7.16. Eigenvalues Of A Symmetric Transformation Obtained As
Values Of Its Quadratic Form
Let x be an eigenvector of norm 1 belonging to an eigenvalue . Then
Q  x    T  x  , x     x, x   
(7.9)
Q  x  x T x   x x  
The set of all x V with
 x, x  
x x  1 is called the unit sphere in V. This
proves the following theorem.
Theorem 7.13.
Let T : V  V be a symmetric transformation on a real Euclidean space V. Then
the eigenvalues of T, if exist, are to be found among the values that
Q  x   T  x  , x 
takes on the unit sphere.
Note that dimV can be infinite.
Example 1.
Let V  R 2 with the usual basis
 i, j
and the usual dot product as inner product.
 4 0
Let T be the symmetric transformation with m T   A  
 . Then the
 0 8
quadratic form of T is diagonal:
Q  x 
2
a x x
i , j 1
ij i
j
 4 x12  8 x22
The eigenvalues of T are 1  4 and 2  8 .
On the unit sphere (circle)
x12  x22  1 , we have
Q  x   4  x12  x22   4 x22  4  4x22
with
whose extrema at x2  0, 1 give the eigenvalues.
x2  1  x12  1
Note that the point x at which
the surface Q  x    intersects the unit sphere is an eigenvector belonging to .
Figure 7.4 shows the unit circle and 2 ellipses given by Q  x   4 and Q  x   8 .
The inner ellipse Q  x   4 intersects the unit circle at x   1,0 . Hence,
u1  t 1,0 is an eigenvector belonging to 1  4 . Similarly, the outer one
Q  x   8 intersects the unit circle at x   0, 1 . Hence, u2  t  0,1 is an
eigenvector belonging to 2  8 .
Theorem 7.14.
Let T : V  V be a symmetric transformation on a real Euclidean space V with a
quadratic form Q  x   T  x  , x  . Assuming that Q does not change sign on V.
Then Q  x   0 implies T  x   O .
In other words, if Q does not change sign on V,
then Q vanishes only on the null space N T  .
Proof
Let t  R and y V . With Q  x   0 , we have
Q  x  ty   T  x  ty  , x  ty   T  x   tT  y  , x  ty 
  T  x  , x   t  T  x  , y   t T  y  , x   t 2 T  y  , y 
 2at  bt 2
where
a  T  x  , y    y , T  x  
and
b  Q y
If Q is nonnegative on V, then b  0 and
2at  bt 2  0
 t R
Consider then the quadratic polynomial p  t   2at  bt 2 .
The condition p  t   0
then requires it to lie entirely in the upper t-p plane. However, p  t  is a parabola.
Hence, it can intersect with the t-axis at no more than one point, otherwise, part of it
will be in the lower t-p plane. Now, the roots of
2at  bt 2  0
are t  0 or t  
the root t  
2a
.
b
Since Q  x   0 , the root t  0 is a necessity. Therefore,
2a
must coincide with t  0 .
b
T  x  , y   O
Thus, we must have a  0 , i.e.,
 y V
Hence, T  x   O as desired.
Proof for the case Q non-positive on V is similar.