Let WC R¹ be a subspace. Recall from Example 17.2 that the reflection of a vector u in W is the linear transformation given by Rw (u) = 2 prw (u) — u. In Practice Questions 7, question 8 you are asked to show that this is an orthogonal linear transformation. The solution there uses the fact that prw (u) - u is orthogonal to everything in W. Here, we will look at the special case where W is 1-dimensional. Sup- pose then that W is a 1-dimensional subspace with orthonormal basis {v}. (a) Use Definition 18.1 and the formula for orthogonal projection from Theorem 17.1 to show that Rw is an orthogonal linear trans- formation. [Note: You are being asked to use the formula for orthogonal projection directly here, therefore you must not use the method described above from the practice questions solution. You should, however, look at that question since the general technique will be useful.] Now let V = span {(1, 2)}. (b) Calculate the standard matrix for Ry.

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Chapter2: Second-order Linear Odes
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please show all working out  therorm 17.1 is also attached below

Let WC R be a subspace. Recall from Example 17.2 that the
reflection of a vector u in W is the linear transformation given by
Rw (u) = 2 prw (u) – u.
In Practice Questions 7, question 8 you are asked to show that this is
an orthogonal linear transformation. The solution there uses the fact
that prw (u) u is orthogonal to everything in W.
Here, we will look at the special case where W is 1-dimensional. Sup-
pose then that W is a 1-dimensional subspace with orthonormal basis
{v}.
(a) Use Definition 18.1 and the formula for orthogonal projection
from Theorem 17.1 to show that Rw is an orthogonal linear trans-
formation.
[Note: You are being asked to use the formula for orthogonal
projection directly here, therefore you must not use the method
described above from the practice questions solution. You should,
however, look at that question since the general technique will be
useful.]
=
= span {(1,2)}.
(b) Calculate the standard matrix for Ry.
Now let V
Transcribed Image Text:Let WC R be a subspace. Recall from Example 17.2 that the reflection of a vector u in W is the linear transformation given by Rw (u) = 2 prw (u) – u. In Practice Questions 7, question 8 you are asked to show that this is an orthogonal linear transformation. The solution there uses the fact that prw (u) u is orthogonal to everything in W. Here, we will look at the special case where W is 1-dimensional. Sup- pose then that W is a 1-dimensional subspace with orthonormal basis {v}. (a) Use Definition 18.1 and the formula for orthogonal projection from Theorem 17.1 to show that Rw is an orthogonal linear trans- formation. [Note: You are being asked to use the formula for orthogonal projection directly here, therefore you must not use the method described above from the practice questions solution. You should, however, look at that question since the general technique will be useful.] = = span {(1,2)}. (b) Calculate the standard matrix for Ry. Now let V
Theorem 17.1. If V is a subspace of R" and wER" there is a unique vector
pry (W) V with the property that w pry (w) is orthogonal to V. Moreover
if V has an orthonormal basis {V₁,...,Vk} then
pry (w) (w v₁)V₁ +
+ (w. Vk)Vk.
Proof. We know already that
pry (w) = (w.V₁) V₁+·
.... + (w. Vk) Vk
satisfies w - pry(w) is orthogonal to V.
Why is pry (w) unique? Assume we also had u V with w
to V. Then
-
u orthogonal
V = = (w - pry (w)) - (w-u) = u-pry (w)
is a vector which is both in V and orthogonal to V. But then vv = 0. So
u = pry (w).
We call pry (w) the orthogonal projection of w onto the subspace V.
Notice that we have
w = pry (w) + (w - pry (w))
the sum of the orthogonal projection of w onto the subspace V and the com-
ponent of w orthogonal to the subspace V. Notice this also shows that the
component of w orthogonal to V is unique.
Transcribed Image Text:Theorem 17.1. If V is a subspace of R" and wER" there is a unique vector pry (W) V with the property that w pry (w) is orthogonal to V. Moreover if V has an orthonormal basis {V₁,...,Vk} then pry (w) (w v₁)V₁ + + (w. Vk)Vk. Proof. We know already that pry (w) = (w.V₁) V₁+· .... + (w. Vk) Vk satisfies w - pry(w) is orthogonal to V. Why is pry (w) unique? Assume we also had u V with w to V. Then - u orthogonal V = = (w - pry (w)) - (w-u) = u-pry (w) is a vector which is both in V and orthogonal to V. But then vv = 0. So u = pry (w). We call pry (w) the orthogonal projection of w onto the subspace V. Notice that we have w = pry (w) + (w - pry (w)) the sum of the orthogonal projection of w onto the subspace V and the com- ponent of w orthogonal to the subspace V. Notice this also shows that the component of w orthogonal to V is unique.
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