김경종
445 lines
24 KiB
Markdown
445 lines
24 KiB
Markdown
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# 2.3 VECTOR SPACES
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In the previous section we defined a vector of order n to be an array of n numbers written in matrix form. We now want to associate a geometric interpretation with the elements of a vector. Consider as an example a column vector of order 3 such as
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$$
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\mathbf {x} = \left[ \begin{array}{l} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right] = \left[ \begin{array}{l} 2 \\ 4 \\ 3 \end{array} \right] \tag {2.42}
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$$
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We know from elementary geometry that x represents a geometric vector in a chosen coordinate system in three-dimensional space. Figure 2.2 shows assumed coordinate axes and the vector corresponding to (2.42) in this system. We should note that the geometric representation of x depends completely on the coordinate system chosen; in other words, if (2.42) would give the components of a vector in a different coordinate system, then the geometric representation of x would be different from the one in Fig. 2.2. Therefore, the coordinates (or components of a vector) alone do not define the actual geometric quantity, but they need to be given together with the specific coordinate system in which they are measured.
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<details>
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<summary>text_image</summary>
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x
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x
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x₁
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x₂
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x₃
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x = [2
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4
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3]
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</details>
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Figure 2.2 Geometric representation of vector x
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The concepts of three-dimensional geometry generalize to a vector of any finite order $n$ . If $n > 3$ , we can no longer obtain a plot of the vector; however, we shall see that mathematically all concepts that pertain to vectors are independent of $n$ . As before, when we considered the specific case $n = 3$ , the vector of order $n$ represents a quantity in a specific coordinate system of an $n$ -dimensional space.
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Assume that we are dealing with a number of vectors all of order n, which are defined in a fixed coordinate system. Some fundamental concepts that we shall find extremely powerful in the later chapters are summarized in the following definitions and facts.
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Definition: A collection of vectors $\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_s$ is said to be linearly dependent if there exist numbers $\alpha_1, \alpha_2, \ldots, \alpha_s$ , which are not all zero, such that
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$$
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\alpha_ {1} \mathbf {x} _ {1} + \alpha_ {2} \mathbf {x} _ {2} + \dots + \alpha_ {s} \mathbf {x} _ {s} = 0 \tag {2.43}
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$$
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If the vectors are not linearly dependent, they are called linearly independent vectors.
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We consider the following examples to clarify the meaning of this definition.
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EXAMPLE 2.17: Let $n = 3$ and determine if the vectors $\mathbf{e}_i, i = 1, 2, 3$ , are linearly dependent or independent.
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According to the definition of linear dependency, we need to check if there are constants $\alpha_{1}$ , $\alpha_{2}$ , and $\alpha_{3}$ , not all zero, that satisfy the equation
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$$
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\alpha_ {1} \left[ \begin{array}{l} 1 \\ 0 \\ 0 \end{array} \right] + \alpha_ {2} \left[ \begin{array}{l} 0 \\ 1 \\ 0 \end{array} \right] + \alpha_ {3} \left[ \begin{array}{l} 0 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{l} 0 \\ 0 \\ 0 \end{array} \right] \tag {a}
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$$
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But the equations in (a) read
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$$
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\left[ \begin{array}{l} \alpha_ {1} \\ \alpha_ {2} \\ \alpha_ {3} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 0 \\ 0 \end{array} \right]
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$$
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which is satisfied only if $\alpha_{i}=0, i=1,2,3$ ; therefore, the vectors $e_{i}$ are linearly independent.
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EXAMPLE 2.18: With $n = 4$ , investigate whether the following vectors are linearly dependent or independent.
