MTH3008 Weekly Problems 1

Original Documents: Problem Sheet / My Handwritten Solutions

Vibes: Fairly low difficulty, just learning how to use new algebraic methods and using new formulae; occasionally deriving them from simple summation properties and hard-to-remember matrix properties.

Used Techniques:

• Converting everything into suffix form by applying free and dummy indices.
• Using $\mathbf{a}\cdot \mathbf{b}=a_i{b}_i$, both ways.
• Using $\mathbf{AB}=A_{ik}+B_{kj}$ and $(A_{ij}{)}^T=A_{ji}$.
• Using ${\delta }_{ij}{a}_j=a_i$.
• Using the definition that ${ϵ}_{ijk}=\{\begin{array}{ll}0& \text{if any of }i,j,k\text{ are equal}\\ +1& \text{if }(i,j,k)=(1,2,3),(2,3,1),\text{or }(3,1,2)\\ -1& \text{if }(i,j,k)=(1,3,2),(2,1,3),\text{or }(3,2,1)\end{array}$, i.e., 1 if $(i,j,k)$ is an even permutation of $(1,2,3)$, -1 if it’s an odd permutation, or 0 if any.
• Using $\mathbf{a}\times \mathbf{b}={ϵ}_{ijk}{a}_j{b}_k$.
• Using the property of ${ϵ}_{ijk}$ that you can swap two indices to switch its polarity between $+1$ and $-1$.

1.1. Relate Dot Product and Angle via Suffix Notation

Question

Write in suffix notation: $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$

$$\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\boxed{\frac{a_i{b}_i}{\sqrt{a_k{a}_k}\sqrt{b_k{b}_k}}}.$$

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1.2. Verify Matrix Product Components Using Suffix Notation

Question

Consider the $3\times 3$ matrix $\mathbf{C}$ given by the product $\mathbf{C}=\mathbf{AB}$, where

$$\mathbf{A}=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right),\mathbf{B}=\left(\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right).$$

Verify that $C_{ij}=A_{ik}{B}_{kj}$.

Let

$$A=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right),B=\left(\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right).$$

Let $C=AB$. Then

$$C_{ij}=A_{i1}{B}_{1j}+A_{i2}{B}_{2j}+\cdots +A_{in}{B}_{nj}=\sum _{k=1}^n{A}_{ik}{B}_{kj}=\boxed{A_{ik}{B}_{kj}}$$

in suffix notation.

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1.3. Demonstrate Non-Commutativity of Matrix Multiplication via Suffix Notation

Question

Let $\mathbf{A}$ and $\mathbf{B}$ be the $3\times 3$ matrices

$$\mathbf{A}=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right),\mathbf{B}=\left(\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right).$$

Show, using suffix notation, that $\mathbf{AB}\ne \mathbf{BA}$, i.e. matrix multiplication does not commute.

Let

$$A=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right),B=\left(\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right).$$

Then

$$AB=A_{i1}{B}_{1j}+A_{i2}{B}_{2j}+\cdots +A_{in}{B}_{nj}=\sum _{k=1}^n{A}_{ik}{B}_{kj}=A_{ik}{B}_{kj},$$

and

$$BA=B_{i1}{A}_{1j}+B_{i2}{A}_{2j}+\cdots +B_{in}{A}_{nj}=\sum _{k=1}^n{B}_{ik}{A}_{kj}=B_{ik}{A}_{kj}=A_{kj}{B}_{ik}.$$

Thus

$$AB=A_{ik}{B}_{kj}\ne A_{kj}{B}_{ik}=BA\because ik\ne kj,kj\ne ik.\boxed{}$$

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1.4. Prove Transpose of a Product Using Suffix Notation

Question

Let $\mathbf{A}$ and $\mathbf{B}$ be the $3\times 3$ matrices

$$\mathbf{A}=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right),\mathbf{B}=\left(\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right).$$

Show, using suffix notation, that $(\mathbf{AB}{)}^T={\mathbf{B}}^T{\mathbf{A}}^T$, where ${\mathbf{A}}^T$ is the transpose of $\mathbf{A}$.

Let

$$A=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right),B=\left(\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right).$$

We have

$$(AB{)}^T=((AB{)}_{ji}),$$

so

$$(AB{)}_{ij}^T=(AB{)}_{ji}=A_{jk}{B}_{ki}.$$

On the other hand,

$$(B^T{A}^T{)}_{ij}=(B^T{)}_{ik}(A^T{)}_{kj}=B_{ki}{A}_{jk}=A_{jk}{B}_{ki}.$$

Therefore

$$(AB{)}_{ij}^T=A_{jk}{B}_{ki}=(B^T{A}^T{)}_{ij},$$

so

$$(AB{)}^T=B^T{A}^T.\boxed{}$$

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1.5. Prove Transpose of Triple Matrix Product via Suffix Notation

Question

Let $\mathbf{L}$, $\mathbf{M}$ and $\mathbf{N}$ be three $3\times 3$ matrices. Show, using suffix notation, that $(\mathbf{LMN}{)}^T={\mathbf{N}}^T{\mathbf{M}}^T{\mathbf{L}}^T$.

