Q66E a. Let聽V聽be the linear space o... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 6: Q66E (page 277)

a. LetVbe the linear space of all functionsF from ${鈩�}^{2 脳 2}$ to $鈩�$ that are linear in both columns. Find a basis of V, and thus determine the dimension ofV.
b. LetWbe the linear space of all functionsDfrom ${鈩�}^{2 脳 2}$ to $鈩�$ that are linear in both columns and alternating on the columns. Find a basis ofW, and thus determine the dimension ofW.

Short Answer

Expert verified

a) $\{F_{1}, F_{2}, F_{3}, F_{4}\}$ is a basis for V, and dimV=4 .

b) ${de t}$ is a basis for W, and dimW=1 .

Step by step solution

(a) To find a basis of V, and thus determine the dimension of V

For $F 鈭� V$ , we have

$F [\begin{matrix} a & b \\ c & d \end{matrix}] = a F [\begin{matrix} 1 & b \\ 0 & d \end{matrix}] + c F [\begin{matrix} 0 & b \\ 1 & d \end{matrix}] F = [\begin{matrix} a & b \\ c & d \end{matrix}] = a b F [\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}] + a d F [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + b c F [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}], 鈭赌 a, b, c, d 鈭� 鈩�$

From this, we can clearly see that V is spanned by functions.

$F_{1} [\begin{matrix} a & b \\ c & d \end{matrix}] = a b, F_{2} [\begin{matrix} a & b \\ c & d \end{matrix}] = a d, F_{3} [\begin{matrix} a & b \\ c & d \end{matrix}] = b c, F_{4} [\begin{matrix} a & b \\ c & d \end{matrix}] = c d,$

Therefore, $\{F_{1}, F_{2}, F_{3}, F_{4}\}$ is a basis for V, and $d i m V = 4$ .

(b) To find a basis of W, and thus determine the dimension of W

Clearly is a subspace of V. However, for $F 鈭� W$ , from the fact F that is alternating on the columns, we have

$F [\begin{matrix} a & b \\ c & d \end{matrix}] = a b F [\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}] + a d F [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + b c F [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] + c d F [\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}] F = [\begin{matrix} a & b \\ c & d \end{matrix}] = a b F [\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}] + a d F [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - b c F [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + c d F [\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}] F = [\begin{matrix} a & b \\ c & d \end{matrix}] = (a b - b c) F [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] F = [\begin{matrix} a & b \\ c & d \end{matrix}] = d e t [\begin{matrix} a & b \\ c & d \end{matrix}]$

Which shows that W is spanned by the determinant.

Therefore, ${\det}$ is a basis for W, and $\dim W = 1$ .

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Short Answer

Step by step solution

(a) To find a basis of V, and thus determine the dimension of V

(b) To find a basis of W, and thus determine the dimension of W

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