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$$
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\mathbf {x} _ {1} = \left[ \begin{array}{l} 1 \\ 1 \\ 0 \\ 0. 5 \end{array} \right]; \quad \mathbf {x} _ {2} = \left[ \begin{array}{c} - 1 \\ 0 \\ 1 \\ 0 \end{array} \right]; \quad \mathbf {x} _ {3} = \left[ \begin{array}{l} 0 \\ - 0. 5 \\ - 0. 5 \\ - 0. 2 5 \end{array} \right]
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$$
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We need to consider the system of equations
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$$
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\alpha_ {1} \left[ \begin{array}{l} 1 \\ 1 \\ 0 \\ 0. 5 \end{array} \right] + \alpha_ {2} \left[ \begin{array}{c} - 1 \\ 0 \\ 1 \\ 0 \end{array} \right] + \alpha_ {3} \left[ \begin{array}{c} 0 \\ - 0. 5 \\ - 0. 5 \\ - 0. 2 5 \end{array} \right] = \left[ \begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \end{array} \right]
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$$
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or, considering each row,
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$$
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\alpha_ {1} - \alpha_ {2} = 0
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$$
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$$
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\alpha_ {1} \quad - 0. 5 \alpha_ {3} = 0
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$$
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$$
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\alpha_ {2} - 0. 5 \alpha_ {3} = 0
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$$
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$$
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0. 5 \alpha_ {1} \quad - 0. 2 5 \alpha_ {3} = 0
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$$
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where we note that the equations are satisfied for $\alpha_{1} = 1$ , $\alpha_{2} = 1$ , and $\alpha_{3} = 2$ . Therefore, the vectors are linearly dependent.
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In the preceding examples, the solution for $\alpha_{1}$ , $\alpha_{2}$ , and $\alpha_{3}$ could be obtained by inspection. We shall later develop a systematic procedure of checking whether a number of vectors are linearly dependent or independent.
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Another way of looking at the problem, which may be more appealing, is to say that the vectors are linearly dependent if any one of them can be expressed in terms of the others. That is, if not all of the $\alpha_{i}$ in (2.43) are zero, say $\alpha_{j} \neq 0$ , then we can write
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$$
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\mathbf {x} _ {j} = - \sum_ {\substack {k = 1 \\ k \neq j}} ^ {s} \frac {\alpha_ {k}}{\alpha_ {j}} \mathbf {x} _ {k} \tag{2.44}
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$$
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Geometrically, when $n \leq 3$ , we could plot the vectors and if they are linearly dependent, we would be able to plot one vector in terms of multiples of the other vectors. For example, plotting the vectors used in Example 2.17, we immediately observe that none of them can be expressed in terms of multiples of the remaining ones; hence the vectors are linearly independent.
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Assume that we are given q vectors of order n, $n \geq q$ , which are linearly dependent, but that we only consider any $(q - 1)$ of them. These $(q - 1)$ vectors may still be linearly dependent. However, by continuing to decrease the number of vectors under consideration, we would arrive at p vectors, which are linearly independent, where, in general, $p \leq q$ . The other $(q - p)$ vectors can be expressed in terms of the p vectors. We are thus led to the following definition.
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Definition: Assume that we have $p$ linearly independent vectors of order $n$ , where $n \geq p$ . These $p$ vectors form a basis for a $p$ -dimensional vector space.
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We talk about a vector space of dimension p because any vector in the space can be expressed as a linear combination of the p base vectors. We should note that the base vectors for the specific space considered are not unique; linear combinations of them can give another basis for the same space. Specifically, if p = n, then a basis for the space considered is $e_{i}, i = 1, \ldots, n$ , from which it also follows that p cannot be larger than n.
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Definition: $q$ vectors, of which $p$ vectors are linearly independent, are said to span a $p$ -dimensional vector space.
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We therefore realize that all the importance lies in the base vectors since they are the smallest number of vectors that span the space considered. All q vectors can be expressed in terms of the base vectors, however large q may be (and indeed q could be larger than n).
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EXAMPLE 2.19: Establish a basis for the space spanned by the three vectors in Example 2.18.
