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olga nikolaevna [1]
3 years ago
9

Find the sum of the arithmetic sequence. -10, -7, -4, -1, 2, 5, 8

Mathematics
1 answer:
Vadim26 [7]3 years ago
4 0
The arithmetic sequence is +3 for every number you move to the right.
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Answer:

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Step-by-step explanation:

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L :V --> W is a linear transformation. Prove each of the following (a) ker L is a subspace of V. (b) range L is a subspace of
iragen [17]

Answer:

a) Assume that x,y\in\ker L, and \alpha is a scalar (a real or complex number).

<em>First. </em>Let us prove that \ker L is not empty. This is easy because L(0_V)=0_W, by linearity. Here, 0_V stands for the zero vector of V, and 0_W stands for the zero vector of W.

<em>Second.</em> Let us prove that \alpha x\in\ker L. By linearity

L(\alpha x) = \alpha L(x)=\alpha 0_W=0_W.

Then, \alpha x\in\ker L.

<em>Third. </em> Let us prove that y+ x\in\ker L. Again, by linearity

L(x+y)=L(x)+L(y) = 0_W + 0_W=0_W.

And the statement readily follows.

b) Assume that u and v are in range of L. Then, there exist x,y\in V such that L(x)=u and L(y)=v.

<em>First.</em> Let us prove that range of L is not empty. This is easy because L(0_V)=0_W, by linearity.

<em>Second.</em> Let us prove that \alpha u is on the range of L.

\alpha u = \alpha L(x) = L(\alpha x) = L(z).

Then, there exist an element z\in V such that L(z)=\alpha u. Thus \alpha u is in the range of L.

<em>Third.</em> Let us prove that u+v is in the range of L.

u+v = L(x)+L(y) = L(x+y)=L(z).

Then, there exist an element z\in V such that L(z)= u +v. Thus u +v is in the range of L.

Notice that in this second part of the problem we used the linearity in the reverse order, compared with the first part of the exercise.

Step-by-step explanation:

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(4,0) i think, it only makes sense since its a square
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