Answered: Suppose that T: R³ → R³ is a linear…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Hi, I need help with this Linear Alebegra exercise, please. Thank you!

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Is definitely not

Might be

Definitely is

T(2,2,2)___(0,0,0)

Might equal

Definitely does not equal

Definitely equals

T(1,2,3)___(0,0,0)

Might equal

Definitely does not equal

Definitely equals

rank(T)__equal to 2

Is definitely not

Might be

Definitely is

Suppose that T: R³ → R³ is a linear transformation and that T(1, 1, 1) = (0,0,0). Choose the option
that correctly completes each of the following statements.
T [ Select ]
v onto.
T(2, 2, 2) [ Select ]
(0,0,0).
T(1, 2, 3) [ Select ]
(0,0,0).
rank(T) [Select ]
equal to 2.

Expert Solution

Step 1

Given $T (1, 1, 1) = (0, 0, 0)$

Let $A = [\begin{matrix} a & d & g \\ b & e & h \\ c & f & i \end{matrix}]$ be the matrix of $T .$

Then, since $T (1, 1, 1) = (0, 0, 0)$ we have,

$\begin{array}{rcl} [\begin{matrix} 1 & 1 & 1 \end{matrix}] [\begin{matrix} a & d & g \\ b & e & h \\ c & f & i \end{matrix}] & = & [\begin{matrix} 0 & 0 & 0 \end{matrix}] \\ a + b + c & = & 0 \\ d + e + f & = & 0 \\ g + h + i & = & 0 \end{array}$

This implies,

$a = - (b + c) d = - (e + f) g = - (h + i)$

T is onto provided span of columns of $A$ is equal to $ℝ^{3}$ .

Now span of columns of matrix $A$ is:

$k [\begin{matrix} - (b + c) \\ b \\ c \end{matrix}] + l [\begin{matrix} - (e + f) \\ e \\ f \end{matrix}] + m [\begin{matrix} - (h + i) \\ h \\ i \end{matrix}] = [\begin{matrix} k (- b - c) + l (- e - f) - m (h + i) \\ k b + l e + m h \\ k c + l f + m i \end{matrix}]$

Let,

$x = k (- b - c) + l (- e - f) - m (h + i) y = k b + l e + m h z = k c + l f + m i$

Then,

$x + y + z = 0$

Therefore,

Span of columns of matrix $A$ ,

$= \{[\begin{matrix} \begin{matrix} x \\ y \\ z \end{matrix} \end{matrix}] | x, y, z \in ℝ and x + y + z = 0\}$

Then span of columns of matrix $A$ is not equal to $ℝ^{3}$ .

Therefore, $T$ is not onto.

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