# Linear transformations

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Let P = {p

p

And let L : P

L(p(t)) = tp(0) + (2 - t)p'(t)

How do I form a matrix representation of L?

_{1}, p_{2}} be a basis of P_{1}given byp

_{1}(t) = 1, p_{2}(t) = tAnd let L : P

_{1}-> P_{1}be the mapping given byL(p(t)) = tp(0) + (2 - t)p'(t)

How do I form a matrix representation of L?

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#2

(Original post by

Let P = {p

p

And let L : P

L(p(t)) = tp(0) + (2 - t)p'(t)

How do I form a matrix representation of L?

**shawn_o1**)Let P = {p

_{1}, p_{2}} be a basis of P_{1}given byp

_{1}(t) = 1, p_{2}(t) = tAnd let L : P

_{1}-> P_{1}be the mapping given byL(p(t)) = tp(0) + (2 - t)p'(t)

How do I form a matrix representation of L?

We want to represent the given linear map L by a matrix (with respect to the basis {p1, p2}). What we mean by this is that if we are given a column vector representing an element of P_1 with respect to our chosen basis, then we want a matrix such that is the column vector representing the image of our original vector under the map L.

Notice in particular that if we consider the effect of this matrix on the vector (the column vector with a 1 in position i and 0's everywhere else), the result is just the ith column of the matrix. e.g. . We can use this observation to compute the values of x, y, z and w:

If we can work out the image of p1 under L and express it as a linear combination of p1 and p2, then these coefficients must form the 1st column of the matrix. Similarly, if we can work out the image of p2 under L and express it as a linear combination of p1 and p2, then these coefficients must form the 2nd column of the matrix.

Recall that p2=t. Then L(p2) = t p2(0) + (2-t) p2'(t). But p2(0)=0, so L(p2) = (2-t)p2'(t). But p2'(t) = 1, so L(p2) = 2-t = 2 p1 - p2.

So the first column of the matrix must be (or in the notation used above, x=2, z=-1).

You can now do the same for L(p2) to find the values of y and w.

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(OK this is a different question but same topic, and I'm stuck on it)

Given a linear transformation L : V -> W, show that ker(L) is a subspace of V. If ker(L) = {0}, show that L is injective.

Given a linear transformation L : V -> W, show that ker(L) is a subspace of V. If ker(L) = {0}, show that L is injective.

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#4

(Original post by

(OK this is a different question but same topic, and I'm stuck on it)

Given a linear transformation L : V -> W, show that ker(L) is a subspace of V. If ker(L) = {0}, show that L is injective.

**shawn_o1**)(OK this is a different question but same topic, and I'm stuck on it)

Given a linear transformation L : V -> W, show that ker(L) is a subspace of V. If ker(L) = {0}, show that L is injective.

Recall the subspace test, can you verify that satisfies it?

To show is injective, take two arbitrary elements from and show that .

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#5

(Original post by

Given a linear transformation L : V -> W, show that ker(L) is a subspace of V.

**shawn_o1**)Given a linear transformation L : V -> W, show that ker(L) is a subspace of V.

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