2}, \dots , e_{i_n}) = \pm 1`` depending on whether the order was interchanged an even or odd number of times: ``det(e_{i_1}, e_{i_2}, \dots , e_{i_n}) = \sigma (i_1, i_2, \dots , i_n).`` Therefore, `` det(A_1, A_2, \dots, A_n) `` ``= \sum_{i_1=1}^{n} \sum_{i_2 = 1}^{n} \cdots \sum_{i_n = 1}^{n} det(e_{i_1}, e_{i_2}, \dots , e_{i_n})A_{i_1, 1}A_{i_2, 2} \cdots A_{i_n, n}`` ``= \sum_{i_1, \dots , i_n \in S_n}^{} \sigma (i_1, i_2, \dots i_n) A_{i_1, 1}A_{i_2, 2} \cdots A_{i_n, n}`` To further simplify the notation, replace arbitrary elements to `\tau = (\tau (1), \dots , \tau (n)).`

 

Theorem Let `F` be a field, and let `det: (F^n)^n \rightarrow F` is defined by the complete expansion formula. Then, det satisfies the definition of the determinant function. 

Further Properties of the Determinant Function

Theorem Let `F` be a field and let `A, B \in F^{n \times n}.` Then ``det(AB) = det(A)det(B).``

 

Theorem Let `\{A_1, A_2, \dots , A_n \}` be a linearly independent subset of `F^n`. Then ``det(A_1, A_2, \dots , A_n) \neq 0``

 

Corollary Let `F` be a field and suppose `A \in F^{n \times n}.` Then `A` is nonsingular iff  `det(A) \neq 0.`

 

Corollary Let `F` be a field and let `A \in F^{n \times n}` be invertible. Then ``det(A^{-1} = \frac{1}{det(A)}.``

 

Theorem Let `F` be a field and let `A \in F^{n \times n}.` Then ``det(A^{T}) = det(A).``

Since columns of `A^{T}` are rows of `A`, the theorem implies that row operations affect `det(A)` in the same manner as column operations do. 

 

Corollary Let `F` be a field and let `A \in F^{n \times n}.` 

1. If `B \in F^{n \times n}` is obtained by interchanging two rows of `A`, then ``det(B) = -det(A).``
2. If `B \in F^{n \times n}` is obtained by multiplying one row of `A` by a constant `\lambda \in F`, then ``det(B) = \lambda det(A).``
3. If `B \in F^{n \times n}` is obtained by adding a multiple of one row of `A` to another, then ``det(B) = det(A)``

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diagonal matrix: a matrix that has zero for every entry except those on the diagonal

eigenvector: a vector that remains parallel to the original vector when applied linear transformation

eigenvalue: the constant factor that scales the eigenvector when applied linear transformation