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10.15 Matrices

Introduction with examples

1

\(\mathbb{Q}\subseteq\mathbb{Q}\left\lbrack\sqrt2\right\rbrack=\{a+\sqrt2b:a,b\in\mathbb{Q}\}\subsetneq\mathbb{R}\)

\(+\) and \(\times\) on \(\mathbb{Q}[\sqrt2]\) are those from \(\R\):

\((a+\sqrt2b)+(a^{\prime}+\sqrt2b^{\prime})=\left(a+a^{\prime}\right)+\sqrt2\left(b+b^{\prime}\right)\in\mathbb{Q}\left\lbrack\sqrt2\right\rbrack\)

\((a+\sqrt2b)\left(a^{\prime}+\sqrt2b^{\prime}\right)=\ldots=\left(aa^{\prime}+2bb^{\prime}\right)+\sqrt2\left(ab^{\prime}+a^{\prime}b\right)\in\mathbb{Q}\left\lbrack\sqrt2\right\rbrack\)​

This is a field, the key is to consider the existence of inverses

Given \(a+\sqrt2 b\neq 0(a\neq 0~or~b\neq 0)\), let's compute: \((a+\sqrt2b)(a-\sqrt2b)=a^2-2b^2\neq 0\) since the non-existence of integer of \(\sqrt2\)

Then \((a+\sqrt2b)(\frac{a}{a^2-2b^2}-\frac{\sqrt2b}{a^2-2b^2})=1\) and \((\frac{a}{a^2-2b^2}-\frac{\sqrt2b}{a^2-2b^2})\) is \((a+\sqrt2b)^{-1}\)

2

A relative family of finite fields

Recall: \(\Z_m=\{0,1,...,m-1\}\) with \(+\) and \(\times\) modulo \(m\)

Example

In \(\Z_6\), \(2\times 3=0\) Thus it is not a field

lack of multiplicative inverse 1

Remark

The sum modulo \(m\) has all the required properties

Example

\(\mathbb{Z}_{12}\) \(7+5=0\)

The product modulo \(m\) is associativity and commutativity and \(1\) is the identity

And \(+\) and \(\times\) satisfy distributivity

Claim

Given \(a\in \Z_m\), \(\exists a^*\in \Z_m\) such that \(aa^*=1\) in \(\Z_m\) (\(aa^*\equiv 1 (mod~m)\))if and only if \((a,m)=1\)

Proof

\(\Rightarrow\)) \(aa^{*}\equiv1(mod~m)\Rightarrow m\mid1-aa^{*}\Rightarrow mc=1-aa^{*}\Rightarrow1=mc+aa^{*}\)

Let \(d=(a,m)\), then \(d\mid a\wedge d\mid m\Rightarrow d\mid1\Rightarrow d=1\)

\(\Leftarrow\)) \((a,m)=1\Rightarrow1=ra+sm,for~some~r,s\in\mathbb{Z}\). Then \(1\equiv ra+sm\equiv ra(mod~m)\)

Take \(a^*\): We have \(1\equiv a^{*}a\Rightarrow1=a^{*}a\) in \(\Z_m\)

Conclusion

\(\Z_m\) is a field if and only if \(m\) is a prime number

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Matrices and polynomials over field

Matrices

Let \(\mathbb{F}\) be a field. \(M_{n \times m}(\mathbb{F}) = \{ n \text{ rows}, m \text{ columns, matrices with entries in } \mathbb{F} \}\)

Examples

\(\left( \begin{matrix} 1 & i \\ 1 + i & 0 \end{matrix} \right) \in M_{2 \times 2}(\mathbb{C}) ,\) \(\left( \begin{matrix} 0 & 1 & 3 \\ 2 & 4 & 0 \end{matrix} \right) \in M_{2 \times 3}(\mathbb{Z}_5) ,\)​\(\left(\begin{matrix}\sqrt2\\ 1+\sqrt2\\ 0\end{matrix}\right)\in M_{3\times1}(\mathbb{Q}\left\lbrack\sqrt2\right\rbrack)\)

Sum

The sum of matrices: it is defined entrywise (using the \(+\) of \(\mathbb{F}\))
(They must be of the same size)

Entry refers to the individual element in the matrices

Property

associative / commutative / 0 matrix (all entries = 0) is the neutral element / \(-A\) is the opposite of \(A\) (\(-A_{ij}=-\left(A_{ij}\right)\))

Examples

1

\(A = \left( \begin{matrix} 0 & 2 & 3 \\ 1 & 4 & 3 \end{matrix} \right), \quad B = \left( \begin{matrix} 4 & 1 & 3 \\ 2 & 2 & 2 \end{matrix} \right) \in M_{2 \times 3}(\mathbb{Z}_5)\)

Then \(A+B=\left(\begin{matrix}4 & 3 & 1\\ 3 & 1 & 0\end{matrix}\right),-A=\left(\begin{matrix}0 & 3 & 2\\ 4 & 1 & 2\end{matrix}\right)\)

2

Is there an \(X\in M_{2\times2}(\mathbb{Q}[\sqrt2])\) such that\(\left( \begin{matrix} \sqrt{2} & 1 - \sqrt{2} \\ 1 + \sqrt{2} & -\sqrt{2} \end{matrix} \right) + X = \left( \begin{matrix} 1 & \sqrt{2} \\ -\sqrt{2} & -1 \end{matrix} \right)\) ?

YES!!

