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 area∧bshould each be equal to 0.5 so that a+b=1=b+a and a−b=b−a=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 v≠0is an eigenvector having eigen value λav=λvimplying|v|=|av|=|λ|∨v∨¿ hence |λ|=1Which gives the value of λ as either 1∨−1Question threeThe quadratic form 2x2+5y2+5z2+4xy−4xz−8yz has an associated matrix that can be orthogonally diagonalized [22−225−4−2−45¿−λ3+12λ2−21λ+10λ1=1hencev1=(−210)λ2=1hencev2=(201)λ3=10hencev3=(−1−22)beingthatλ1=λ2,v1isperpendicular¿v3∧v2¿v3∧v1¿v2 this can be solved in a linear combination of e1=1√5(201)∧e3=13∗(−1−22)
So e2=e3∗e1=[−23√5√5343√5]Then [x'y'z']=(2√5−23√5−130√53−231√543√533)−1(xyz)=(2√501√5−23√5√5343√5−13−2323)(xyz)(2√5−23√5−130√53−231√543√523)(22−225−4−2−45)(2√501√5−23√5√5343√5−13−2323)=(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.
A>0,AB>F24∧det(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 1627−827−827=0position and length of principal axisgetting matrix M=[434343434343434343]Now we obtain the eigen values
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