02. Inverses and Transposes

Square Matrix

Equal number of rows and columns ($m=n$)

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Transpose Matrix of $n\times n$ matrix $A$

Transpose $A^T$ defined such that

$$(i,j)\text{th entry of }{A}^T=(j,i)\text{th entry of A}$$

Essentially reflecting across the main diagonal of the matrix

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Properties of Transpose

1. Addition and Scalar Multiplication Rules (for $A,B$ being $m\times n$ matrices and $\lambda \in \mathbb{R}$)

$$(A+B{)}^T=A^T+B^T,(\lambda A{)}^T=\lambda A^T$$

2. Product Rule (for $A$ being a $m\times n$ matrix and $B$ being a $n\times p$ matrix)

$$(AB{)}^T=B^T{A}^T$$

3. Involution Rule (for $A$ being a $m\times n$ matrix)

$$(A^T{)}^T=A$$

4. Inverse Rule ($A$ invertible $⟺$ $A^T$ invertible)

$$(A^T{)}^{-1}=(A^{-1}{)}^T$$

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Types of Square Matrices ( $A=(a_{ij})$)

1. Symmetric

$$A^T=A$$

2. Skew-Symmetric (or antisymmetric)

$$A^T=-A$$

Entries along the main diagonal are 0
3) Upper Triangular

$$a_{ij}=0⟺i>j$$

Entries below the diagonal are zero
4) Strictly Upper Triangular

$$a_{ij}=0⟺i\ge j$$

Entries on or below the main diagonal are zero
5) Triangular

$$\text{Either upper or lower triangular}$$

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Determining Invertibiilty

Let $A$ be a $n\times n$ matrix

Suppose there exists a sequence of EROs such that $A$ is taken to it’s RRE form $R$
Applying the same sequence of EROs to the augmented matrix $(A\mid I_n)$ to take it to $(R\mid P)$ for some matrix $P$

1. $R=I_n$ then $A$ is invertible and $P=A^{-1}$

2. $R\ne I_n$ then $A$ is singular

Proof

Suppose the finite sequence of EROs that reduce $A$ to $R$ are $E_1,E_2,\cdots ,E_k$
Applying it to $(A|I_n)$ s.t.

$$(E_k{E}_{k-1}\cdots E_{1}A\mid E_k{E}_{k-1}\cdots E_1)=(R\mid P)$$

With $R=PA$ and $P=E_K{E}_{k-1}\cdots E_1$

If $R=I_n$ then as elementary matrices are invertible then

$$PA=(E_k{E}_{k-1}\cdots E_1)A=I_n⟹A^{-1}=E_k{E}_{k-1}\cdots E_1=P$$

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If $R\ne I_n$ then $R$ contains at least one zero row
As in RRE form the zero row is at the bottom then

$$(0,0,\cdots ,1)(PA)=0\text{(selects the last row)}$$

Then as $P$ is invertible and if we assume $A$ is also invertible (i.e. $A^{-1}$) exists then

$$(0,0,\cdots ,1)=0\cdot A^{-1}{P}^{-1}=0$$

This is a contradiction so $A$ is not invertible! Therefore $A$ is singular

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Orthogonal $n\times n$ matrix

$$A^T=A^{-1}$$

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Properties of Orthogonal $n\times n$ Matrices

1. $AB$ and $A^{-1}$ are orthogonal
   Linking it to Term 2 Groups the $n\times n$ matrices form a group $O(n)$

2. $A$ is orthogonal $⟺$ the columns (or rows) are $n$ unit length, mutually perpendicular vectors

3. $A$ preserves the dot product i.e. If $\mathbf{x},\mathbf{y}\in {\mathbb{R}}_{col}^n$ then $A\mathbf{x}\cdot A\mathbf{y}=x\cdot y$

Proof

$(AB{)}^T=B^T{A}^T=B^{-1}{A}^{-1}=(AB{)}^{-1}$
$(A^{-1}{)}^T=(A^T{)}^T=A=(A^{-1}{)}^{-1}$
2

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Preservation of Dot Product $⟺$ Orthogonal

$$A\text{ is orthogonal}⟺A\mathbf{x}\cdot A\mathbf{y}=x\cdot y\text{ for all }\mathbf{x},\mathbf{y}\in {\mathbb{R}}_{\mathrm{col}}^n$$

Proof

Using the fact that ${\mathbf{x}}^T\mathbf{y}=x\cdot y$

$$\begin{array}{rl}A\mathbf{x}\cdot \mathbf{A}\cdot \mathbf{y}& =\mathbf{x}\cdot \mathbf{y}\\ ⟺(A\mathbf{x}{)}^T(A\mathbf{y})& ={\mathbf{x}}^T\mathbf{y}\\ ⟺{\mathbf{x}}^T{A}^{T}A\mathbf{y}& =x^{T}y\end{array}$$

This is true if we can show that $A^{T}A=I_n$

By setting $\mathbf{x}={\mathbf{e}}_i$ and $\mathbf{y}={\mathbf{e}}_j$ (standard basis of ${\mathbb{R}}_{col}^n$)
Then $x^T{A}^{T}Ay$ selects the $(i,j)$th entry of $A^{T}A$
We also have $x^{T}y={\delta }_{ij}$ (Kronecker Delta)
Hence

$$(i,j)\text{th entry of }{A}^{T}A={\delta }_{ij}=(i,j)\text{th entry of }{I}_n$$

Therefore $A^{T}A=I_n$ hence ${\mathbf{x}}^T{A}^{T}A\mathbf{y}={\mathbf{x}}^T\mathbf{y}$

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