\(X = \left( \begin{matrix} 1 & \sqrt{2} \\ -\sqrt{2} & -1 \end{matrix} \right) - \left( \begin{matrix} \sqrt{2} & 1 - \sqrt{2} \\ 1 + \sqrt{2} & -\sqrt{2} \end{matrix} \right)\)​\(= \left( \begin{matrix} 1 - \sqrt{2} & -1 + 2\sqrt{2} \\ -1 - \sqrt{2} & -1 + \sqrt{2} \end{matrix} \right)\)

Product

Examples

1

\(\left( \begin{matrix} i & i & 1 + i \\ 0 & 1 & -i \end{matrix} \right) \cdot \left( \begin{matrix} -i \\ 1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} 1 + i \\ 1 \end{matrix} \right)\)

\(~~~~~~~~2\times3\quad~~~~~\quad3\times1\quad~~\quad2\times1\)

2

\(\left( \begin{matrix} 8 & 1 \\ 0 & 4 \end{matrix} \right) \left( \begin{matrix} 3 & 7 \\ 2 & 0 \end{matrix} \right) = \left( \begin{matrix} 4 & 1 \\ 8 & 0 \end{matrix} \right) \quad (\mathbb{F} = \mathbb{Z}_{11})\)

\(\left( \begin{matrix} 8 & 1 \\ 0 & 4 \end{matrix} \right) \left( \begin{matrix} 3 & 7 \\ 2 & 0 \end{matrix} \right) = \left( \begin{matrix} 0 & 4 \\ 8 & 0 \end{matrix} \right) \quad (\mathbb{F} = \mathbb{Z}_{13})\)

Properties

  • Commutativity(not even defined always): If \(A\) and \(B\) are square of the same square, \(AB\) and \(BA\) is defined.

In general, \(AB\neq BA\)​ * Associativity holds * If \(A\in M_{n\times n}\left(\mathbb{F}\right)\), then \(I_nA=A=AI_n\), where ​\(I_{n}=\left(\begin{matrix}1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 1\end{matrix}\right)\in M_{n\times n}(\mathbb{F})\)​ * In general given \(A\), \(\nexists A^*\) such that \(AA^*=I_n\) or\(A^*A=I_n\) for square matrices

Examples

1

\(\left(\begin{matrix}1 & 1\\ 1 & 1\end{matrix}\right)\left(\begin{matrix}0 & 1\\ 2 & 1\end{matrix}\right)=\left(\begin{matrix}2 & 2\\ 2 & 2\end{matrix}\right)\quad\text{ they are not equal over every field}\quad\)

\(\left( \begin{matrix} 0 & 1 \\ 2 & 1 \end{matrix} \right) \left( \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \right) = \left( \begin{matrix} 1 & 1 \\ 3 & 3 \end{matrix} \right)\)

If \(1 = 2 \implies 0 = 1 \quad \text{ABSURD}\)

2

\(A = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right)\) \(\exists B\) such that \(AB = I\) (of same size)? \(\exists C\) such that \(CA = I\) (of same size)?

\(B\) must be \(3 \times 2\): \(B = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{matrix} \right)\) ✔

\(C\) must be \(3 \times 2\). No because the identity matrix must be \(3\times 3\), then the third column will be 0 ✖️

Remark

If \(A=\left(\begin{matrix}0 & \cdots & 0\\ \ldots & a_2 & \ldots\\ \ldots & \ldots & \ldots\\ \ldots & a_{n} & \ldots\end{matrix}\right)\), then \(AB = \left( \begin{matrix} 0 & \cdots & 0 \\ & \ast & \\ \end{matrix} \right)\)

\(B=\left(\begin{matrix}0 & . & \cdots & .\\ . & \beta_2 & \cdots & \beta_{m}\\ 0 & . & \cdots & .\end{matrix}\right)\), then \(AB=\left(\begin{matrix}0 & & \\ \ldots & \ast & \\ 0 & & \end{matrix}\right)\)

Inverses of square matrices

Example

\(\lambda I = \left( \begin{matrix} \lambda & 0 \\ 0 & \lambda \end{matrix} \right)\)

We have that \(\lambda I \cdot \lambda^{-1} I = I\)

\(\lambda^{-1} I \cdot \lambda I = I \quad (\lambda \neq 0)\)

Notation

\(\lambda A=\lambda I\cdot A\), \((\lambda A)_{ij} = \lambda (A_{ij})\)

Question

Q: Does \(A-\lambda X=B\) have a solution (in \(M_{n \times n}(\mathbb{F})\))?

A: YES!!

\(\lambda X=B-A\implies(\lambda I)\cdot X=(B-A)\Rightarrow\left(\lambda^{-1}I\right)\left(\lambda I\right)X=\left(\lambda^{-1}I\right)\left(B-A\right)\)

\(\implies I\cdot X=\lambda^{-1}(B-A)\implies X=\lambda^{-1}(B-A)\)

Example

When does \(A = \left( \begin{matrix} a & b \\ c & d \end{matrix} \right)\) have a one-side or two-side inverse?

If \(A\) has a left inverse, then \(AB=I\Rightarrow\begin{pmatrix}a & b\\ c & d\end{pmatrix}B=I,\text{in particular }\begin{pmatrix}a & b\\ c & d\end{pmatrix}\neq0.\)

Now, consider
​\(\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}\cdot I=\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}\cdot AB=\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}\\\Rightarrow\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}\begin{pmatrix}a & b\\ c & d\end{pmatrix}B=\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}\\\Rightarrow \begin{pmatrix}ad-bc & 0\\ 0 & ad-bc\end{pmatrix}B=\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}\neq0\\\Rightarrow ad-bc\neq0.\) which is the \(det(A)\neq 0\)

\(\Rightarrow\) two-sided inverse.

If \(A\) has a right inverse, similarly...

Final Remark

\(M_{m\times n}(\mathbb{F})\) and \(M_{n\times m}(\mathbb{F})\) are not fields

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