Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

This exercise shows one way to define the quaternions,discovered in 1843 by the Irish mathematician Sir W.R. Hamilton (1805-1865).Consider the set H of all $4 \times 4$ matrices M of the form
$M = [\begin{matrix} p & - q & - r & - s \\ q & p & s & - r \\ r & - s & p & q \\ s & r & - q & p \end{matrix}]$
**where p,q,r,s are arbitrary real numbers.We can write M more sufficiently in partitioned form as
$M = (\begin{matrix} A & - B^{T} \\ B & A^{T} \end{matrix})$

where A and B are rotation-scaling matrices.
a.Show that H is closed under addition:If M and N are in H then so is $M + N$
M+Nb.Show that H is closed under scalar multiplication .If M is in H and K is an arbitrary scalar then kM is in H.c.Parts (a) and (b) Show that H is a subspace of the linear space $R^{4 \times 4}$ .Find a basis of H and thus determine the dimension of H.
d.Show that H is closed under multiplication If M and N are in H then so is MN.
e.Show that if M is in H,then so is $M^{T}$ .
f.For a matrix M in H compute $M^{T} M$ .
g.Which matrices M in H are invertible.If a matrix M in H is invertible is $M^{- 1}$ necessarily in H as well?
h. If M and N are in H,does the equation $M N = N M$ always hold?**

(a) It is proved that M+N is also closed under addition.

(b) kM also closed under scalar multiplication.

(d) MN is also closed under addition.

(e) It is proved that M is in H and $M^{T}$ is also in H.

(f) $M^{T} M = (\begin{matrix} A^{2} + {B^{T}}^{2} & A B - A^{T} B^{T} \\ A B - A^{T} B^{T} & B^{2} + A^{T 2} \end{matrix})$

(g) M is invertible then M^-1be also in the invertible as well.

(h) Yes, $M N = N M$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1 Explanation for the dimension of the space

If a linear space has a basis with nelements, then all other bases of V, consists of n elements as well.

Also, we say that n is the dimension of V: $d i m (V) = n$

(a)Step2:Matrix under addition

Consider the set H of all $4 \times 4$ matrices M of the form.

$M = [\begin{array}{l} p - q - r - s \\ q p s - r \\ r - s p q \\ s r - q p \end{array}]$

where p,q,r,s are arbitrary real numbers.We can write M more sufficiently in partitioned form as:

$M = (\begin{matrix} A & - B^{T} \\ B & A^{T} \end{matrix})$

where A and B are rotation-scaling matrices.

Assume that H is the subspace which is closed under addition.

We know that M has the partitioned form and N also be considered in the same partitioned form.

Hence M and N are in the subspace H.

Since we know that the subspace H is closed under addition,then M+N is also closed under addition.

(b)Step:3 Matrix under scalar multiplication

Consider the set H of all $4 \times 4$ matrices M of the form.

$M = [\begin{array}{l} p - q - r - s \\ q p s - r \\ r - s p q \\ s r - q p \end{array}]$

where p,q,r,s are arbitrary real numbers .We can write M more sufficiently in partitioned form as

$M = (\begin{matrix} A & - B^{T} \\ B & A^{T} \end{matrix})$

where A and B are rotation-scaling matrices.

Assume that H is the subspace which is closed under scalar multiplication.

We know that M has the partitioned form and k also be considered in the arbitrary scalar

Hence M is in the subspace H.

Since we know that the subspace H is closed under scalar multiplication,then the arbitrary k is kM also closed under scalar multiplication.

(c)Step 4: Dimension of the subspace H

Assume that a linear space has a basis with n elements then all other bases of H, consists of n elements as well.

Also we say that n is the dimension of H: $\dim (H) = n$

Here n is two then the dimension in the subspace of the linear space H is 2.

$\dim (H) = 2$

Hence the solution

(d)Step 6: Matrix under multiplication

Consider the set H of all $4 \times 4$ matrices M of the form.

$M = [\begin{array}{l} p - q - r - s \\ q p s - r \\ r - s p q \\ s r - q p \end{array}]$

where p,q,r,s are arbitrary real numbers .We can write M more sufficiently in partitioned form as:

$M = (\begin{matrix} A & - B^{T} \\ B & A^{T} \end{matrix})$

where A and B are rotation-scaling matrices.

Assume that H is the subspace which is closed under multiplication.

We know that M has the partitioned form and N also be considered in the same partitioned form.

Hence M and N are in the subspace H.

Since we know that the subspace H is closed under multiplication,then MN is also closed under addition.

(e)Step5: Transpose of a matrix

Consider the set H of all $4 \times 4$ matrices M of the form.

$M = [\begin{array}{l} p - q - r - s \\ q p s - r \\ r - s p q \\ s r - q p \end{array}]$

where p,q,r,s are arbitrary real numbers .We can write M more sufficiently in partitioned form as:

$M = (\begin{matrix} A & - B^{T} \\ B & A^{T} \end{matrix})$

where A and B are rotation-scaling matrices.

Assume that H is the subspace

We know that M has the partitioned form.

Now we have to find the transpose of the matrix M.

$M^{T} = (\begin{matrix} A & B \\ - B^{T} & A^{T} \end{matrix})$

(f)

The matrix and the transpose of the matrix be as follows

$\begin{matrix} M^{T} M = (\begin{matrix} A & B \\ - B^{T} & A^{T} \end{matrix}) (\begin{matrix} A & - B^{T} \\ B & A^{T} \end{matrix}) \\ = (\begin{matrix} A^{2} + {B^{T}}^{2} & A B - A^{T} B^{T} \\ A B - A^{T} B^{T} & B^{2} + A^{T 2} \end{matrix}) \end{matrix}$

Thus, the solution.

(g)Step6: Invertible of a matrix

Consider the set H of all $4 \times 4$ matrices M of the form.

$M = [\begin{array}{l} p - q - r - s \\ q p s - r \\ r - s p q \\ s r - q p \end{array}]$

where p,q,r,s are arbitrary real numbers .We can write M more sufficiently in partitioned form as:

$M = (\begin{matrix} A & - B^{T} \\ B & A^{T} \end{matrix})$

where A and B are rotation-scaling matrices.

Assume that H is the subspace

We know that M has the partitioned form and M^-1 be the invertible matrix

Hence M and M^-1be the matrix in the subspace H

Thus, the M is invertible then M^-1be also in the invertible as well

(h)

Using the matrix M and N as the partitioned form,then $M N - N M = 0$ be the equation.

Hence the equation $M N = N M$ holds

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

Step 1 Explanation for the dimension of the space

(a)Step2:Matrix under addition

(b)Step:3 Matrix under scalar multiplication

(c)Step 4: Dimension of the subspace H

(d)Step 6: Matrix under multiplication

(e)Step5: Transpose of a matrix

(g)Step6: Invertible of a matrix

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

Step 1 Explanation for the dimension of the space

(a)Step2:Matrix under addition

(b)Step:3 Matrix under scalar multiplication

(c)Step 4: Dimension of the subspace H

(d)Step 6: Matrix under multiplication

(e)Step5: Transpose of a matrix

(g)Step6: Invertible of a matrix

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