Suggested languages for you:

Americas

Europe

Q64E

Expert-verified
Found in: Page 235

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# 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 ${\mathbf{4}}{\mathbf{×}}{\mathbf{4}}$ matrices M of the form${\mathbit{M}}{\mathbf{=}}\left[\begin{array}{cccc}p& -q& -r& -s\\ q& p& s& -r\\ r& -s& p& q\\ s& r& -q& p\end{array}\right]$where p,q,r,s are arbitrary real numbers.We can write M more sufficiently in partitioned form as ${\mathbit{M}}{\mathbf{=}}\left(\begin{array}{cc}A& -{B}^{T}\\ B& {A}^{T}\end{array}\right)$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 ${\mathbit{M}}{\mathbf{+}}{\mathbit{N}}$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 ${{\mathbit{R}}}^{\mathbf{4}\mathbf{×}\mathbf{4}}$ .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 ${{\mathbit{M}}}^{{\mathbf{T}}}$.f.For a matrix M in H compute ${{\mathbit{M}}}^{{\mathbf{T}}}{\mathbit{M}}$.g.Which matrices M in H are invertible.If a matrix M in H is invertible is ${{\mathbit{M}}}^{\mathbf{-}\mathbf{1}}$ necessarily in H as well?h. If M and N are in H,does the equation ${\mathbit{M}}{\mathbit{N}}{\mathbf{=}}{\mathbit{N}}{\mathbit{M}}$ always hold?

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

(b) kM also closed under scalar multiplication.

(c) $\mathrm{dim}\left(H\right)=2$

(d) MN is also closed under addition.

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

(f) ${M}^{T}M=\left(\begin{array}{cc}{A}^{2}+{{B}^{T}}^{2}& AB-{A}^{T}{B}^{T}\\ AB-{A}^{T}{B}^{T}& {B}^{2}+{A}^{T2}\end{array}\right)$

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

(h) Yes, $MN=NM$ .

See the step by step solution

## 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: ${d}{i}{m}\left(V\right){=}{n}$

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

$M=\left[\begin{array}{l}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-s\\ q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-r\\ r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}q\\ s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\end{array}\right]$

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

$M=\left(\begin{array}{cc}A& -{B}^{T}\\ B& {A}^{T}\end{array}\right)$

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 $4×4$ matrices M of the form.

$M=\left[\begin{array}{l}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-s\\ q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-r\\ r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}q\\ s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\end{array}\right]$

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

$M=\left(\begin{array}{cc}A& -{B}^{T}\\ B& {A}^{T}\end{array}\right)$

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: $\mathrm{dim}\left(H\right)=n$

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

$\mathrm{dim}\left(H\right)=2$

Hence the solution

## (d)Step 6: Matrix under multiplication

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

$M=\left[\begin{array}{l}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-s\\ q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-r\\ r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}q\\ s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\end{array}\right]$

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

$M=\left(\begin{array}{cc}A& -{B}^{T}\\ B& {A}^{T}\end{array}\right)$

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 $4×4$ matrices M of the form.

$M=\left[\begin{array}{l}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-s\\ q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-r\\ r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}q\\ s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\end{array}\right]$

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

$M=\left(\begin{array}{cc}A& -{B}^{T}\\ B& {A}^{T}\end{array}\right)$

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.

${M}^{T}=\left(\begin{array}{cc}A& B\\ -{B}^{T}& {A}^{T}\end{array}\right)$

(f)

The matrix and the transpose of the matrix be as follows

$\begin{array}{c}{M}^{T}M=\left(\begin{array}{cc}A& B\\ -{B}^{T}& {A}^{T}\end{array}\right)\left(\begin{array}{cc}A& -{B}^{T}\\ B& {A}^{T}\end{array}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\begin{array}{cc}{A}^{2}+{{B}^{T}}^{2}& AB-{A}^{T}{B}^{T}\\ AB-{A}^{T}{B}^{T}& {B}^{2}+{A}^{T2}\end{array}\right)\end{array}$

Thus, the solution.

## (g)Step6: Invertible of a matrix

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

$M=\left[\begin{array}{l}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-s\\ q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-r\\ r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}q\\ s\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}r\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-q\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p\end{array}\right]$

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

$M=\left(\begin{array}{cc}A& -{B}^{T}\\ B& {A}^{T}\end{array}\right)$

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-1 be the matrix in the subspace H

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

(h)

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

Hence the equation $MN=NM$ holds