Diagonalization of Quadratic Forms - PDF document

Diagonalization of Quadratic Forms Recall in days past when you were given an equation which looked like122++yx and you were asked to sketch the set of points which satisfy thisequation. It was necessary to complete the square so that the equation looked like the(h,k) form of an ellipse. That is, () ) ) () 52212221222122211++=+++++xyyxyyxThen, now that we have rewritten the equation into a form we recognize, we can see thatthis is a circle (which is an ellipse) of radius 45 centered at the point ( ) 21,0. But we hadto do some work to get to this point. Similarly, we need to be able to rewrite1222++ into a new form so that we can read off all of the informationnecessary to sketch the set.Suppose that we are given an equation of the following form,1222++Then, as it turns out, we can rewrite this equation using matrix notation. Indeed, [] ûùêëéúûùêëé++xcbaxcy22.The nice thing about this representation is that the matrix úûùêëéba has a property calledsymmetry. Specifically the matrix is symmetric about the diagonal. Formerly,úûùêëéûùêëébaba. There are many nice results about symmetric matrices. One of themis the fact that any symmetric matrix is unitarily diagonalizable. Okay, so I’m throwingout a lot of fancy words, so let me try to elaborate on exactly what this means.Definition : (Unitary Matrix) An nxn matrix U is said to be unitary if the columns of U form an orthonormal basis for n : (Unitarily Diagonalizable) An nxn matrix A is said to be unitarily if there exists a unitary matrix U so that 1U where L is a diagonal matrix.Recall : A set of vectors { } nee,,21K is an orthonormal basis for n if and only if any vector n has a (unique) representation nneex aa ++= 2211 andîíì¹=ijiei0. Then with these facts it should become clear as mud that TU Care to see theproof? Too bad, I’m doing it anyway.Proposition : If U is an nxn unitary matrix then, nU, where n denotes the nxn identity matrix. Thus, 1UProof : Let U be an nxn unitary matrix. Then observe the following computation.úúúúûùêêêêëé××××××××úúûùêêêëéúúúúûùêêêêëé-------=nnnnnnnuuuuuuuuuuuuuuuuuuuuuuuUvvvvMvvvv v v vv v 122212121112121||||||||Then since the i ’s form an orthonormal basis we have,nnnnnnnuuuuuuuuuuuuuuuuuuúúúûùêêêêëéúúúûùêêêêëé××××××××10000010001212221212111LLvvvvMvvvv v v as desired.So, to translate this into slightly more understandable language, this means that a unitarymatrix is a matrix which moves around the coordinate axes to a new set of orthogonalaxes (i.e. a change of basis). Or, we could say that the axes have been rotated (orreflected) to a new set of perpendicular axes.Allow me to go back to a previous example to try to lower the level ofintimidation a little bit. Recall the ‘ol complete the square trick to find the (h,k) form foran equation representing an ellipse. This representation corresponds to shifting the originto the point (h,k) and drawing the ellipse around that point, right? So, if you feelreasonably comfortable shifting around the coordinate axes, which is exactly what youare doing when completing the square, then you should feel comfortable twirling the axesaround. After all it’s the same idea. The only difference is that this new tool we havedeveloped is more computation intensive.Let us now set up a set of procedures to perform this task.1) Write 1222++ in the matrix form [] ûùêëéúûùêëéxcbax Find the eigenvalues 1 and 2 for the matrix úûùêëéba. Find the eigenvectors 1 and 2 corresponding to the eigenvalues 1 and 2 Draw lines passing through the vectors 1 and 2 . (These are the new axes.)5) Rewrite the new equation as 1222211+w l where 1 and 2 are variablesrepresenting the distance from the origin in the 1 and 2 directions respectively. Just like x and y represent the distance from the origin in thedirection 1 and 2 Draw the darn picture.I can hear you now, “Whoa hoss! You didn’t do that unitary diagonalization thingyou were talking about!” You’re right. To sketch the picture I didn’t need to gothat far. However, notice that I have all the information to do so if I wish. Let úûùêëé=11111uuv v v and úûùêëé=21222uuv v v. Then,úûùêëéúûùêëéúûùêëéûùêëé211211212212211100uuuuuuuuba l is the desired unitary diagonalization. Example 1: Sketch the image of the set of points which satisfy the equation33182722+-xyxSolution 1:1) [] ûùêëéúûùêëé-=+-xyxyxyx99273182722 039927÷øöççèæúûùêëé--- l ) ,003003008130810813272122=---+---- lllllllll úûùêëé---399027 ûùêëéúûùêëéûùêëé-0190099271 ûùêëé311 is an eigenvector for 01 ûùêëéúûùêëé-ûùêëé---ûùêëé---0310939933993027ûùêëé=32 is an eigenvector for 302 4) Sketch on a piece of paper lines passing through úûùêëé311 and úûùêëé=32These represent the new axes. Notice 021v The new equation we have is then 33002221+w. Notice this is a bit of aweird animal. What does it say? Well, doing some algebra we have, 10122222213303300=+wwwThis gives us a set of parallel lines which intersect the 2 axis at 101 Draw the darn picture. Example 2: Sketch the image of the set of points which satisfy the equation2513101322+-xyxSolution 2:1) [] ûùêëéúûùêëé-=+-xyxyxyx551313101322 0135513÷øöççèæúûùêëé--- l )() ,801880144260252616902513132122=--+--+---- lllllllll úûùêëé---1355813ûùêëéûùêëéûùêëé-011055555ûùêëé11 is an eigenvector for 81 ûùêëéúûùêëé-ûùêëé---ûùêëé---01105555513551813ûùêëé=12 is an eigenvector for 182 4) Sketch on a piece of paper lines passing through úûùêëé11 and úûùêëé=12These represent the new axes. Notice 021v The new equation we have is then 251882221+w, which after somecleaning up looks like () () 223522222521+w. This then is an ellipse thatintersects the 1 axis at 522 and intersects the 2 axis at 523 Draw the darn picture. Example 3: Sketch the image of the set of points which satisfy the equation12422-+-xyxSolution 3:1) [] ûùêëéúûùêëé-=-+-xyxyxyx1142422 01114÷øöççèæúûùêëé--- l 5, 2 135021352135035015401142122-=+-=÷øöççèæ--÷øöççèæ--++-++----- lllllllll Ick. How can we possible find eigenvalues corresponding to these monsters?Well, you can if you want but you don’t need to. Because if we skip the wholefind the eigenvectors step notice what will happen. That is, assume for themoment that we found the eigenvectors for these eigenvalues. Then we have thenew representation for the equation as follows, 22212221213521351121352135www÷øöççèæ+÷øöççèæ=-÷øöççèæ-+÷øöççèæ-Do you see anything wrong with this picture. Notice, 02135÷øöççèæ 02135÷øöççèæ021, and 022 but the equation says that I can multiply them together and addthem up to obtain a negative number. There are certainly no real numbers which cansatisfy this equation. So, the sketch of the set of points which satisfy the equation 1213521352221÷øöççèæ-+÷øöççèæ-w is empty.To summarize the cases we have the following,i) If 01 and 02 we have an ellipse.ii) If 021 l (so either 01 or 02 , not both) we have ahyperbola. If 01 or 02 (not both zero) we have a set of parallel lines.iv) If 021= l we have the empty set.v) If 01 and 02 we have the empty set.