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In this case $q = 3$ and $n = 4$ . We find by inspection that the two vectors $\mathbf{x}_1$ and $\mathbf{x}_2$ are linearly independent. Hence $\mathbf{x}_1$ and $\mathbf{x}_2$ can be taken as base vectors of the two-dimensional space spanned by $\mathbf{x}_1, \mathbf{x}_2,$ and $\mathbf{x}_3$ . Also, using the result of Example 2.18, we have $\mathbf{x}_3 = -\frac{1}{2}\mathbf{x}_2 - \frac{1}{2}\mathbf{x}_1$ .
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Assume that we are given a p-dimensional vector space which we denote as $E_{p}$ , for which $x_{1}, x_{2}, \ldots, x_{p}$ are chosen base vectors, p > 1. Then we might like to consider only
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all those vectors that can be expressed in terms of $x_{1}$ and $x_{2}$ . But the vectors $x_{1}$ and $x_{2}$ also form the basis of a vector space that we call $E_{2}$ . If p = 2, we note that $E_{p}$ and $E_{2}$ coincide. We call $E_{2}$ a subspace of $E_{p}$ , the concise meaning of which is defined next.
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Definition: A subspace of a vector space is a vector space such that any vector in the subspace is also in the original space. If $x_{1}, x_{2}, \ldots, x_{p}$ are the base vectors of the original space, any subset of these vectors forms the basis of a subspace; the dimension of the subspace is equal to the number of base vectors selected.
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EXAMPLE 2.20: The three vectors $x_{1}$ , $x_{2}$ , and $x_{3}$ are linearly independent and therefore form the basis of a three-dimensional vector space $E_{3}$ :
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$$
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\mathbf {x} _ {1} = \left[ \begin{array}{l} 1 \\ 2 \\ 1 \\ 0 \end{array} \right]; \quad \mathbf {x} _ {2} = \left[ \begin{array}{l} 1 \\ 0 \\ 0 \\ 0 \end{array} \right]; \quad \mathbf {x} _ {3} = \left[ \begin{array}{l} 0 \\ - 1 \\ 0 \\ 1 \end{array} \right] \tag {a}
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$$
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Identify some possible two-dimensional subspaces of $E_3$ .
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Using the base vectors in (a), a two-dimensional subspace is formed by any two of the three vectors; e.g., $x_{1}$ and $x_{2}$ represent a basis for a two-dimensional subspace; $x_{1}$ and $x_{3}$ are the basis for another two-dimensional subspace; and so on. Indeed, any two linearly independent vectors in $E_{3}$ form the basis of a two-dimensional subspace, and it follows that there are an infinite number of two-dimensional subspaces in $E_{3}$ .
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Having considered the concepts of a vector space, we may now recognize that the columns of any rectangular matrix A also span a vector space. We call this space the column space of A. Similarly, the rows of a matrix span a vector space, which we call the row space of A. Conversely, we may assemble any q vectors of order n into a matrix A of order $n \times q$ . The number of linearly independent vectors used is equal to the dimension of the column space of A. For example, the three vectors in Example 2.20 form the matrix
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$$
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\mathbf {A} = \left[ \begin{array}{c c c} 1 & 1 & 0 \\ 2 & 0 & - 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right] \tag {2.45}
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$$
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Assume that we are given a matrix A and that we need to calculate the dimension of the column space of A. In other words, we want to evaluate how many columns in A are linearly independent. The number of linearly independent columns in A is neither increased nor decreased by taking any linear combinations of them. Therefore, in order to identify the column space of A, we may try to transform the matrix, by linearly combining its columns, to obtain unit vectors $e_{i}$ . Because unit vectors $e_{i}$ with distinct i are linearly independent, the dimension of the column space of A is equal to the number of unit vectors that can be obtained. While frequently we are not able to actually obtain unit vectors $e_{i}$ (see Example 2.21), the process followed in the transformation of A will always lead to a form that displays the dimension of the column space.
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EXAMPLE 2.21: Calculate the dimension of the column space of the matrix A formed by the vectors $x_{1}$ , $x_{2}$ , and $x_{3}$ considered in Example 2.20.
