Quadratic Forms In the stud of linear algebra, we have been concerned with linear epressions, such as, + 9z, 7 + + etc. These are sometimes called linear forms. The most general linear form is a + a + + a k k. These are the simplest algebraic epressions. The net simplest are quadratic forms, such as +, + z, + 6z + z z. So a quadratic form in variables,,..., n is a sum of terms involving products of two variables, a ij i j. We can generate a quadratic form starting with an n n matri. For eample, let A =, 6 and for = (, ), form the product T A = [ ] [ 6 ][ = + + 6 = + 8 6, which is a quadratic form in and. ] = + 6 In general, if A is an n n matri and = (,,..., n ), then the matri product n n T A = a ij i j i= j= produces a quadratic form in,,..., n. Since =, there are man other matrices that produce the same quadratic form, for eample, for + 8 6 we can take A to be an one of 8,,,, 8 6 6 6 6 and millions more. Of those listed, the last one,, is special, because it is smmetric, 6 and we now know that smmetric matrices have nice properties. Furthermore, b averaging the a ij and a ji entries, we can convert an matri into a smmetric one that produces the same quadratic form. This is equivalent to replacing A b (A + AT ). For eample, ( + 6 ) = 8 =. 6 8 6 This means that we can alwas use smmetric matrices to generate quadratic forms. Definition Given a smmetric n n matri A, the quadratic form is the form generated b A. Q A () = T A = n i= j= n a ij i j Eample: For each quadratic form, epress it as A, where A is a smmetric matri: (a) 9 + 7 (b) + + 7 6 Solution: The diagonal entries are the coefficients of the squared terms, and the off-diagonal entries are half the coefficients of the cross terms: 9 / (a) A = (b) A = 7/ / 7 7/ Make sure ou collect collect terms before identifing coefficients. For eample, if we are given the quadratic form + + + 8 + +, we must first simplif it to 6 + 6 + 6, and then we see that the corresponding matri is 6 A = / / 6 /. / 6 Change of Variable If we make a linear substitution, or change of variable in a quadratic form, we get another quadratic form. For eample, given Q() = +, if we introduce new variables u and v and make the linear substitution then we get = u + v, = u v, ( u + v) ( u + v)(u v) + (u v) = (u uv + v ) ( u + 7uv 6v ) + (u uv + 9v ) = ( + + )u + ( 7 )uv + ( + 6 + 9)v = 7u uv + 9v
This particular substitution, chosen more or less at random, isn t particularl useful. However suppose we tr = u + v and = u v. Then + = (u + v) (u + v)(u v) + (u v) = (u + uv + v ) + (u v ) + (u uv + v ) = u + v This form is simpler, in that the cross term uv term is missing. A quadratic form whose cross terms are all zero is sometimes said to be a diagonal form. This is because it comes from a diagonal matri. Theorem Suppose D is a diagonal n n matri whose diagonal entries are λ, λ,..., λ n. Then the quadratic form generated b D is in diagonal form: λ Q D () = T λ D = [ n ].... λ n n = λ + λ + + λ n n. Another name for this is principal aes form. 7 So the following question comes to mind. Given a quadratic form, is there a substitution that will convert it to diagonal form, as in our last eample? If so, how do we find it? To answer these questions, we need to epress our linear substitutions in matri form. For eample, our first substitution, can be written as = u + v, = u v, u =, ][ v and our second substitution as [ u =, ] ][ v 8 Linear Substitution Definition A linear substitution of variables,,..., n, for,,..., n is an equation = P ( ) where P is an n n matri. If P is invertible, this is called a change of variable, and if P is orthogonal, it is an orthogonal change of variable. If we make the substitution ( ) in the quadratic form T A, we get T A = (P) T A(P) = T (P T AP) So the effect of the change of variable is to replace A b P T AP. 9 Now if A is smmetric, it is orthogonall diagonalizable, so we can find an orthogonal matri P that diagonalizes A. Then we have P T AP = D, where D is a diagonal matri with the eigenvalues of A along its diagonal, and we have λ T A = T λ D = [ n ].... λ n n = λ + λ + + λ n n We can sum up the situation in the following theorem. Theorem (The Principal Aes Theorem) If A is a smmetric n n matri, and the matri P orthogonall diagonalizes A, then substituting the change of variable = P into the quadratic form T A produces a quadratic form T D in principal aes form: λ + λ + + λ n n. The coefficient λ j is the eigenvalue corresponding to the j-th column of P.
