[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