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Chapter 5. Eigenvalues and Eigenvectors
82
Recall that in Chapter 3 we discussed the matrix of a linear map
from one vector space to another vector space. This matrix depended
on a choice of a basis for each of the two vector spaces. Now that we are
studying operators, which map a vector space to itself, we need only
one basis. In addition, now our matrices will be square arrays, rather
than the more general rectangular arrays that we considered earlier.
Specifically, let T ∈L(V). Suppose (v 1 ,...,v n ) is a basis of V. For
each k = 1,...,n, we can write
Tv k = a 1,k v 1 + ··· + a n,k v n ,
The k th column of the where a j,k ∈ F for j = 1,...,n. The n-by-n matrix
matrix is formed from
a 1,1 ... a 1,n
the coefficients used to
. .
write Tv k as a linear 5.11 . . . .
combination of the v’s. a n,1 ... a n,n
is called the matrix of T with respect to the basis (v 1 ,...,v n ); we de-
note it by M T, (v 1 ,...,v n ) or just by M(T) if the basis (v 1 ,...,v n )
is clear from the context (for example, if only one basis is in sight).
n
If T is an operator on F and no basis is specified, you should assume
that the basis in question is the standard one (where the j th basis vector
is 1 in the j th slot and 0 in all the other slots). You can then think of
the j th column of M(T) as T applied to the j th basis vector.
A central goal of linear algebra is to show that given an operator
T ∈L(V), there exists a basis of V with respect to which T has a
reasonably simple matrix. To make this vague formulation (“reasonably
simple” is not precise language) a bit more concrete, we might try to
make M(T) have many 0’s.
If V is a complex vector space, then we already know enough to
show that there is a basis of V with respect to which the matrix of T
has 0’s everywhere in the first column, except possibly the first entry.
In other words, there is a basis of V with respect to which the matrix
of T looks like
We often use ∗ to λ
denote matrix entries 0 ∗
;
.
that we do not know .
.
about or that are
0
irrelevant to the
here the ∗ denotes the entries in all the columns other than the first
questions being
discussed. column. To prove this, let λ be an eigenvalue of T (one exists by 5.10)