Editing Trace (linear algebra) (section)

== Properties ==

=== Basic properties ===
The trace is a [[linear operator|linear mapping]]. That is,<ref name=":1" /><ref name=":2" />
<math display="block">\begin{align}
\operatorname{tr}(\mathbf{A} + \mathbf{B}) &= \operatorname{tr}(\mathbf{A}) + \operatorname{tr}(\mathbf{B}) \\
\operatorname{tr}(c\mathbf{A}) &= c \operatorname{tr}(\mathbf{A})
\end{align}</math>
for all square matrices {{math|'''A'''}} and {{math|'''B'''}}, and all [[scalar (mathematics)|scalar]]s {{mvar|c}}.<ref name="LipschutzLipson"/>{{rp|34}}

A matrix and its [[transpose]] have the same trace:<ref name=":1" /><ref name=":2" /><ref name="LipschutzLipson"/>{{rp|34}}
<math display="block">\operatorname{tr}(\mathbf{A}) = \operatorname{tr}\left(\mathbf{A}^\mathsf{T}\right).</math>

This follows immediately from the fact that transposing a square matrix does not affect elements along the main diagonal.

=== Trace of a product ===
The trace of a square matrix which is the product of two matrices can be rewritten as the sum of entry-wise products of their elements, i.e. as the sum of all elements of their [[Hadamard product (matrices)|Hadamard product]]. Phrased directly, if {{math|'''A'''}} and {{math|'''B'''}} are two {{math|''m'' × ''n''}} matrices, then:
<math display="block">
\operatorname{tr}\left(\mathbf{A}^\mathsf{T}\mathbf{B}\right) =
\operatorname{tr}\left(\mathbf{A}\mathbf{B}^\mathsf{T}\right) =
\operatorname{tr}\left(\mathbf{B}^\mathsf{T}\mathbf{A}\right) =
\operatorname{tr}\left(\mathbf{B}\mathbf{A}^\mathsf{T}\right) =
\sum_{i=1}^m \sum_{j=1}^n a_{ij}b_{ij} \; .
</math>

If one views any real {{math|''m'' × ''n''}} matrix as a vector of length {{mvar|mn}} (an operation called [[Vectorization (mathematics)|vectorization]]) then the above operation on {{math|'''A'''}} and {{math|'''B'''}} coincides with the standard [[dot product]]. According to the above expression, {{math|tr('''A'''<sup>⊤</sup>'''A''')}} is a sum of squares and hence is nonnegative, equal to zero if and only if {{math|'''A'''}} is zero.<ref name="HornJohnson">{{cite book |title=Matrix Analysis |edition=2nd |first1=Roger A. |last1=Horn |first2=Charles R. |last2=Johnson |isbn=9780521839402 |publisher=Cambridge University Press|year=2013}}</ref>{{rp|7}} Furthermore, as noted in the above formula, {{math|tr('''A'''<sup>⊤</sup>'''B''') {{=}} tr('''B'''<sup>⊤</sup>'''A''')}}. These demonstrate the positive-definiteness and symmetry required of an [[inner product]]; it is common to call {{math|tr('''A'''<sup>⊤</sup>'''B''')}} the [[Frobenius inner product]] of {{math|'''A'''}} and {{math|'''B'''}}. This is a natural inner product on the [[vector space]] of all real matrices of fixed dimensions. The [[norm (mathematics)|norm]] derived from this inner product is called the [[Frobenius norm]], and it satisfies a submultiplicative property, as can be proven with the [[Cauchy–Schwarz inequality]]:
<math display="block">0 \leq
\left[\operatorname{tr}(\mathbf{A} \mathbf{B})\right]^2 \leq
\operatorname{tr}\left(\mathbf{A}^\mathsf{T} \mathbf{A}\right) \operatorname{tr}\left(\mathbf{B}^\mathsf{T} \mathbf{B}\right) ,</math>
if {{math|'''A'''}} and {{math|'''B'''}} are real matrices such that {{math|'''A''' '''B'''}} is a square matrix. The Frobenius inner product and norm arise frequently in [[matrix calculus]] and [[statistics]].

