A First Course in Linear Algebra, 2017a

7.4. Orthogonality 423
We now turn our attention to the focus of this section. Our goal is to start with a quadratic form q
as given above and find a way to rewrite it to eliminate the x i x j terms. This is done through a change of
variables. In other words, we wish to find y i such that
⎡
Letting ⃗y = ⎢
⎣
⎤
y 1
y 2 ⎥
.
y n
q = d 11 y 2 1 + d 22 y 2 2 + ···+ d nn y 2 n
⎦ and D = [ d ij
]
, we can write q =⃗y T D⃗y where D is the matrix of coefficients from q.
There is something special about this matrix D that is crucial. Since no y i y j terms exist in q, it follows
that d ij = 0foralli ≠ j. Therefore, D is a diagonal matrix. Through this change of variables, we find the
principal axes y 1 ,y 2 ,···,y n of the quadratic form.
This discussion sets the stage for the following essential theorem.
Theorem 7.92: Diagonalizing a Quadratic Form
Let q be a quadratic form in the variables x 1 ,···,x n . It follows that q can be written in the form
q =⃗x T A⃗x where
⎡ ⎤
x 1
x 2...
⃗x = ⎢ ⎥
⎣ ⎦
x n
and A = [ ]
a ij is the symmetric matrix of coefficients of q.
New variables y 1 ,y 2 ,···,y n can be found such that q =⃗y T D⃗y where
⎡ ⎤
y 1
y 2...
⃗y = ⎢ ⎥
⎣ ⎦
y n
and D = [ d ij
]
is a diagonal matrix. The matrix D contains the eigenvalues of A and is found by
orthogonally diagonalizing A.
While not a formal proof, the following discussion should convince you that the above theorem holds.
Let q be a quadratic form in the variables x 1 ,···,x n . Then, q can be written in the form q = ⃗x T A⃗x for a
symmetric matrix A. By Theorem 7.54 we can orthogonally diagonalize the matrix A such that U T AU = D
for an orthogonal matrix U and diagonal matrix D.
⎡ ⎤
y 1
y 2 Then, the vector⃗y = ⎢ ⎥
⎣ . ⎦ is found by⃗y = U T ⃗x. To see that this works, rewrite⃗y = U T ⃗x as⃗x = U⃗y.
y n
Letting q =⃗x T A⃗x, proceed as follows:
q
= ⃗x T A⃗x