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9966 [12]
3 years ago
14

Let​ T: set of real numbers R Superscript nℝnright arrow→set of real numbers R Superscript mℝm be a linear​ transformation, and

let ​{v1​, v2​, v3​} be a linearly dependent set in set of real numbers R Superscript nℝn. Explain why the set ​{T(v1​), ​T(v2​), ​T(v3​)} is linearly dependent.
Mathematics
1 answer:
Klio2033 [76]3 years ago
6 0

Answer:

\{T(v_1), T(v_2), T(v_3)\} is linearly dependent set.

Step-by-step explanation:

Given:  \{v_1,v_2,v_3\} is a linearly dependent set in set of real numbers R

To show: the set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

Solution:

If \{v_1,v_2,v_3,...,v_n\} is a set of linearly dependent vectors then there exists atleast one k_i:i=1,2,3,...,n such that k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0

Consider k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0

A linear transformation T: U→V satisfies the following properties:

1. T(u_1+u_2)=T(u_1)+T(u_2)

2. T(au)=aT(u)

Here, u,u_1,u_2∈ U

As T is a linear transformation,

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0\\T(k_1v_1)+T(k_2v_2)+T(k_3v_3)=0\\T(k_1v_1+k_2v_2+k_3v_3)=0\\

As \{v_1,v_2,v_3\} is a linearly dependent set,

k_1v_1+k_2v_2+k_3v_3=0 for some k_i\neq 0:i=1,2,3

So, for some k_i\neq 0:i=1,2,3

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0

Therefore, set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

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Part A: the value of h(4) - m(16) is -4

Part B: The y-intercepts are 4 units apart

Part C: m(x) can not exceed h(x) for any value of x

Step-by-step explanation:

Let us use the table to find the function m(x)

There is a constant difference between each two consecutive values of x and also in y, then the table represents a linear function

The form of the linear function is m(x) = a x + b, where

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- Subtract 4 from both sides

∴ -2 = b

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Now let us answer the questions

Part A:

∵ h(x) = \frac{1}{2} (x - 2)²

∴ h(4) = \frac{1}{2} (4 - 2)²

∴ h(4) = \frac{1}{2} (2)²

∴ h(4) =  \frac{1}{2}(4)

∴ h(4) = 2

∵ m(x) = \frac{1}{2} x - 2

∴ m(16) =  \frac{1}{2} (16) - 2

∴ m(16) = 8 - 2

∴ m(16) = 6

- Find now h(4) - m(16)

∵ h(4) - m(16) = 2 - 6

∴ h(4) - m(16) = -4

Part B:

The y-intercept is the value of h(x) at x = 0

∵ h(x) = \frac{1}{2} (x - 2)²

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∴ h(0) = \frac{1}{2} (0 - 2)²

∴ h(0) =  \frac{1}{2} (-2)² =  

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∴ The y-intercept of h(x) is 2

∵ m(x) = \frac{1}{2} x - 2

∵ x = 0

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∴ m(0) = -2

∴ The y-intercept of m(x) is -2

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∵ h(8) = \frac{1}{2} (8 - 2)² = 18

∵ h(14) = \frac{1}{2} (14 - 2)² = 72

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∴ m(x) can not exceed h(x) for any value of x

<em>Look to the attached graph for more understand</em>

The blue graph represents h(x)

The green graph represents m(x)

The blue graph is above the green graph for all values of x, then there is no value of x make m(x) exceeds h(x)

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