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Orthogonal matrices Part 1 Question one Orthogonal matrices

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Added on  2019-10-30

Orthogonal matrices Part 1 Question one Orthogonal matrices

   Added on 2019-10-30

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Part 1Question oneOrthogonal matrixFor a matrix to be orthogonal; Transpose¿inverseThe conditions areabshould each be equal to 0.5 so that a+b=1=b+a and ab=ba=0 this makes the matrix an identity which is orthogonal.If it is an identity matrix it will as well be a diagonal matrix which is also an orthogonal matrix.The matrix will only be orthogonal if a and b have real values.Question twoIf A is symmetric orthogonal then ±1 are the only possible eigen valuesConsider |Av|=¿v¿ for any v,suppose a is orthogonal and v0is an eigenvector having eigen value λav=λvimplying|v|=|av|=|λ|v¿ hence |λ|=1Which gives the value of λ as either 11Question threeThe quadratic form 2x2+5y2+5z2+4xy4xz8yz has an associated matrix that can be orthogonally diagonalized [222254245¿λ3+12λ221λ+10λ1=1hencev1=(210)λ2=1hencev2=(201)λ3=10hencev3=(122)beingthatλ1=λ2,v1isperpendicular¿v3v2¿v3v1¿v2 this can be solved in a linear combination of e1=15(201)e3=13(122)
Orthogonal matrices Part 1 Question one Orthogonal matrices_1
So e2=e3e1=[23553435]Then [x'y'z']=(2523513053231543533)1(xyz)=(2501523553435132323)(xyz)(2523513053231543523)(222254245)(2501523553435132323)=(1000100010)Now expressing q in terms of the new variables givesQ=xr2+yr2+10zr2Question fourTo show that a quadratic equation represent ellipsoid43x2+43y2+43z2+43xy+43xz+43yz=1The general form of an ellipsoid is Ax2+Bx2+cz2+Dyz+Ezx+Fxy+Hy+Jz+k=0The matrixM=[AF2D2F2BD2E2D2C]should be positive definite i.e.
Orthogonal matrices Part 1 Question one Orthogonal matrices_2
A>0,AB>F24det(M)=0Now proving our equationM=[434646464346464643]Now A=43>0AB=(43)2=169While F24=49=49Hence AB>F24The det(M)=43|46464643|46|46464643|+46|46434646|This gives 1627827827=0position and length of principal axisgetting matrix M=[434343434343434343]Now we obtain the eigen values
Orthogonal matrices Part 1 Question one Orthogonal matrices_3

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