The Frobenius inner product may be extended to a [[hermitian inner product]] on the [[complex vector space]] of all complex matrices of a fixed size, by replacing {{math|'''B'''}} by its [[complex conjugate]].

The symmetry of the Frobenius inner product may be phrased more directly as follows: the matrices in the trace of a product can be switched without changing the result. If {{math|'''A'''}} and {{math|'''B'''}} are {{math|''m'' × ''n''}} and {{math|''n'' × ''m''}} real or complex matrices, respectively, then<ref name=":1" /><ref name=":2" /><ref name="LipschutzLipson"/>{{rp|34}}<ref group="note">This is immediate from the definition of the [[matrix product]]:
<math display="block">\operatorname{tr}(\mathbf{A}\mathbf{B}) = \sum_{i=1}^m \left(\mathbf{A}\mathbf{B}\right)_{ii} = \sum_{i=1}^m \sum_{j=1}^n a_{ij} b_{ji} = \sum_{j=1}^n \sum_{i=1}^m b_{ji} a_{ij} = \sum_{j=1}^n \left(\mathbf{B}\mathbf{A}\right)_{jj} = \operatorname{tr}(\mathbf{B}\mathbf{A}).</math>
</ref>

{{Equation box 1
|indent=:
|title=
|equation = <math>\operatorname{tr}(\mathbf{A}\mathbf{B}) = \operatorname{tr}(\mathbf{B}\mathbf{A})</math>
|cellpadding= 6
|border
|border colour = #0073CF
|background colour=#F5FFFA
}}

This is notable both for the fact that {{math|'''AB'''}} does not usually equal {{math|'''BA'''}}, and also since the trace of either does not usually equal {{math|tr('''A''')tr('''B''')}}.<ref group="note">For example, if
<math display="block">
\mathbf{A} = \begin{pmatrix}
  0 & 1 \\
  0 & 0
\end{pmatrix},\quad
\mathbf{B} = \begin{pmatrix}
  0 & 0 \\
  1 & 0
\end{pmatrix},
</math>

then the product is
<math display="block">\mathbf{AB} = \begin{pmatrix}
  1 & 0 \\
  0 & 0
\end{pmatrix},</math>
and the traces are {{math|tr('''AB''') {{=}} 1 ≠ 0 ⋅ 0 {{=}} tr('''A''')tr('''B''')}}.</ref> The [[similarity invariance|similarity-invariance]] of the trace, meaning that {{math|tr('''A''') {{=}} tr('''P'''<sup>−1</sup>'''AP''')}} for any square matrix {{math|'''A'''}} and any invertible matrix {{math|'''P'''}} of the same dimensions, is a fundamental consequence. This is proved by
<math display="block">
\operatorname{tr}\left(\mathbf{P}^{-1}(\mathbf{A}\mathbf{P})\right) =
\operatorname{tr}\left((\mathbf{A} \mathbf{P})\mathbf{P}^{-1}\right) =
\operatorname{tr}(\mathbf{A}).
</math>
Similarity invariance is the crucial property of the trace in order to discuss traces of [[linear transformation]]s as below.

Additionally, for real column vectors <math>\mathbf{a}\in\mathbb{R}^n</math> and <math>\mathbf{b}\in\mathbb{R}^n</math>, the trace of the outer product is equivalent to the inner product:
{{Equation box 1
|indent=:
|title=
|equation = <math>\operatorname{tr}\left(\mathbf{b}\mathbf{a}^\textsf{T}\right) = \mathbf{a}^\textsf{T}\mathbf{b}</math>
|cellpadding= 6
|border
|border colour = #0073CF
|background colour=#F5FFFA
}}

=== Cyclic property ===
More generally, the trace is ''invariant under [[circular shift]]s'', that is,

{{Equation box 1
|indent=:
|title=
|equation = <math>\operatorname{tr}(\mathbf{A}\mathbf{B}\mathbf{C}\mathbf{D}) = \operatorname{tr}(\mathbf{B}\mathbf{C}\mathbf{D}\mathbf{A}) = \operatorname{tr}(\mathbf{C}\mathbf{D}\mathbf{A}\mathbf{B}) = \operatorname{tr}(\mathbf{D}\mathbf{A}\mathbf{B}\mathbf{C}).</math>
|cellpadding= 6
|border
|border colour = #0073CF
|background colour=#F5FFFA}}

This is known as the ''cyclic property''.