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The matrix considered is
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$$
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\mathbf {A} = \left[ \begin{array}{c c c} 1 & 1 & 0 \\ 2 & 0 & - 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]
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$$
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Writing the second and third columns as the first and second columns, respectively, we obtain
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$$
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\mathbf {A} _ {1} = \left[ \begin{array}{c c c} 1 & 0 & 1 \\ 0 & - 1 & 2 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]
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$$
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Subtracting the first column from the third column, adding twice the second column to the third column, and finally multiplying the second column by $(-1)$ , we obtain
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$$
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\mathbf {A} _ {2} = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 2 \end{array} \right]
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$$
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But we have now reduced the matrix to a form where we can identify that the three columns are linearly independent; i.e., the columns are linearly independent because the first three elements in the vectors are the columns of the identity matrix of order 3. However, since we obtained $A_{2}$ from A by interchanging and linearly combining the original columns of A and thus in the solution process have not increased the space spanned by the columns of the matrix, we find that the dimension of the column space of A is 3.
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In the above presentation we linearly combined the vectors $x_{1}, \ldots, x_{q}$ , which were the columns of A, in order to identify whether they were linearly independent. Alternatively, to find the dimension of the space spanned by a set of vectors $x_{1}, x_{2}, \ldots, x_{q}$ , we could use the definition of vector linear independence in (2.43) and consider the set of simultaneous homogeneous equations
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$$
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\alpha_ {1} \mathbf {x} _ {1} + \alpha_ {2} \mathbf {x} _ {2} + \dots \alpha_ {q} \mathbf {x} _ {q} = \mathbf {0} \tag {2.46}
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$$
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which is, in matrix form,
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$$
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\mathbf {A} \boldsymbol {\alpha} = \mathbf {0} \tag {2.47}
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$$
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where $\alpha$ is a vector with elements $\alpha_{1},\ldots,\alpha_{q}$ , and the columns of A are the vectors $x_{1},x_{2},\ldots,x_{q}$ . The solution for the unknowns $\alpha_{1},\ldots,\alpha_{q}$ is not changed by linearly combining or multiplying any of the rows in the matrix A. Therefore, we may try to reduce A by multiplying and combining its rows into a matrix in which the columns consist only of unit vectors. This reduced matrix is called the row-echelon form of A. The number of unit column vectors in the row-echelon form of A is equal to the dimension of the column space of A and, from the preceding discussion, is also equal to the dimension of the row space of A. It follows that the dimension of the column space of A is equal to the dimension of the
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row space of A. In other words, the number of linearly independent columns in A is equal to the number of linearly independent rows in A. This result is summarized in the definition of the rank of A and the definition of the null space (or kernel) of A.
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Definition: The rank of a matrix $\mathbf{A}$ is equal to the dimension of the column space and equal to the dimension of the row space of $\mathbf{A}$ .
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Definition: The space of vectors $\alpha$ such that $A\alpha = 0$ is the null space (or kernel) of A.
