Trace (linear algebra)纪念袁隆平
In linear algebra, the trace of an n-by-n square matrix A is defined to be the sum of the elements on the main diagonal (the diagonal from the upper left to the lower right) of A, i.e.,
where a
ii
reprents the entry on the i th row and i th column of A. Equivalently, the trace of a matrix is the sum of its eigenvalues, making it an invariant with respect to a change of basis. This characterization can be ud to define the trace for a linear operator in general. Note that the trace is only defined for a square matrix (i.e. n×n). Geometrically, the trace can be interpreted as the infinitesimal change in volume (as the derivative of the determinant), which is made preci in Jacobi's formula.
The u of the term trace aris from the German term Spur (cognate with the English spoor), which, as a function in mathematics, is often abbreviated to "Sp".
Examples
Let T be a linear operator reprented by the matrix
Then tr(T) = −2 + 1 − 1 = −2.
The trace of the identity matrix is the dimension of the space; this leads to generalizations of dimension using trace. The trace of a projection (i.e., P2 = P) is the rank of the projection. The trace of a nilpotent matrix is zero. The product of a symmetric matrix and a skew-symmetric matrix has zero trace.
More generally, if f(x) = (x− λ
1)d1···(x− λ
k
)d k is the characteristic polynomial of a matrix A, then
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If A and B are positive mi-definite matrices of the same order then
[1]
Properties
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The trace is a linear map. That is,
for all square matrices A and B, and all scalars c.
If A is an m×n matrix and B is an n×m matrix, then
[2]
Converly, the above properties characterize the trace completely in the n as follows. Let be a linear functional on the space of square matrices satisfying . Then and tr are proportional.[3]
The trace is similarity-invariant, which means that A and P−1AP have the same trace. This is becau
A matrix and its transpo have the same trace:
.
Let A be a symmetric matrix, and B an anti-symmetric matrix. Then
.
When both A and B are n by n, the trace of the (ring-theoretic) commutator of A and B vanishes: tr([A, B]) = 0; one can state this as "the trace is a map of Lie algebras from operators to scalars", as the commutator of scalars is trivial (it is an abelian Lie algebra). In particular, using similarity invariance, it follows that the identity matrix is never similar to the commutator of any pair of matrices.
Converly, any square matrix with zero trace is the commutator of some pair of matrices.[4] Moreover, any square matrix with zero trace is unitarily equivalent to a square matrix with diagonal consisting of all zeros.
The trace of any power of a nilpotent matrix is zero. When the characteristic of the ba field is zero, the conver
also holds: if for all , then is nilpotent.
Note that order does matter in taking traces: in general,
In other words, we can only interchange the two halves of the expression, albeit repeatedly. This means that the trace is invariant under cyclic permutations, i.e.,
However, if products of three symmetric matrices are considered, any permutation is allowed. (Proof: tr(ABC) = tr(A T B T C T) = tr((CBA)T) = tr(CBA).) For more than three factors this is not true. This is known as the cyclic property.
Unlike the determinant, the trace of the product is not the product of traces. What is true is that the trace of the tensor product of two matrices is the product of their traces:
The trace of a product can be rewritten as the sum of all elements from a Hadamard product (entry-wi product):
.
This should be more computationally efficient, since the matrix product of an matrix with an one
(first and last dimensions must match to give a square matrix for the trace) has multiplications and additions, whereas the computation of the Hadamard version (entry-wi product) requires only multiplications followed by additions.
The exponential trace
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Expressions like , where A is a square matrix, occur so often in some fields (e.g. multivariate statistical theory), that a shorthand notation has become common:
This is sometimes referred to as the exponential trace function.
Trace of a linear operator
Given some linear map f : V→ V (V is a finite-dimensional vector space) generally, we can define the trace of this map by considering the trace of matrix reprentation of f, that is, choosing a basis for V and describing f as a matrix relative to this basis, and taking the trace of this square matrix. The result will not depend on the basis chon, since different bas will give ri to similar matrices, allowing for the possibility of a basis independent definition for the trace of a linear map.
Such a definition can be given using the canonical isomorphism between the space End(V) of linear maps on V and V⊗V*, where V* is the dual space of V. Let v be in V and let f be in V*. Then the trace of the decomposable element
v⊗f is defined to be f(v); the trace of a general element is defined by linearity. Using an explicit basis for V and the corresponding dual basis for V*, one can show that this gives the same definition of the trace as given above.
Eigenvalue relationships
If A is a square n-by-n matrix with real or complex entries and if λ
1,...,λ
n
are the (complex and distinct) eigenvalues
of A (listed according to their algebraic multiplicities), then
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This follows from the fact that A is always similar to its Jordan form, an upper triangular matrix having λ
1,...,λ
n
on
the main diagonal. In contrast, the determinant of is the product of its eigenvalues; i.e.,
More generally,
Derivatives
The trace is the derivative of the determinant: it is the Lie algebra analog of the (Lie group) map of the determinant. This is made preci in Jacobi's formula for the derivative of the determinant (e under determinant). As a particular ca, : the trace is the derivative of the determinant at the identity. From this (or from the connection between the trace and the eigenvalues), one can derive a connection between the trace function, the exponential map between a Lie algebra and its Lie group (or concretely, the matrix exponential function), and the determinant: det(exp(A)) = exp(tr(A)).