Eample: Recall the eample A from lecture : A =. This is a smmetric matri, and gives the quadratic form T A = + Recall that A has eigenvalues and, and we found an orthogonal eigenbasis B = { (,, ), (,, ), (,, ) }. These vectors have norms,, and 6, respectivel, so an orthogonal matri that diagonalizes A is P = 6 6 6 Eercise: Put D =. Verif that P T AP = D. So the substitution = P gives the principal aes form T A = T D = +. For another eample, let Q = + 7. The smmetric matri that produces this form is 6 A = 6 7 The eigenvalues of A are and, with corresponding eigenvectors (, ) and (, ). (Verif this!) Since A is smmetric, these are orthogonal eigenvectors, as epected. An orthonormal eigenbasis is then So we set b = [ ], b = [ ]. P = =. Then we have P T AP = P AP = = D, and the change of variable = P gives the principal aes form Q = T A = T D =. Now the change of variables given b = P is = =, + or = ( ) and = ( + ). Eercise: Verif directl that making this substitution converts + 7 into. 7 Applications There are man uses for quadratic forms, and these can usuall be simplified b reduction to principal aes form. For eample, consider the following question. For n = or n =, what is the geometric interpretation (=graph) of the equation for a constant k? T A = k For geometric problems, it is convenient to change our notation slightl. For n =, we let = (, ), and for n = we let = (,, z). And when we make a change of variable, we will use = (, ), etc. So for n =, equation ( ) takes the form a + b + c = k. The curve (if an) represented b ( ) is called a central conic. ( ) 6 8 The reason we assume k in ( ) is that if k =, we get a degenerate case that is of little geometric interest. If b =, that is if the quadratic form in principal aes form, then ( ) is in standard position, and, ecept for degenerate cases, it can be classified as one of a small number of eamples. Since we assume k, we can after multipling b k, assume we have the form a + c = ( ) If both a < and b <, then equation ( ) is actuall impossible (for real numbers and ), so there is no curve at all.
Avoiding this case, we see that we can choose α = a and β = b in such a wa that we have one of the following possibilities: α + β =, α β =, β α = The first of these represents an ellipse and the others are hperbolas. For the ellipse, there are two cases, depending on which of α and β is larger. (And if α = β, we have a circle of radius α.) 9 Generall, orthogonal change of variables converts the equation T A = into the form λ + λ = where λ and λ are the eigenvalues of A. Then the curve is an ellipse if both λ > and λ >. The curve is a hperbola if both λ and λ have opposite signs. There is no graph if both λ < and λ <. The ellipse or hperbola is in standard position relative to a set of aes pointing along the eigenvectors of A. Finall, note that the parameters α and β in the standard forms are given b α = λ, β = λ. A general central conic is the result of rotating one in standard form through some angle θ. This causes the appearance of a cross term, b. Removing the cross term b an orthogonal substitution amounts to appling the reverse rotation, to bring the conic back to standard form. An alternative interpretation is that the diagonalizing change of variable gives a rotated coordinate sstem, and the conic is in standard position relative to that sstem. With the change of variables = P, where = (, ) we can think of the variables and as coordinates with respect to coordinate aes pointing along the eigenvectors that make up the columns of P. To illustrate this, let s look at our last eample again, where we considered the quadratic form Q = + 7, or, in terms of and, Q = + 7. What is the graph of the equation + 7 =? Now after the change of variables, the equation of the conic has the form = To recognize this in our list of eamples, we rewrite it as =. Then we put α = and β =, and we have the equation α β =. So in the, -plane, this represents a hperbola in standard position. = (, )
Recall that eigenvectors of A were u = (, ) and u = (, ). Plotting these in the, -plane tells us the directions of the, -aes: u u So we transfer the previous, -graph to these rotated aes: 6 And finall, all we need to do is erase all the scaffolding : + 7 = For another eample, consider + = 9. The matri A is A = The eigenvalues are easil found to be λ = and λ =, with corresponding eigenvectors u = (, ) and u = (, ). Again, these are orthogonal, as epected. Since both eigenvalues are positive, we know that the graph will be an ellipse. The principal aes form of the equation is then + = 9 So we will have α = and β =. or 9 + =, 7 In the, -plane, the graph is 8 The and aes point along u and u. So the rotated graph looks like + 9 = 9
And here it is without the primed aes: + = 9