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EXAMPLE 2.22: Consider the following three vectors:
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$$
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\mathbf {x} _ {1} = \left[ \begin{array}{l} 1 \\ 2 \\ 1 \\ 3 \\ 4 \\ 3 \end{array} \right]; \quad \mathbf {x} _ {2} = \left[ \begin{array}{c} 3 \\ 1 \\ - 2 \\ 4 \\ 2 \\ - 1 \end{array} \right]; \quad \mathbf {x} _ {3} = \left[ \begin{array}{l} 2 \\ 3 \\ 1 \\ 5 \\ 6 \\ 4 \end{array} \right]
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$$
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Use these vectors as the columns of a matrix $\mathbf{A}$ and reduce the matrix to row-echelon form. We have
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$$
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\mathbf {A} = \left[ \begin{array}{c c c} 1 & 3 & 2 \\ 2 & 1 & 3 \\ 1 & - 2 & 1 \\ 3 & 4 & 5 \\ 4 & 2 & 6 \\ 3 & - 1 & 4 \end{array} \right]
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$$
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Subtracting multiples of the first row from the rows below it in order to obtain the unit vector $e_{1}$ in the first column, we obtain
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$$
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\mathbf {A} _ {1} = \left[ \begin{array}{r r r} 1 & 3 & 2 \\ 0 & - 5 & - 1 \\ 0 & - 5 & - 1 \\ 0 & - 5 & - 1 \\ 0 & - 1 0 & - 2 \\ 0 & - 1 0 & - 2 \end{array} \right]
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$$
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Dividing the second row by $(-5)$ and then subtracting multiples of it from the other rows in order to reduce the second column to the unit vector $e_{2}$ , we obtain
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$$
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\mathbf {A} _ {2} = \left[ \begin{array}{l l l} 1 & 0 & \frac {7}{5} \\ 0 & 1 & \frac {1}{5} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right]
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$$
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Hence we can give the following equivalent statements:
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1. The solution to $\mathbf{A}\alpha = \mathbf{0}$ is
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$$
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\alpha_ {1} = - \frac {7}{5} \alpha_ {3}
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$$
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$$
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\alpha_ {2} = - \frac {1}{5} \alpha_ {3}
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$$
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2. The three vectors $x_{1}$ , $x_{2}$ , and $x_{3}$ are linearly dependent. They form a two-dimensional vector space. The vectors $x_{1}$ and $x_{2}$ are linearly independent, and they form a basis of the two-dimensional space in which $x_{1}$ , $x_{2}$ , and $x_{3}$ lie.
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3. The rank of A is 2.
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4. The dimension of the column space of $\mathbf{A}$ is 2.
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5. The dimension of the row space of $\mathbf{A}$ is 2.
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6. The null space (kernel) of $\mathbf{A}$ has dimension 1 and a basis is the vector
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$$
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\left[ \begin{array}{c} - \frac {7}{5} \\ - \frac {1}{5} \\ 1 \end{array} \right]
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$$
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Note that the rank of $A^{T}$ is also 2, but that the kernel of $A^{T}$ has dimension 4.
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# 2.4 DEFINITION OF TENSORS
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In engineering analysis, the concept of tensors and their matrix representations can be important. We shall limit our discussion to tensors in three-dimensional space and primarily be concerned with the representation of tensors in rectangular Cartesian coordinate frames.
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Let the Cartesian coordinate frame be defined by the unit base vectors $e_{i}$ (see Fig. 2.3). A vector u in this frame is given by
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$$
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\mathbf {u} = \sum_ {i = 1} ^ {3} u _ {i} \mathbf {e} _ {i} \tag {2.48}
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$$
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<details>
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<summary>text_image</summary>
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x'₃
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x₃
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e'₃
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e₃
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e'₂
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x'₂
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e₁
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e'₁
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e₂
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x₂
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x₁
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x'₁
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</details>
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Figure 2.3 Cartesian coordinate systems for definition of tensors
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where the $u_{i}$ are the components of the vector. In tensor algebra it is convenient for the purpose of a compact notation to omit the summation sign in (2.48); i.e., instead of (2.48) we simply write
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$$
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\mathbf {u} = u _ {i} \mathbf {e} _ {i} \tag {2.49}
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$$
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where the summation on the repeated index i is implied (here i = 1, 2, 3). Since i could be replaced by any other subscript without changing the result (e.g., k or j), it is also called a dummy index. This convention is referred to as the summation convention of indicial notation (or the Einstein convention) and is used with efficiency to express in a compact manner relations involving tensor quantities (see Chapter 6 where we use this notation extensively).
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Considering vectors in three-dimensional space, vector algebra is employed effectively.
