Math 307 Spring, 2005 Hentzel Time: 10:00 to 10:50 MWF Room: 205 Carver Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu Assignment: Write the matrix of the equation 2 2 2 x + 2 y + 3 z + 2 x y + 4 x z + 6 y z = 13 Make the substitution | x | | 1/3 2/3 -2/3 | | x' | | y | = | 2/3 1/3 2/3 | | y' | | z | | -2/3 2/3 1/3 | | z' | March 21 4.6 Rank Main Idea: Review of various bases and dimensions of a matrix and change of basis. Key Words: Rank, Row Rank, Column Rank, Row Space, Column Space, Null Space. Goal: Understand the definitions of the spaces, their ranks, and the relationship to RCF. Page 263 Example 1 | -2 -5 8 0 -17 | A = | 1 3 -5 1 5 | | 3 11 -19 7 1 | | 1 7 -13 5 -3 | Find bases for the Row Space, the Column Space, the Null Space. The Row Space of A is the space spanned by the row of A. The Column Space of A is the space spanned by the columns of A. The Null Space of A is the space of all Vectors X such that AX = 0. One gathers all the information one needs from the Row Canonical form of A. ---------------------------------------------------------- | -2 -5 8 0 -17 | A = | 1 3 -5 1 5 | | 3 11 -19 7 1 | | 1 7 -13 5 -3 | | 1 3 -5 1 5 | | -2 -5 8 0 -17 | | 3 11 -19 7 1 | | 1 7 -13 5 -3 | | 1 3 -5 1 5 | | 0 1 -2 2 -7 | | 0 2 -4 4 -14 | | 0 4 -8 4 -8 | | 1 0 1 -5 26 | | 0 1 -2 2 -7 | | 0 0 0 0 0 | | 0 0 0 -4 20 | | 1 0 1 -5 26 | | 0 1 -2 2 -7 | | 0 0 0 0 0 | | 0 0 0 1 -5 | x y z=a w u=b | 1 0 1 0 1 | | 0 1 -2 0 3 | | 0 0 0 1 -5 | | 0 0 0 0 0 | The Row Rank of A = The Column Rank of A = The Rank of A = 3 Columns 1 2 and 4 of A are a basis of the Column Space of A. | -2 | | -5 | | 0 | Basis of Column Space of A | 1 | | 3 | | 1 | | 3 |,| 11 |,| 7 | | 1 | | 7 | | 5 | Basis of the Null Space of A | x | | -1 | | -1 | | y | | 2 | | -3 | | z | = a | 1 | + b | 0 | | w | | 0 | | 5 | | u | | 0 | | 1 | The dimension of the Null Space of A is 2. A basis of the Null Space of A is: | -1 | | -1 | | 2 | | -3 | | 1 | , | 0 | | 0 | | 5 | | 0 | | 1 | Columns 3 and 5 are not used in the basis. They have to be linear combinations of the columns that come before. These dependence relations are given by the basis of the null space. 1*Column 1 - 2*Column 2 = Column 3 Column 1 + 3*Column 2 - 5 Column 4 = Column 5 1 -2 = # | -2 -5 8 0 -17 | A = | 1 3 -5 1 5 | | 3 11 -19 7 1 | | 1 7 -13 5 -3 | 1 3 -5 = # | -2 -5 8 0 -17 | A = | 1 3 -5 1 5 | | 3 11 -19 7 1 | | 1 7 -13 5 -3 | ------------------------------------------------------- We know that the dimension of the row space is 3. We know that the three non zero rows of the Row Canonical Form are a basis of the Row Space. But, if we want a basis of the row space chosen from the original rows of A, we have to do more work. We transpose the matrix of A and pick a basis of the T columns of A | -2 -5 8 0 -17 | A = | 1 3 -5 1 5 | | 3 11 -19 7 1 | | 1 7 -13 5 -3 | T | -2 1 3 1 | A = | -5 3 11 7 | | 8 -5 -19 -13 | | 0 1 7 5 | |-17 5 1 -3 | | -2 1 3 1 | | 1 0 2 4 | | 8 -5 -19 -13 | | 0 1 7 5 | |-17 5 1 -3 | | 1 0 2 4 | | -2 1 3 1 | | 8 -5 -19 -13 | | 0 1 7 5 | |-17 5 1 -3 | | 1 0 2 4 | | 0 1 7 9 | | 0 -5 -35 -45 | | 0 1 7 5 | | 0 5 35 65 | | 1 0 2 4 | | 0 1 7 9 | | 0 0 0 0 | | 0 0 0 -4 | | 0 0 0 20 | x y z=a w | 1 0 2 0 | | 0 1 7 0 | | 0 0 0 0 | | 0 0 0 1 | | 0 0 0 0 | So Rows 1 2 4 of A are linearly independent A basis of the row space of A is: ( -2 -5 8 0 -17 ), ( 1 3 -5 1 5 ), ( 1 7 -13 5 -3 ) Row 3 of A is a linear combination of rows 1 and 2. x y z=a w | 1 0 2 0 | | 0 1 7 0 | | 0 0 0 0 | | 0 0 0 1 | | 0 0 0 0 | | x | | -2 | | y | = a | -7 | | z | | 1 | | 2 | | 0 | 2 Row 1 + 7 Row 2 = Row 3. 