# tat05 - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2010 Tutorial 5 — Solutions 1. Find a geometric interpretation for the following linear transformations. a ) T : R 2 −→ R 2 ; T ( a b ) mapsto−→ ( a + b b ) . b ) T : R 2 −→ R 2 ; T ( a b ) mapsto−→ parenleftBigg 1 2 a- √ 3 2 b √ 3 2 a + 1 2 b parenrightBigg . Solution a ) Since T ( a b ) = ( a + b b ) = ( 1 a +1 b a +1 b ) , we see that T ( a b ) = M ( a b ) , where M = ( 1 1 0 1 ) . Geometrically, this transformation can be thought of as a shear along the x − axis with shear factor 1. To see the effect of the shear, plot the four vertices of a unit square ( ) , ( 1 ) , ( 1 1 ) , ( 1 ) Now calculate their images under T and plot these. (Remember that matrix transformations map lines to lines.) You will see that the square is transformed into a parallelogram. All points on the x − axis are fixed, as T ( x ) = parenleftBig x parenrightBig for all values of x . b ) Since T ( a b ) = parenleftBigg 1 2 a- √ 3 2 b √ 3 2 a + 1 2 b parenrightBigg , we see that M = parenleftBigg 1 2- √ 3 2 √ 3 2 1 2 parenrightBigg = parenleftBig cos π 3- sin π 3 sin π 3 cos π 3 parenrightBig . This is the matrix corresponding to an anticlockwise rotation through an angle of π 3 radians about the origin. 2. Let T : R 3 −→ R 3 be the matrix transformation given by T parenleftBig x 1 x 2 x 3 parenrightBig = parenleftBig 4- 2 1 2 0 1 2- 2 3 parenrightBigparenleftBig x 1 x 2 x 3 parenrightBig . Find all the lines through the origin which are fixed by T . Solution Observe that T parenleftBig x 1 x 2 x 3 parenrightBig = parenleftBig 4- 2 1 2 0 1 2- 2 3 parenrightBigparenleftBig x 1 x 2 x 3 parenrightBig = M parenleftBig x 1 x 2 x 3 parenrightBig , where M = parenleftBig 4- 2 1 2 0 1 2- 2 3 parenrightBig . Therefore, a line { r v | r ∈ R } in R 3 , with v negationslash = 0 , is fixed by T if and only if T ( v ) = r v = M v ; that is, if and only if v is an eigenvector for M . To find the eigenvalues we solve the equation | M − λI | = 0 : vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 4 − λ − 2 1 2 − λ 1 2 − 2 3 − λ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = (4 − λ )[ λ ( λ − 3) + 2] + 2[2(3 − λ ) − 2] + 1[ − 4 + 2 λ ] = (4 − λ )( λ − 2)( λ − 1) + 4(2 − λ ) − 2(2 − λ ) = ( λ − 2)( − λ 2 + 5 λ − 6) = ( λ − 2) 2 (3 − λ ) . Hence the eigenvalues of M are 2 (with multiplicity 2) and 3. Math 2061: Tutorial 5 — Solutions S.B and A.M 25/1/2010 Linear Mathematics Tutorial 5 — Solutions Page 2 When λ = 3 , we want to find those vectors v ∈ R 3 such that ( M − 3 I ) v = . The augmented matrix for this system of equations is parenleftBigg 1 − 2 1 2 − 3 1 2 − 2 parenrightBigg Row reduce −−−−−→ parenleftBigg 1 − 1 1 − 1 parenrightBigg ....
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