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The scalar (or dot) product of the vectors u and v, denoted by u · v is given by
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$$
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\mathbf {u} \cdot \mathbf {v} = | \mathbf {u} | | \mathbf {v} | \cos \theta \tag {2.50}
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$$
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where $|u|$ is equal to the length of the vector u, $|u| = \sqrt{u_{i}u_{i}}$ . The dot product can be evaluated using the components of the vectors,
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$$
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\mathbf {u} \cdot \mathbf {v} = u _ {i} v _ {i} \tag {2.51}
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$$
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The vector (or cross) product of the vectors u and v produces a new vector w = u × v
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$$
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\mathbf {w} = \det \left[ \begin{array}{l l l} \mathbf {e} _ {1} & \mathbf {e} _ {2} & \mathbf {e} _ {3} \\ u _ {1} & u _ {2} & u _ {3} \\ v _ {1} & v _ {2} & v _ {3} \end{array} \right] \tag {2.52}
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$$
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Figure 2.4 illustrates the vector operations performed in (2.50) and (2.52). We should note that the direction of the vector w is obtained by the right-hand rule; i.e., the right-hand thumb points in the direction of w when the fingers curl from u to v.
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<details>
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<summary>text_image</summary>
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cos θ = \frac{U_i V_i}{|u| |v|}
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|w| = |u| |v| sin θ
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x3
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w
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e3
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v
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e2
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θ
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x2
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e1
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u
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x1
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</details>
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Figure 2.4 Vectors used in products
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These vector algebra procedures are frequently employed in finite element analysis to evaluate angles between two given directions and to establish the direction perpendicular to a given plane.
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EXAMPLE 2.23: Assume that the vectors u and v in Fig. 2.4 are
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$$
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\mathbf {u} = \left[ \begin{array}{l} 3 \\ 3 \\ 0 \end{array} \right]; \quad \mathbf {v} = \left[ \begin{array}{l} 0 \\ 2 \\ 2 \end{array} \right]
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$$
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Calculate the angle between these vectors and establish a vector perpendicular to the plane that is defined by these vectors.
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Here we have
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$$
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| \mathbf {u} | = 3 \sqrt {2}
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$$
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$$
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| \mathbf {v} | = 2 \sqrt {2}
|
||
$$
|
||
|
||
Hence $\cos \theta = \frac{1}{2}$
|
||
|
||
and $\theta = 60^{\circ}$
|
||
|
||
A vector perpendicular to the plane defined by u and v is given by
|
||
|
||
$$
|
||
\mathbf {w} = \det \left[ \begin{array}{l l l} \mathbf {e} _ {1} & \mathbf {e} _ {2} & \mathbf {e} _ {3} \\ 3 & 3 & 0 \\ 0 & 2 & 2 \end{array} \right]
|
||
$$
|
||
|
||
hence $\mathbf{w} = \begin{bmatrix} 6 \\ -6 \\ 6 \end{bmatrix}$
|
||
|
||
Using $|\mathbf{w}| = \sqrt{w_i w_i}$ , we obtain
|
||
|
||
$$
|
||
| \mathbf {w} | = 6 \sqrt {3}
|
||
$$
|
||
|
||
which is also equal to the value obtained using the formula given in Fig. 2.4.
|
||
|
||
Although not specifically stated, the typical vector considered in (2.48) is a tensor. Let us now formally define what we mean by a tensor.
|
||
|
||
For this purpose, we consider in addition to the unprimed Cartesian coordinate frame a primed Cartesian coordinate frame with base vectors $e_{j}^{\prime}$ which spans the same space as the unprimed frame (see Fig. 2.3).
|
||
|
||
An entity is called a scalar, a vector (i.e., a tensor of first order or rank 1), or a tensor (i.e., a tensor of higher order or rank) depending on how the components of the entity are defined in the unprimed frame (coordinate system) and how these components transform to the primed frame.
