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Contact [7]
3 years ago
7

Suppose that L1 : V → W and L2 : W → Z arelinear transformations and E, F, and G are orderedbases for V, W, and Z, respectively.

Show that, if Arepresents L1 relative to E and F and B representsL2 relative to F and G, then the matrix C = BA representsL2 ◦ L1: V → Z relative to E and G. Hint:Show that BA[v]E = [(L2 ◦ L1)(v)]G for all v ∈ V.

Mathematics
1 answer:
daser333 [38]3 years ago
3 0

Answer:

a) v ∈ ker(<em>L</em>) if only if [V]_{E} ∈ <em>N</em>(<em>A</em>)

b) w ∈ <em>L</em>(<em>v</em>) if and only if [W]_{F} is in the column space of <em>A</em>

<em />

<em>See attached</em>

Step-by-step explanation:

See attached the proof Considering the vector spaces <em>V</em> and <em>W</em> with other bases <em>E</em> and <em>F</em> respectively.

Let <em>L</em> be the Linear transformation form <em>V</em> and <em>W</em> and A is the matrix representing <em>L</em> relative to<em> E</em> and <em>F</em>

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egoroff_w [7]

Answer:

(1, 3)

Step-by-step explanation:

You are given the h coordinate of the vertex as 1, but in order to find the k coordinate, you have to complete the square on the parabola.  The first few steps are as follows.  Set the parabola equal to 0 so you can solve for the vertex.  Separate the x terms from the constant by moving the constant to the other side of the equals sign.  The coefficient HAS to be a +1 (ours is a -2 so we have to factor it out).  Let's start there.  The first 2 steps result in this polynomial:

-2x^2+4x=-1.  Now we factor out the -2:

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-2(x^2-2x+1)=-1-2.  Simplifying gives us this:

-2(x^2-2x+1)=-3

On the left we have created a perfect square binomial which reflects the h coordinate of the vertex.  Stating this binomial and moving the -3 over by addition and setting the polynomial equal to y:

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5 0
3 years ago
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Colt1911 [192]
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s344n2d4d5 [400]

Answer:

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