(a) By plugging in different possibilities for v and w, show that if Σr=ΣWas holds for all u, w then ijk show that Σrkirkj= bij. (b) Show the converse of (a), namely: given that Σrurks = bi Σταντάμενος της Στ for all e, w. (e) Show that the expression ₁ raj= 6; can be rewritten in matrix form as: RTR=I.
(a) By plugging in different possibilities for v and w, show that if Σr=ΣWas holds for all u, w then ijk show that Σrkirkj= bij. (b) Show the converse of (a), namely: given that Σrurks = bi Σταντάμενος της Στ for all e, w. (e) Show that the expression ₁ raj= 6; can be rewritten in matrix form as: RTR=I.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please do Exercise 11.6.3 part ABC and please show step by step and explain.
![So let's go back to the first condition for rotation matrices, namely that
they preserve inner products: Rv-Rw = v-w. Let's rewrite this in coordinate
and
notation. First, note that [Ru] and [Rw) can be written as Σ,
Ejrkju, respectively, where raj is the (k. j) entry of R. Therefore we have:
Re-Rw == =Σ[Re]k[Rw]k
- Σ( ) ( )
-Σ(Σ
Tkili
TkjWj
Recall our rotation condition: Rv Rw=vw, which must be true for any
two vectors and w. In summation notation, this becomes:
ΣrkirkjVw₁ = vmwm
i.j.k
Now let's consider different possibilities for v and w. For example we may
let v = = [1,0,0]T. this means that v;= 81 and w; = 61, where & is our
old friend the Kronecker delta. Plugging this into our summation notation
expression gives:
Σrkirkj818j1 = 8m18m1.
ij.k
Because of the 5's, when we sum over i, j, and m the only terms that con-
tribute will be i=j= m = 1. In summary, we obtain:
-Σ(rivi)(rkjWj)
ijk
- Erwerkenne
i.j.k
Σ
Using this strategy, we can obtain a whole bunch of identities:
holds for all v, w then
Exercise 11.6.3.
(a) By plugging in different possibilities for v and w, show that if
ΣrkirkjV;Wj = ΣmWm
show that
ijk
Tk1rkl = 1.
i.j.k
Στ
(b) Show the converse of (a), namely: given that
Σksk;= bij
m
Tkirkj= dij.
Turkity= ImWm
for alle, w.
(e) Show that the expression E = dj can be rewritten in matrix
form as:
RTR = I.
We summarize these results in a proposition:
Proposition 11.6.4. A 3 x 3 matrix R is rotation matrix if and only if
det (R) > 0 and RTR = I.
Please do Exercise 11.6.3 part ABC and please
show step by step and explain](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892e817a-9b32-4eeb-b8fc-5dd7ffde6479%2F4e5e84db-e98b-4daa-9fe2-ac6905803b88%2Fli251xbt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:So let's go back to the first condition for rotation matrices, namely that
they preserve inner products: Rv-Rw = v-w. Let's rewrite this in coordinate
and
notation. First, note that [Ru] and [Rw) can be written as Σ,
Ejrkju, respectively, where raj is the (k. j) entry of R. Therefore we have:
Re-Rw == =Σ[Re]k[Rw]k
- Σ( ) ( )
-Σ(Σ
Tkili
TkjWj
Recall our rotation condition: Rv Rw=vw, which must be true for any
two vectors and w. In summation notation, this becomes:
ΣrkirkjVw₁ = vmwm
i.j.k
Now let's consider different possibilities for v and w. For example we may
let v = = [1,0,0]T. this means that v;= 81 and w; = 61, where & is our
old friend the Kronecker delta. Plugging this into our summation notation
expression gives:
Σrkirkj818j1 = 8m18m1.
ij.k
Because of the 5's, when we sum over i, j, and m the only terms that con-
tribute will be i=j= m = 1. In summary, we obtain:
-Σ(rivi)(rkjWj)
ijk
- Erwerkenne
i.j.k
Σ
Using this strategy, we can obtain a whole bunch of identities:
holds for all v, w then
Exercise 11.6.3.
(a) By plugging in different possibilities for v and w, show that if
ΣrkirkjV;Wj = ΣmWm
show that
ijk
Tk1rkl = 1.
i.j.k
Στ
(b) Show the converse of (a), namely: given that
Σksk;= bij
m
Tkirkj= dij.
Turkity= ImWm
for alle, w.
(e) Show that the expression E = dj can be rewritten in matrix
form as:
RTR = I.
We summarize these results in a proposition:
Proposition 11.6.4. A 3 x 3 matrix R is rotation matrix if and only if
det (R) > 0 and RTR = I.
Please do Exercise 11.6.3 part ABC and please
show step by step and explain
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