10.29 Subpaces

Subspace

Definition

Direct sum

Given s subspaces \(W_1,W_2,...,W_k,\) the sum \(W_1+W_2+...+W_k\) is a direct sum if every vector in it can be written as \(w_1+...+w_k\) with \(w_i\in W_i\) in a unique way

Natation

In this case, we write \(W_1\oplus W_2\oplus...\oplus W_k\)

Example

\(V=\R^3,W_1=\{(x,0,x+y):x,y\in \R\},W_2=\{(x,x,0):x\in\R\}\)

\(W_1+W_2=V\), moreover \(W_1\oplus W_2=V\)

\((a,b,c)=(x,0,x+y)+(z,z,0)=(x+z,z,x+y)\Rightarrow z=b,x+b=a\Rightarrow z=b,x=a-b,a-b+y=c\Rightarrow y=c-a+b\)
\(V=\R^3. W_3=\{(x,x+y,0):x,y\in \R\}\)

\(W_1+W_3=V\), but the sum is not direct

\((1,1,0)=\begin{cases}\left(1,0,0\right)+\left(0,1,0\right)\\ (0,0,0)+(1,1,0)\end{cases}\Rightarrow\) the sum is not direct

Proposition

The sum \(W_1+W_2+...+W_k\) is direct iff \(0\) can be written in a unique way as \(w_1+w_2+...+w_k\) with \(w_i\in W_i(w_i=0,\forall i=1,...,k)\)

Proof

\(\Rightarrow\)) Follows from definition

\(\Leftarrow\)) Assume that \(w=w_1+w_2+...+w_{k}=w_1^{\prime}+w_2^{\prime}+...+w_{k}^{\prime}\).

Hence \(0=w-w=(w_1-w_1')+(w_2-w_2')+...+(w_k-w_k')\)

Since \(w_1-w_1^{\prime}\in W_1,...,w_{k}-w_{k}^{\prime}\in W_{k}\), then \(w_1-w_1^{\prime}=0,...,w_{k}-w_{k}^{\prime}=0\Rightarrow w_1=w_1^{\prime},...,w_{k}=w_{k}^{\prime}\)

Analysis: We have \(0\) can be written in a unique way, we want any \(v\) can be written in a unique way

Remark

\(W_1+W_2\) is direct iff \(W_1\cap W_2=\{0\}\)

\(\Rightarrow\)) Analysis: Since \(W_1+W_2\) is direct, then \(0\) can be written as a unique way which is two vector \(v,w\) has both to be \(0\)

NTP \(W_1\cap W_2\) has \(0\), then \(v\) in it then \(v\) is \(0\)

Choose \(v\in W_1\cap W_2\). We have \(v+(-v)=0\). Since \(v\in W_1\cap W_2\subset W_1,-v\in W_1\cap W_2\subset W_2\). Then apply the proposition we have \(v=0\)

\(\Leftarrow\)) Analysis: Since \(W_1\cap W_2=\{0\}\), then \(v\) in it then \(v\) is \(0\)

NTP \(W_1+W_2\) is direct sum, which is to prove if \(0=v+w\), then \(v=w=0\)

Assume that \(v+w=0\), with \(v\in W_1\) and \(w\in W_2\)

Then \(v+w=0\Rightarrow w=-v\Rightarrow v\in W_2\wedge w\in W_1\), thus \(v\in W_1\cap W_2\wedge w\in W_1\cap W_2\)

Therefore, \(v=0=w\), then the sum is direct.

How about \(W_1+W_2+W_3\)

Example

\(W_1=\langle\left(0,1\right)\rangle,W_2=\langle\left(1,0\right)\rangle,W_3=\langle\left(1,1\right)\rangle\)

\(W_1+W_2+W_3=\R^2\), not direct

However, \(W_1\cap W_2\cap W_3=\{0\}\)

Subspaces form systems of linear equation

A system of linear equation is a finite set of linear equation on finitely many unknowns or variables with coefficients on a field \(\mathbb{F}\)

Examples

\(\begin{cases}2x-y+z=\frac12\\ -3y+7z=\frac35\end{cases},\mathbb{F}=\mathbb{Q}\)

\(\begin{cases}x+y-z=0\\ x-y+z=0\\ x+y+z=0\\ -x-y+z=0\end{cases}\)\(\mathbb{F}=\Z_3\). \(-1\) should be \(2\), but it is also ok

Thus in general, every existing system of linear equation can be written as:

\[ AX=b,\text{ for some matrices }A,X\text{ and }b \]

