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Linear transformations

Let P = {p₁, p₂} be a basis of P₁ given by

p₁(t) = 1, p₂(t) = t

And let L : P₁ -> P₁ be the mapping given by

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

How do I form a matrix representation of L?

Reply 1

Mark13

Original post by shawn_o1

So with respect to your chosen basis {p_1, p_2}, the vector

\begin{pmatrix}a\\b\end{pmatrix}

refers to the polynomial a+bt [it's worth making sure that you're completely happy with this idea - specifying a basis means that you can refer to elements of your vector space using column vectors, and the elements of the column vectors are the components of the vector wrt your chosen basis].

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

\begin{pmatrix}a\\b\end{pmatrix}

representing an element of P_1 with respect to our chosen basis, then we want a matrix

\begin{pmatrix}x & y \\ z & w \end{pmatrix}

such that

\begin{pmatrix}x & y \\ z & w \end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix} xa + yb \\ za + wb \end{pmatrix}

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

e_i

(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.

\begin{pmatrix}x & y \\ z & w \end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}x \\ z \end{pmatrix}

. 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

\begin{pmatrix} 2 \\ -1 \end{pmatrix}

(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.

Reply 2

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.

Reply 3

Ateo

Original post by 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.

You want to make use of the fact that

L

is a linear transformation, so it has the property

L(\lambda u + \mu v) = \lambda L(u) + \mu L(v)

for

u,v \in V, \; \; \lambda, \mu \in \mathbb{R}

.

Recall the subspace test, can you verify that

ker(L)

satisfies it?

To show

L

is injective, take two arbitrary elements from

V

and show that

L(v) = L(u) \Rightarrow v = u

(edited 10 years ago)

Reply 4

Smaug123

Original post by shawn_o1

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

By the way, this is an equivalent definition of "subspace": a set K is a subspace of V iff it is the kernel of some linear map L: V->W (some W). Cf normal subgroups being precisely kernels of homomorphisms, ideals being precisely kernels of homomorphisms: they're special cases of "submodules are precisely the kernels of R-homomorphisms". (A linear map is a homomorphism on a field-module = vector space.)