|
||
|
||
Definition: An entity is called a scalar if it has only a single component $\phi$ in the coordinates $x_{i}$ measured along $\mathbf{e}_{i}$ and this component does not change when expressed in the coordinates $x_{i}^{\prime}$ measured along $\mathbf{e}_{i}^{\prime}$ :
|
||
|
||
$$
|
||
\phi (x _ {1}, x _ {2}, x _ {3}) = \phi^ {\prime} (x _ {1} ^ {\prime}, x _ {2} ^ {\prime}, x _ {3} ^ {\prime}) \tag {2.53}
|
||
$$
|
||
|
||
A scalar is also a tensor of order 0. As an example, temperature at a point is a scalar.
|
||
|
||
<!-- source-page: 60 -->
|
||
|
||
Definition: An entity is called a vector or tensor of first order if it has three components $\xi_{i}$ in the unprimed frame and three components $\xi_{i}^{\prime}$ in the primed frame, and if these components are related by the characteristic law (using the summation convention)
|
||
|
||
$$
|
||
\xi_ {i} ^ {\prime} = p _ {i k} \xi_ {k} \tag {2.54}
|
||
$$
|
||
|
||
where $p_{ik} = \cos \left( e_{i}^{\prime}, e_{k} \right)$ (2.55)
|
||
|
||
The relation (2.54) can also be written in matrix form as
|
||
|
||
$$
|
||
\boldsymbol {\xi} ^ {\prime} = \mathbf {P} \boldsymbol {\xi} \tag {2.56}
|
||
$$
|
||
|
||
where $\pmb{\xi}^{\prime}$ , $\mathbf{P}$ , and $\pmb{\xi}$ contain the elements of (2.54).
|
||
|
||
The transformation in $(2.54)$ corresponds to a change of basis in the representation of the vector. To arrive at $(2.54)$ we recognize that the same vector is considered in the two different bases; hence we have
|
||
|
||
$$
|
||
\xi_ {j} ^ {\prime} \mathbf {e} _ {j} ^ {\prime} = \xi_ {k} \mathbf {e} _ {k} \tag {2.57}
|
||
$$
|
||
|
||
Using the fact that the base vectors in each coordinate frame are orthogonal to each other and are of unit length, we can take the dot products [see (2.50)] on both sides of (2.57) with $e_{i}^{\prime}$ and obtain (2.54). Of course, analogously we could also take the dot product on both sides with $e_{m}$ to obtain the inverse transformation
|
||
|
||
$$
|
||
\xi_ {m} = \cos (\mathbf {e} _ {m}, \mathbf {e} _ {j} ^ {\prime}) \xi_ {j} ^ {\prime} \tag {2.58}
|
||
$$
|
||
|
||
or in matrix form, $\pmb{\xi} = \mathbf{P}^T\pmb{\xi}'$ (2.59)
|
||
|
||
Hence we note that $\mathbf{P}^{-1} = \mathbf{P}^T$ , and this leads us to the following definition.
|
||
|
||
Definition: A matrix $\mathbf{Q}$ is an orthogonal matrix if $\mathbf{Q}^T\mathbf{Q} = \mathbf{Q}\mathbf{Q}^T = \mathbf{I}$ . Therefore, for an orthogonal matrix, we have $\mathbf{Q}^{-1} = \mathbf{Q}^T$ .
|
||
|
||
Hence the matrix P defined in (2.55) and (2.56) is an orthogonal matrix, and because the elements of P produce a rotation, we also refer to P as a rotation matrix.
|
||
|
||
We demonstrate the preceding discussion in the following example.
|
||
|
||
EXAMPLE 2.24: The components of a force expressed in the unprimed coordinate system shown in Fig. E2.24 are
|
||
|
||
$$
|
||
\mathbf {R} = \left[ \begin{array}{c} 0 \\ 1 \\ \sqrt {3} \end{array} \right]
|
||
$$
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
x'₃
|
||
x₃
|
||
√3
|
||
R
|
||
θ
|
||
x₁, x'₁
|
||
1
|
||
x₂
|
||
x'₂
|
||
</details>
|
||
|
||
Figure E2.24 Representation of a force in different coordinate systems
|