l Orthogonal changes of coordinates x U x affect the symmetric matrix Q of a

L orthogonal changes of coordinates x u x affect the

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(l) Orthogonal changes of coordinates x = U x affect the symmetric matrix Q of a quadratic form by similarity transformations U t QU = U 1 QU and hence preserve the characteristic polynomial of Q . According to Orthogonal Diagonaliza- tion Theorem the quadratic form can be transformed by such changes of coordi- nates to one of the normal forms AX 2 1 + CX 2 2 with the characteristic polynomial ( λ A )( λ C ). Thus the coefficients A,C in the normal form are roots of the characteristic polynomial vextendsingle vextendsingle vextendsingle vextendsingle λ a b b λ c vextendsingle vextendsingle vextendsingle vextendsingle = λ 2 ( a + c ) λ + ( ac b 2 ) .
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30 1. GEOMETRY ON THE PLANE We complete this section with a proof of Orthogonal Diagonalization Theorem. It has several reformulations (see Principal Axes Theorem in 1 . 2 . 2, Example 1 . 3 . 3( a ) and Section 1 . 3 . 4( IV )) equivalent to the following one: For any symmetric matrix Q there exists an orthogonal matrix U such that U 1 QU is diagonal. Proof. The discriminant of the characteristic polynomial of Q equals ( a + c ) 2 4( ac b 2 ) = a 2 + 2 ac + c 2 4 ac + 4 b 2 = ( a c ) 2 + 4 b 2 and is non-negative. This means that the characteristic polynomial of a symmetric matrix has a real root λ 1 and hence — a real eigenvector, a non-trivial solution to the homogeneous linear system Q v 1 = λ 1 v 1 . Let v 2 be a non-zero vector orthogonal to v 1 . Then ( Q v 2 , v 1 ) = ( v 2 ,Q v 1 ) = ( v 2 1 v 1 ) = λ 1 ( v 2 , v 1 ) = 0 . Thus Q v 2 is also orthogonal to v 1 and is therefore proportional to v 2 : Q v 2 = λ 2 v 2 with some real coefficient λ 2 . Normalizing the perpendicular eigenvectors v 1 and v 2 to the unit length we get an orthonormal basis u 1 = v 1 / | v 1 | , u 2 = v 2 / | v 2 | of eigenvectors. Let U denote the matrix whose columns represent u 1 and u 2 . It is orthogonal since U t U = bracketleftbigg u t 1 u 1 u t 1 u 2 u t 2 u 1 u t 2 u 2 bracketrightbigg = bracketleftbigg 1 0 0 1 bracketrightbigg . The columns of QU represent λ 1 u 1 and λ 2 u 2 . Thus QU = U Λ, where Λ = bracketleftbigg λ 1 0 0 λ 2 bracketrightbigg is the diagonal matrix of eigenvalues, and hence U 1 QU = Λ. Exercises 1 . 5 . 2 . (a) Verify the multiplicative property of determinants. (b) Give an example of two matrices which are not similar but have the same characteristic polynomial. (c) Show that tr( A + B ) = tr A + tr B , tr( AB BA ) = 0. (d) Find the intersection point of the lines 3 x 1 + 2 x 2 = 1 and x 1 + 2 x 2 = 3. Find the coordinates of the intersection point in the basis f 1 = (1 , 3) , f 2 = (3 , 1). (e) Find out which of the following quadratic curves is an ellipse and which is a hyperbola and compute their semiaxes (that is the coefficients a 1 ,a 2 in the normal forms x 2 1 /a 2 1 ± x 2 2 /a 2 2 = 1): 2 x 2 1 + 3 x 1 x 2 + x 2 2 = 1 , 3 x 2 1 3 x 1 x 2 + x 2 2 = 1 .
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