For example, consider the one-parameter family of linear transformations given by rotation through angle θ,
The transformations all have determinant 1, so they prerve area. The derivative of this family at θ = 0 is the antisymmetric matrix
which clearly has trace zero, indicating that this matrix reprents an infinitesimal transformation which prerves area.
A related characterization of the trace applies to linear vector fields. Given a matrix A, define a vector field F on R n by F(x) = A x. The components of this vector field are linear functions (given by the rows of A). The divergence div F is a constant function, who value is equal to tr(A). By the divergence theorem, one can interpret this in terms of flows: if F(x) reprents the velocity of a fluid at the location x, and U is a region in R n, the net flow of the fluid out of U is given by tr(A)· vol(U), where vol(U) is the volume of U.
The trace is a linear operator, hence its derivative is constant:
Applications
The trace is ud to define characters of group reprentations. Two reprentations of a
group are equivalent (up to change of basis on ) if for all .
The trace also plays a central role in the distribution of quadratic forms.
Lie algebra
The trace is a map of Lie algebras from the Lie algebra gl
of operators on a n-dimensional space (
n
matrices) to the Lie algebra k of scalars; as k is abelian (the Lie bracket vanishes), the fact that this is a map
of Lie algebras is exactly the statement that the trace of a bracket vanishes:
The kernel of this map, a matrix who trace is zero, is said to be traceless or tracefree, and the matrices form the simple Lie algebra sl
, which is the Lie algebra of the special linear group of matrices with determinant 1. The
n
special linear group consists of the matrices which do not change volume, while the special linear algebra is the matrices which infinitesimally do not change volume.
In fact, there is a internal direct sum decomposition of operators/matrices into traceless operators/m
atrices and scalars operators/matrices. The projection map onto scalar operators can be expresd in terms of the trace, concretely as:
Formally, one can compo the trace (the counit map) with the unit map of "inclusion of scalars" to
obtain a map mapping onto scalars, and multiplying by n. Dividing by n makes this a projection, yielding the formula above.
In terms of short exact quences, one has
which is analogous to
for Lie groups. However, the trace splits naturally (via times scalars) so but the splitting of the determinant would be as the n th root times scalars, and this does not in general define a function, so the determinant does not split and the general linear group does not decompo:
Bilinear forms
The bilinear form
is called the Killing form, which is ud for the classification of Lie algebras.
The trace defines a bilinear form:
(x, y square matrices).
The form is symmetric, non-degenerate[5] and associative in the n that:
In a simple Lie algebra (e.g., ), every such bilinear form is proportional to each other; in particular, to the Killing form.
Two matrices x and y are said to be trace orthogonal if
Inner product
For an m-by-n matrix A with complex (or real) entries and * being the conjugate transpo, we have
with equality if and only if A = 0. The assignment
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The norm induced by the above inner product is called the Frobenius norm. Indeed it is simply the Euclidean norm if the matrix is considered as a vector of length mn.
Generalization
The concept of trace of a matrix is generalid to the trace class of compact operators on Hilbert spaces, and the analog of the Frobenius norm is called the Hilbert-Schmidt norm.
The partial trace is another generalization of the trace that is operator-valued.
If A is a general associative algebra over a field k, then a trace on A is often defined to be any map tr: A→ k which vanishes on commutators: tr([a, b]) = 0 for all a, b in A. Such a trace is not uniquely defined; it can always at least be modified by multiplication by a nonzero scalar.
A supertrace is the generalization of a trace to the tting of superalgebras.
The operation of tensor contraction generalizes the trace to arbitrary tensors.
Coordinate-free definition
We can identify the space of linear operators on a vector space V with the space , where . We also have a canonical bilinear function that consists of applying an element of to an element of to get an element of in symbols
This induces a linear function on the tensor product (by its universal property) which, as it turns out, when that tensor product is viewed as the space of operators, is equal to the trace.
This also clarifies why and why as composition of operators (multiplication of matrices) and trace can be interpreted as the same pairing. Viewing one may interpret the composition map as
coming from the pairing on the middle terms. Taking the trace of the product then comes from pairing on the outer terms, while taking the product in the opposite order and then taking the trace just switches which pairing is applied first. On the other hand, taking the trace of A and the trace of B corresponds to applying the pairing on the left terms and on the right terms (rather than on inner and outer), and is thus different.
In coordinates, this corresponds to indexes: multiplication is given by so
and which is the same, while
which is different.
For V finite-dimensional, with basis and dual basis , then is the entry of the matrix of the operator with respect to that basis. Any operator is therefore a sum of the form . With
defined as above, . The latter, however, is just the Kronecker delta, being 1 if i=j and 0 otherwi. This shows that is simply the sum of the coefficients along the diagonal. This method, however, makes coordinate invariance an immediate conquence of the definition.