Let $L,M,N$ be $3\times 3$ matrices. Then

$$(LMN{)}_{ij}=(LM{)}_{ik}{N}_{kj}=(L_{i\ell }{M}_{\ell k})N_{kj}=L_{i\ell }{M}_{\ell k}{N}_{kj}.$$

For the right-hand side,

$$(N^T{M}^T{L}^T{)}_{ij}=(N^T{)}_{ik}(M^T{)}_{k\ell }(L^T{)}_{\ell j}=N_{ki}{M}_{\ell k}{L}_{j\ell }.$$

Now rename the dummy indices to match the pattern of $(LMN{)}_{ij}$. Swapping dummy labels and reordering scalar factors,

$$N_{ki}{M}_{\ell k}{L}_{j\ell }=L_{j\ell }{M}_{\ell k}{N}_{ki}=L_{i\ell }{M}_{\ell k}{N}_{kj}.$$

Thus

$$(LMN{)}_{ij}^T=L_{i\ell }{M}_{\ell k}{N}_{kj}=(N^T{M}^T{L}^T{)}_{ij},$$

so

$$(LMN{)}^T=N^T{M}^T{L}^T.\boxed{}$$

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1.6. Simplify Kronecker Delta Expression and Rewrite in Vector Form

Question

Simplify the suffix notation expression ${\delta }_{ij}{a}_j{b}_{\ell }{c}_k{\delta }_{i\ell }$ and write the result in vector form.

$${\delta }_{ij}{a}_j{b}_{\ell }{c}_k{\delta }_{i\ell }=a_i{b}_{\ell }{c}_k{\delta }_{i\ell }=a_i{b}_i{c}_k.$$

Now $a_i{b}_i=\mathbf{a}\cdot \mathbf{b}$, so

$$a_i{b}_i{c}_k=(\mathbf{a}\cdot \mathbf{b})c_k.$$

In vector form this is

$$\boxed{(\mathbf{a}\cdot \mathbf{b})\mathbf{c}}.$$

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1.7. Evaluate Alternating Tensor Components

Question

Recall the alternating tensor ${\epsilon }_{ijk}$. Evaluate the following in vector notation.

1. ${\epsilon }_{122}$
2. ${\epsilon }_{321}$
3. ${\epsilon }_{223}+{\epsilon }_{111}$

For ${\epsilon }_{ijk}$:

$$\begin{array}{rl}1.& {\epsilon }_{123}=\boxed{1},\\ 2.& {\epsilon }_{321}=\boxed{-1},\\ 3.& {\epsilon }_{112}+{\epsilon }_{122}=0+0=\boxed{0}.\end{array}$$

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1.8. Prove Antisymmetry of the Cross Product Using Suffix Notation

Question

Use suffix notation to show that $\mathbf{a}\times \mathbf{b}=-\mathbf{b}\times \mathbf{a}$.

LHS:

$$\mathbf{a}\times \mathbf{b}=(\mathbf{a}\times \mathbf{b}{)}_i={\epsilon }_{ijk}{a}_j{b}_k.$$

RHS:

$$-\mathbf{b}\times \mathbf{a}=-(\mathbf{b}\times \mathbf{a}{)}_i=-{\epsilon }_{ijk}{b}_j{a}_k.$$

Since scalar multiplication commutes and ${\epsilon }_{ijk}$ is antisymmetric,

$$-{\epsilon }_{ijk}{b}_j{a}_k=-{\epsilon }_{ikj}{a}_j{b}_k={\epsilon }_{ijk}{a}_j{b}_k.$$

Therefore

$$\text{LHS}=\text{RHS},\text{i.e.}\mathbf{a}\times \mathbf{b}=-\mathbf{b}\times \mathbf{a}.\boxed{}$$

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1.9. Convert Dot–Cross Vector Equation to Suffix Notation

Question

Write the vector equation $\mathbf{a}\times \mathbf{b}+(\mathbf{a}\cdot \mathbf{d})\mathbf{c}=\mathbf{e}$ in suffix notation.

$$\mathbf{a}\times \mathbf{b}+(\mathbf{a}\cdot \mathbf{d})\mathbf{c}=\mathbf{e}.$$

In components,

$$(\mathbf{a}\times \mathbf{b}{)}_i+(\mathbf{a}\cdot \mathbf{d})c_i=e_i.$$

Using suffix notation,

$$(\mathbf{a}\times \mathbf{b}{)}_i={\epsilon }_{ijk}{a}_j{b}_k,\mathbf{a}\cdot \mathbf{d}=a_j{d}_j,$$

so

$$\boxed{{\epsilon }_{ijk}{a}_j{b}_k+a_j{d}_j{c}_i=e_i}.$$