Notation

If \(b=0\), the system is said to be homogeneous

If \(b\neq0\), the system is said to be non-homogeneous

Remark

Notice that \(M_{n\times 1}(\mathbb{F})\) is naturally equivalent to \(\mathbb{F}^n\)

A general system of m linear equations in n variables

\[ \begin{aligned} & \begin{cases}a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_{n}=b_1,\\ a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_{n}=b_2,\\ \vdots\\ a_{m1}x_1+a_{m2}x_2+\cdots+a_{mn}x_{n}=b_{m},\end{cases}\Rightarrow\begin{pmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{pmatrix}\begin{pmatrix}x_1\\ x_2\\ \vdots\\ x_{n}\end{pmatrix}=\begin{pmatrix}b_1\\ b_2\\ \vdots\\ b_{m}\end{pmatrix}\end{aligned} \]

\[ \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix} \in M_{m \times n}(\mathbb{F})~~~~\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} \in M_{n \times 1}(\mathbb{F})~~~~~\begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} \in M_{m \times 1}(\mathbb{F}). \]

Given a system \(AX=b\), a solution of it is an \(x_0\in\mathbb{F}^n\) such that \(A\cdot X_0=b\)

That is \(x_0=\left(x_0^1,x_0^2,x_0^3,\ldots,x_0^{n}\right)\)such that \(\begin{cases}a_{11}x_0^1+a_{12}x_0^2+\cdots+a_{1n}x_0^{n}=b_1\\ a_{21}x_0^1+a_{22}x_0^2+\cdots+a_{2n}x_0^{n}=b_2\\ \vdots\\ a_{m1}x_0^1+a_{m2}x_0^2+\cdots+a_{mn}x_0^{n}=b_{m}\end{cases}\)

Example

\(\mathbb{F} = \mathbb{C}\), \(\begin{cases} ix - y = 1, \\ x + y = 0 \end{cases} \Rightarrow \begin{pmatrix} i & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) with a solution \(x_0 = \frac{1 - i}{2}, y_0 = \frac{i - 1}{2}\).

\(\mathbb{F} = \mathbb{Z}_3\) \(\begin{cases} x + y + z = 0 \\ 2x + y + 2z = 0 \end{cases} \Rightarrow \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & 2 \end{pmatrix} \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

\(\mathbb{F} = \mathbb{Q}\) \(\begin{cases} x + y = 1 \\ y + z = 1 \end{cases} \Rightarrow \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\)

\((0,1,0) \text{ and } (3,-2,3) \text{ are solutions. Moreover, } (3,-2,3) = (0,1,0) + (3,-3,3).\\ \text{ CLAIM: } (0,1,0) + (a,-a,a) \text{ is a solution.}\)

\(\text{Check: }\begin{pmatrix}1 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}a\\ -a\\ \alpha\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix},\begin{pmatrix}1 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}\left\lbrack\begin{pmatrix}0\\ 1\\ 0\end{pmatrix}+\begin{pmatrix}a\\ -a\\ a\end{pmatrix}\right\rbrack=\begin{pmatrix}1\\ 1\end{pmatrix}+a\begin{pmatrix}0\\ 0\end{pmatrix}\)

\(A \cdot (C + D) = AC + AD\)

Proposition

Given a homogeneous system of linear equations with \(n\) unknowns over \(\mathbb{F}\), \(AX = 0\), the set of all solutions of it is a subspace of \(\mathbb{F}^n\).

Proof Let \(W = \{ \text{solutions of } AX = 0 \}\).

(a) Given \(X_1, X_2 \in W \implies A(X_1 + X_2) = AX_1 + AX_2 = 0 + 0 = 0 \implies X_1 + X_2 \in W\).

(b) Given \(X_1\in W,\lambda\in\mathbb{F}\implies A\left(\lambda X_1)=A\left(\left(\lambda I\right)X_1\right)=\left(\lambda I\right.\right)AX_1=\lambda I\cdot0=0\implies\lambda X_1\in W\).

Questions

Q: Given \(AX = b\) and \(X_0\) a solution (\(AX_0 = b\)), is it true that all solutions of \(AX = b\) are of the form \(X = X_0 + X_1\), where \(AX_1 = 0\)? yes

\(A(X_0+X_1)=AX_0+AX_1=b+0=b\)

Q: Is the set of all solutions of \(AX = b \neq 0\) a subspace? no

Take \(X_1, X_2\) solutions. \(AX_1 = b \, ( \neq 0 )\) and \(AX_2 = b \, ( \neq 0 )\).

Compute: \(A \cdot (X_1 + X_2) = A \cdot X_1 + A \cdot X_2 = b + b = 2b \neq b\)

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