2 | -2 -5 8 0 -17 | 7 | 1 3 -5 1 5 | # | 3 11 -19 7 1 | | 1 7 -13 5 -3 | ============================================ Change of Basis. This is conic x^2 + y^2 + 6xy = 8 with respect to the doted axis. What is its equation with respect to the standard axis? | ellipse?,parabola?,hyperbola? . | . | * * . |* . * . * | . *. | . * * . | . * . |. * ----------*-----------.+------------------------ . | . * * . | . . | * . * . |* . . * | . * * | . In general the equation: 2 2 ax + bxy + cy = d can be written in matrix form as | x y | | a b/2 | | x | | b/2 c | | y | T or X A X = d. Now if X = PX' is a change of basis matrix, then T (PX') A PX' = d T X' P A P X' = d and the matrix of the equation for the new system is T P A P. Example: If x^2 + y^2 + 6xy = 8 is the equation with respect to the skew axis. The basis of the skew axis is | 1/Sqrt[2] -1/Sqrt[2] | | 1/Sqrt[2] 1/Sqrt[2] | skew basis The standard basis is | 1 0 | | 0 1 | The change of basis matrix is | 1 0 | = | 1/Sqrt[2] -1/Sqrt[2] | | 1/Sqrt[2] 1/Sqrt[2] | | 0 1 | | 1/Sqrt[2] 1/Sqrt[2] | |-1/Sqrt[2] 1/Sqrt[2] | standard Skew Coefficients basis basis Multiply on right by |x'| |y'| | 1 0 ||x'| = | 1/Sqrt[2] -1/Sqrt[2] | | 1/Sqrt[2] 1/Sqrt[2] | |x'| | 0 1 ||y'| | 1/Sqrt[2] 1/Sqrt[2] | |-1/Sqrt[2] 1/Sqrt[2] | |y'| Skew Basis <----------------------------> point in The point is Skew Basis standard basis The change of basis matrix is | x | = | 1/Sqrt[2] 1/Sqrt[2] | |x'| | y | |-1/Sqrt[2] 1/Sqrt[2] | |y'| skew Change of Basis Matrix standard T X A X = d becomes |x' y'| |1/Sqrt[2] -1/Sqrt[2] | | 1 3 | 1/Sqrt[2] 1/Sqrt[2] | |x'| = 8 |1/Sqrt[2] 1/Sqrt[2] | | 3 1 |-1/Sqrt[2] 1/Sqrt[2] | |y'| |x' y'| | -2/Sqrt[2] 2/Sqrt[2] | | 1/Sqrt[2] 1/Sqrt[2] | | x' | = 8 | 4/Sqrt[2] 4/Sqrt[2] | |-1/Sqrt[2] 1/Sqrt[2] | | y' | |x' y' | | -4/2 0 | | x' | = 8 | 0 8/2 | | y' | 2 2 -2 (x') + 4/Sqrt[2] (y') = 8 2 2 (x') - 2 (y') = -4 <==== New Equation ----------------------------------------------------------- ans = Solve[x^2 + y^2 + 6x y == 8,y]; y1[x_] = y /. ans[[1]]; y2[x_] = y /. ans[[2]]; a = Plot[y1[x],{x,-10,1.5}] b = Plot[y2[x],{x, -1.5,10}] c = Show[a,b,PlotLabel->"x^2 + y^2 + 6 x y = 8 standard axis", AspectRatio->Automatic,PlotRange->All]; Display["c.ps",c]; aa = ParametricPlot[ x{1/Sqrt[2],1/Sqrt[2]}+y1[x]{-1/Sqrt[2],1/Sqrt[2]},{x,-10,1.5}]; bb = ParametricPlot[ x{1/Sqrt[2],1/Sqrt[2]}+y2[x]{-1/Sqrt[2],1/Sqrt[2]},{x, -1.5, 10}]; cc = ParametricPlot[ x{ 1/Sqrt[2],1/Sqrt[2]},{x,-10,10},PlotStyle->{RGBColor[1,0,0]}]; dd = ParametricPlot[ y{-1/Sqrt[2],1/Sqrt[2]},{y,-10,10},PlotStyle->{RGBColor[1,0,0]}]; ee = Show[aa,bb,cc,dd,PlotLabel->"x^2 + y^2 + 6 x y = 8 skew axis", AspectRatio->Automatic,PlotRange->All]; Display["ee.ps",ee]; r = Plot[ Sqrt[ (x^2 + 2 Sqrt[2])/2 ],{x,-10,10}]; s = Plot[ -Sqrt[ (x^2 + 2 Sqrt[2])/2 ],{x,-10,10}]; t = Show[r,s, PlotLabel->" (x')^2 - 2 (y')^2 = -2 Sqrt[2] on standard axis", PlotRange->All, AspectRatio->Automatic]; Display["t.ps",t]; tt = Show[ee,t,PlotLabel->"x^2+y^2+6x y = 8 and (x')^2 - 2(y')^2 = -2 Sqrt[2]", AspectRatio->Automatic,PlotRange->All]; u = Display["u.ps",tt]; --------------------------------------------------- The y' intercept is at 2 2 (x') - 2 (y') = -2 Sqrt[2] the point (0, Sqrt[2]) in the standard system. This corresponds to / | 1/Sqrt[2] 1/Sqrt[2] | | 0 | = | 1 | |-1/Sqrt[2] 1/Sqrt[2] | |Sqrt[2]) | | 1 | skew This should satisfy the original x,y equation. x^2 + y^2 + 6xy = 8 When x = 1 and y = 1 it does satisfy the original equation. ============================================================ Assignment: Write the matrix of the equation 2 2 2 x + 2 y + 3 z + 2 x y + 4 x z + 6 y z = 13 Make the substitution | x | | 1/3 2/3 -2/3 | | x' | | y | = | 2/3 1/3 2/3 | | y' | | z | | -2/3 2/3 1/3 | | z' |