7—Operators and Matrices 166The final equation comes from the preceding line. The coefficients of~ekmust agree on the two sides
of the equation. ∑
jSjifkj=
∑
jfji′Skj
Now rearrange this in order to place the indices in their conventionalrow,columnorder.
∑jSkjfji′ =
∑
jfkjSji
(
S 11 S 12
S 21 S 22
)(
f 11 ′ f 12 ′
f 21 ′ f 22 ′
)
=
(
f 11 f 12
f 21 f 22
)(
S 11 S 12
S 21 S 22
) (7.57)
In turn, this matrix equation is usually written in terms of the inverse matrix ofS,
(S)(f′) = (f)(S) is (f′) = (S)−^1 (f)(S) (7.58)
and this is called a similarity transformation. For the example Eq. (7.55) this is
~e′ 1 = 2xˆ+ 0. 5 ˆy=S 11 ~e 1 +S 21 ~e 2
which determines the first column of the matrix(S), then~e′ 2 determines the second column.
(S) =
(
2 0. 5
0. 5 2
)
then (S)−^1 =
1
3. 75
(
2 − 0. 5
− 0. 5 2
)
Eigenvectors
In defining eigenvalues and eigenvectors I pointed out the utility of having a basis in which the com-
ponents of an operator form a diagonal matrix. Finding the non-zero solutions to Eq. (7.50) is then
the way to find the basis in which this holds. Now I’ve spent time showing that you can find a matrix
in a new basis by using a similarity transformation. Is there a relationship between these two subjects?
Another way to ask the question: I’ve solved the problem to find all the eigenvectors and eigenvalues,
so what is the similarity transformation that accomplishes the change of basis (and why is it necessary
to know it if I already know that the transformed, diagonal matrix is just the set of eigenvalues, and I
already know them)?
For the last question, the simplest answer is that youdon’tneed to know the explicit transforma-
tion once you already know the answer. It is however useful to know that it exists and how to construct
it.If it exists — I’ll come back to that presently. Certain manipulations are more easily done in terms
of similarity transformations, so you ought to know how they are constructed, especially because almost
all the work in constructing them is done when you’ve found the eigenvectors.
The equation (7.57) tells you the answer. Suppose that you want the transformed matrix to be
diagonal. That means thatf 12 ′ = 0andf 21 ′ = 0. Write out the first column of the product on the
right. (
f 11 f 12
f 21 f 22
)(
S 11 S 12
S 21 S 22
)
−→
(
f 11 f 12
f 21 f 22
)(
S 11
S 21
)
This equals the first column on the left of the same equation