Q64E In all parts of this problem, le... [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 7: Q64E (page 325)

In all parts of this problem, let V be the linear space of all 2 脳 2 matrices for which $[\begin{matrix} 1 \\ 2 \end{matrix}]$ is an eigenvector.
(a) Find a basis of V and thus determine the dimension of V.
(b) Consider the linear transformation T (A) = A $[\begin{matrix} 1 \\ 2 \end{matrix}]$ from V to $R^{2}$ . Find a basis of the image of Tand a basis of the kernel of T. Determine the rank of T .
(c) Consider the linear transformation L(A) = A $[\begin{matrix} 1 \\ 3 \end{matrix}]$ from V to $R^{2}$ . Find a basis of the image of L and a basis of the kernel of L. Determine the rank of L.

Short Answer

Expert verified

Solution for

(a) V = 3

(b) A = 2

Step by step solution

Solving for (a):

We solve:

$\begin{aligned} [\begin{array}{ll} a & b \\ c & d \end{array}] [\begin{array}{l} 1 \\ 2 \end{array}] = [\begin{array}{c} 位 \\ 2 位 \end{array}] \\ a + 2 b = 位, c + 2 d = 2 位 \\ a = 位鈭� 2 b, c = 2 位鈭� 2 d \end{aligned}$

Thus, al such matrices are in form

$A = [\begin{array}{cc} 位鈭� 2 b & b \\ 2 位鈭� 2 b & d \end{array}] = b [\begin{array}{cc} 鈭� 2 & 1 \\ 0 & 0 \end{array}] + d [\begin{array}{cc} 0 & 0 \\ 鈭� 2 & 1 \end{array}] + 位 [\begin{array}{ll} 1 & 0 \\ 2 & 0 \end{array}] V = span ([\begin{array}{cc} 鈭� 2 & 1 \\ 0 & 0 \end{array}], [\begin{array}{cc} 0 & 0 \\ 鈭� 2 & 1 \end{array}], [\begin{array}{ll} 1 & 0 \\ 2 & 0 \end{array}])$

And thus dim V = 3.

Solving for (b):

We easily compute:

$[\begin{matrix} - 2 & 1 \\ 0 & 0 \end{matrix}] [\begin{matrix} 1 \\ 2 \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] [\begin{matrix} 0 & 0 \\ 2 & 1 \end{matrix}] [\begin{matrix} 1 \\ 2 \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] [\begin{matrix} 1 & 0 \\ 2 & 0 \end{matrix}] [\begin{matrix} 1 \\ 2 \end{matrix}] = [\begin{matrix} 1 \\ 2 \end{matrix}]$

Im A = span $([\begin{matrix} 1 \\ 2 \end{matrix}])$ , dim im A = 1

And

Ker A = span $([\begin{matrix} - 2 & 1 \\ 0 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ - 2 & 1 \end{matrix}])$ , dim ker A = 2

Solving for (c):

We easily compute:

Im A = span $([\begin{matrix} 1 \\ 0 \end{matrix}], [\begin{matrix} 0 \\ 1 \end{matrix}], [\begin{matrix} 1 \\ 2 \end{matrix}])$ , dim im A = 3

And

Ker A = span $\{0\}$ , dim ker A = 0

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Short Answer

Step by step solution

Solving for (a):

Solving for (b):

Solving for (c):

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