Vectors and Vector Operations .edu
4.15 The Eigenvalues of a 2 ( 2 Matrix
In the version of the QR algorithm discussed in the next section we use as a shift the eigenvalue of the 2 ( 2 submatrix in the lower right corner that is closest to the element in the lower right corner. In this section we derive a formula for that eigenvalue.
Let
A =
be a two by two matrix. We want a formula for the eigenvalue ( of A closest to d.
The eigenvalues of A are the solutions of the equation
0 = det( A - (I ) = = (a - ()(d - () - bc
= (2 – (a + d)( + ad - bc
So the eigenvalues are
( = =
= =
So
(1 = + ( and (2 = - (
where
( = and ( =
We want to pick out the one of (1 and (2 that is closest to d. If we think of complex numbers as points or vectors in the plane, then the point is half way between a and d and also halfway between (1 and (2. The vector ( goes from to a and the vector ( goes from to (1. If the angle between ( and ( is less than 90( then (1 is closest to a and (2 is closest to d. It the angle is greater than 90( it is just the reverse. If the angle is exactly 90( then the two eigenvalues are equally close to d and we can take either one.
The angle between ( and ( is less than 90( if the dot product ( . ( is positive and the angle is less than 90( if this dot product is negative. So
( = - sign(( . ()(
where sign(t) = 1 if t ( 0 and sign(t) = -1 if t < 0.
We can make this formula a little better by expressing the dot product ( . ( in terms of the usual operations on complex numbers. Suppose ( = x + iy and ( = u + iv. Then
( = (x + iy)(u - iv) = (xu + yv) + i(yu – xv)
So
( . ( = xu + yv = Re(( )
where Re(z) denotes the real part of the complex number z. Thus
( = - sign(Re(( ))(
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