{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 2: Vector Spaces"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.1_1: Vector_Spaces_and_subspaces.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 70\n",
"clear;\n",
"close;\n",
"clc;\n",
"disp('Consider all vectors in R^2 whose components are positive or zero')\n",
"disp('The subset is first Quadrant of x-y plane,the co-ordinates satisfy x>=0 and y>=0.It is not a subspace.')\n",
"v=[1,1];\n",
"disp(v,'If the Vector=');\n",
"disp('Taking a scalar,c=-1')\n",
"c=-1; //scalar\n",
"disp(c*v,'c*v=') \n",
"disp('It lies in third Quadrant instead of first,Hence violating the rule(ii).')\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.1_2: Vector_Spaces_and_subspaces.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 71\n",
"clear;\n",
"close;\n",
"clc;\n",
"disp('Take vector space of 3X3 matrices')\n",
"disp('One possible subspace is the set of lower triangular matrices,Another is set of symmetric matrices')\n",
"disp('A+B,cA are both lower triangular if A and B are lower triangular ,and are symmetric if A and B are symmetric and Zero matrix is in both subspaces')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3_1: Linear_Independence.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 92\n",
"clear;\n",
"close;\n",
"clc;\n",
"disp('For linear independence,C1V1+C2V2+......CkVk=0')\n",
"disp('If we choose V1=zero vector,then the set is linearly dependent.We may choose C1=3 and all other Ci=0;this is a non-trival solution that produces zero.')\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3_2: Linear_Independence.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 92\n",
"clear;\n",
"close;\n",
"clc;\n",
"A=[1 3 3 2;2 6 9 5;-1 -3 3 0];\n",
"disp('Given matrix:')\n",
"disp(A)\n",
"B=A;\n",
"disp('C2->C2-3*C1')\n",
"A(:,2)=A(:,2)-3*A(:,1);\n",
"disp(A)\n",
"disp('Here,C2=3*C1,Therefore the columns are linearly dependent.')\n",
"disp('R3->R3-2*R2+5*R1')\n",
"B(3,:)=B(3,:)-2*B(2,:)+5*B(1,:);\n",
"disp(B)\n",
"disp('Here R3=R3-2*R2+5*R1,therefore the rows are linearly dependent.')\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3_3: Linear_Independence.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"close;\n",
"clc;\n",
"A=[3 4 2;0 1 5;0 0 2];\n",
"disp(A,'A=');\n",
"disp('The columns of the triangular matrix are linearly independent,it has no zeros on the diagonal');\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3_4: Linear_Independence.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 93\n",
"clear;\n",
"close;\n",
"clc;\n",
"disp('The columns of the nxn identity matrix are independent.')\n",
"n=input('Enter n:');\n",
"I=eye(n,n);\n",
"disp(I,'I=');\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3_5: Linear_Independence.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 93\n",
"clear;\n",
"close;\n",
"clc;\n",
"disp('Three columns in R2 cannot be independent.')\n",
"A=[1 2 1;1 2 3];\n",
"disp(A,'Given matrix:')\n",
"[L,U]=lu(A);\n",
"disp(U,'U=');\n",
"disp('If c3 is 1 ,then back-substitution Uc=0 gives c2=-1,c1=1,With these three weights,the first column minus the second plus the third equals zero ,therefore linearly dependent.')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3_6: Linear_Independence.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 93\n",
"clear;\n",
"close;\n",
"clc;\n",
"disp('The vectors w1=(1,0,0),w2=(0,1,0),w3=(-2,0,0) span a plane (x-y plane) in R3. The first two vectors also span this plane, whereas w1 and w3 span only a line.');\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3_7: Linear_Independence.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 93\n",
"clear;\n",
"close;\n",
"clc;\n",
"disp('The column space of A is excatly the space that is spanned by its columns.The row space is spanned by the rows.The definition is made to order.Multiplying A by any x gives a combination of columns; it is a vector Ax in the column space. The coordinate vectors e_1,....e_n coming from the identity matrix span Rn. Every vector b=(b_1.....,b_n) is a combination of those columns.In this example the weights are the components b_i themselves:b=b_1e_1+.....+b_ne_n.But the columns of other matrices also span R_.')\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3_8: Basis_for_a_vector_space.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 93\n",
"clear;\n",
"close;\n",
"clc;\n",
"disp('Here,the vector v1 by itself is linearly independent , but it fails to span R2.The three vectors v1,v2,v3 certainly span R2, but are not independent. Any two of these vectors say v1 and v2 have both properties -they span and they are independent.So they form a basis.(A vector space does not have a unique basis)')\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3_9: Basis_for_a_vector_space.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 96\n",
"clear;\n",
"close;\n",
"clc;\n",
"disp('These four columns span the column space U,but they are not independent.')\n",
"U=[1 3 3 2;0 0 3 1;0 0 0 0];\n",
"disp(U,'U=');\n",
"disp('The columns that contains pivots (here 1st & 3rd) are a basis for the column space. These columns are independent, and it is easy to see that they span the space.In fact,the column space of U is just the x-y plane withinn R3. C(U) is not the same as the column space C(A) before elimination-but the number of independent columns did not change.')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.4_1: The_four_fundamental_subspaces.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 107\n",
"clear;\n",
"close;\n",
"clc;\n",
"A=[1 2;3 6];\n",
"disp(A,'A=');\n",
"[m,n]=size(A);\n",
"disp(m,'m=');\n",
"disp(n,'n=');\n",
"[v,pivot]=rref(A);\n",
"r=length(pivot);\n",
"disp(r,'rank=')\n",
"cs=A(:,pivot);\n",
"disp(cs,'Column space=');\n",
"ns=kernel(A);\n",
"disp(ns,'Null space=');\n",
"rs=v(1:r,:)';\n",
"disp(rs,'Row space=')\n",
"lns=kernel(A');\n",
"disp(lns,'Left null sapce=');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.4_2: Inverse_of_a_mxn_matrix.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 108\n",
"clear;\n",
"close;\n",
"clc;\n",
"A=[4 0 0;0 5 0];\n",
"disp(A,'A=');\n",
"[m,n]=size(A);\n",
"disp(m,'m=');\n",
"disp(n,'n=')\n",
"r=rank(A);\n",
"disp(r,'rank=');\n",
"disp('since m=r=2 ,there exists a right inverse .');\n",
"C=A'*inv(A*A');\n",
"disp(C,'Best right inverse=')\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.5_1: Networks_and_discrete_applied_mathematics.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//page 121\n",
"clear;\n",
"close;\n",
"clc;\n",
"disp('Applying current law A''y=f at nodes 1,2,3:')\n",
"A=[-1 1 0;0 -1 1; -1 0 1;0 0 -1;-1 0 0];\n",
"disp(A','A''=');\n",
"C=diag(rand(5,1)); //Taking some values for the resistances.\n",
"b=zeros(5,1);\n",
"b(3,1)=rand(1);//Taking some value of the battery.\n",
"f=zeros(3,1);\n",
"f(2,1)=rand(1);//Taking some value of the current source.\n",
"B=[b;f];\n",
"disp('The other equation is inv(C)y+Ax=b.The block form of the two equations is:')\n",
"C=[inv(C) A;A' zeros(3,3)];\n",
"disp(C);\n",
"X=['y1';'y2';'y3';'y4';'y5';'x1';'x2';'x3'];\n",
"disp(X,'X=')\n",
"X=C\B;\n",
"disp(X,'X=');\n",
"//end"
   ]
   }
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