Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

MATRICES AND VECTOR SPACES


8.26 Show that the quadratic surface


5 x^2 +11y^2 +5z^2 − 10 yz+2xz− 10 xy=4
is an ellipsoid with semi-axes of lengths 2, 1 and 0.5. Find the direction of its
longest axis.
8.27 Find the direction of the axis of symmetry of the quadratic surface


7 x^2 +7y^2 +7z^2 − 20 yz− 20 xz+20xy=3.

8.28 For the following matrices, find the eigenvalues and sufficient of the eigenvectors
to be able to describe the quadratic surfaces associated with them:


(a)



51 − 1


151


−11 5



, (b)



122


212


221



, (c)



121


242


121



.


8.29 This exercise demonstrates the reverse of the usual procedure of diagonalising a
matrix.


(a) Rearrange the resultA′=S−^1 ASof section 8.16 to express the original
matrixAin terms of the unitary matrixSand the diagonal matrixA′. Hence
show how to construct a matrixAthat has given eigenvalues and given
(orthogonal) column matrices as its eigenvectors.
(b) Find the matrix that has as eigenvectors (1 2 1)T,(1 −11)Tand
(1 0 −1)T, with corresponding eigenvaluesλ,μandν.
(c) Try a particular case, sayλ=3,μ=−2andν= 1, and verify by explicit
solution that the matrix so found does have these eigenvalues.

8.30 Find an orthogonal transformation that takes the quadratic form


Q≡−x^21 − 2 x^22 −x^23 +8x 2 x 3 +6x 1 x 3 +8x 1 x 2

into the form
μ 1 y^21 +μ 2 y^22 − 4 y 32 ,

and determineμ 1 andμ 2 (see section 8.17).
8.31 One method of determining the nullity (and hence the rank) of anM×Nmatrix
Ais as follows.



  • Write down an augmented transpose ofA, by adding on the right anN×N
    unit matrix and thus producing anN×(M+N) arrayB.

  • Subtract a suitable multiple of the first row ofBfrom each of the other lower
    rows so as to makeBi 1 =0fori>1.

  • Subtract a suitable multiple of the second row (or the uppermost row that
    does not start withMzero values) from each of the other lower rows so as to
    makeBi 2 =0fori>2.

  • Continue in this way until all remaining rows have zeros in the firstMplaces.
    The number of such rows is equal to the nullity ofA,andtheNrightmost
    entries of these rows are the components of vectors that span the null space.
    They can be made orthogonal if they are not so already.
    Use this method to show that the nullity of


A=








−13 2 7


310 − 617


− 1 − 22 − 3


23 − 44


40 − 8 − 4







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