Arbitrary permutations are not allowed: in general,
<math display="block">\operatorname{tr}(\mathbf{A}\mathbf{B}\mathbf{C}\mathbf{D}) \ne \operatorname{tr}(\mathbf{A}\mathbf{C}\mathbf{B}\mathbf{D}) ~.</math>

However, if products of ''three'' [[symmetric matrix|symmetric]] matrices are considered, any permutation is allowed, since:
<math display="block">\operatorname{tr}(\mathbf{A}\mathbf{B}\mathbf{C}) = \operatorname{tr}\left(\left(\mathbf{A}\mathbf{B}\mathbf{C}\right)^{\mathsf T}\right) = \operatorname{tr}(\mathbf{C}\mathbf{B}\mathbf{A}) = \operatorname{tr}(\mathbf{A}\mathbf{C}\mathbf{B}),</math>
where the first equality is because the traces of a matrix and its transpose are equal. Note that this is not true in general for more than three factors.

=== Trace of a Kronecker product ===
The trace of the [[Kronecker product]] of two matrices is the product of their traces:
<math display="block">\operatorname{tr}(\mathbf{A} \otimes \mathbf{B}) = \operatorname{tr}(\mathbf{A})\operatorname{tr}(\mathbf{B}).</math>

===Characterization of the trace===
The following three properties:
<math display="block">\begin{align}
\operatorname{tr}(\mathbf{A} + \mathbf{B}) &= \operatorname{tr}(\mathbf{A}) + \operatorname{tr}(\mathbf{B}), \\
\operatorname{tr}(c\mathbf{A}) &= c \operatorname{tr}(\mathbf{A}), \\
\operatorname{tr}(\mathbf{A}\mathbf{B}) &= \operatorname{tr}(\mathbf{B}\mathbf{A}),
\end{align}</math>
characterize the trace [[up to]] a scalar multiple in the following sense: If <math>f</math> is a [[linear functional]] on the space of square matrices that satisfies <math>f(xy) = f(yx),</math> then <math>f</math> and <math>\operatorname{tr}</math> are proportional.<ref group="note">Proof: Let <math>e_{ij}</math> the standard basis and note that <math>f\left(e_{ij}\right) = f\left(e_{i} e_{j}^\top\right) = f\left(e_i e_1^\top e_1 e_j^\top\right) = f\left(e_1 e_j^\top e_i e_1^\top\right) = f\left(0\right) = 0</math> if  <math>i \neq j</math> and <math>f\left(e_{jj}\right) = f\left(e_{11}\right)</math>
<math display="block">f(\mathbf{A}) = \sum_{i, j} [\mathbf{A}]_{ij} f\left(e_{ij}\right) = \sum_i [\mathbf{A}]_{ii} f\left(e_{11}\right) = f\left(e_{11}\right) \operatorname{tr}(\mathbf{A}).</math>

More abstractly, this corresponds to the decomposition
<math display="block">\mathfrak{gl}_n = \mathfrak{sl}_n \oplus k,</math>
as <math>\operatorname{tr}(AB) = \operatorname{tr}(BA)</math> (equivalently, <math>\operatorname{tr}([A, B]) = 0</math>) defines the trace on <math>\mathfrak{sl}_n,</math> which has complement the scalar matrices, and leaves one degree of freedom: any such map is determined by its value on scalars, which is one scalar parameter and hence all are multiple of the trace, a nonzero such map.</ref>

For <math>n\times n</math> matrices, imposing the normalization <math>f(\mathbf{I}) = n</math> makes <math>f</math> equal to the trace.

===Trace as the sum of eigenvalues===
Given any {{math|''n'' × ''n''}} matrix {{math|'''A'''}}, there is

{{Equation box 1
|indent=:
|title=
|equation = <math>\operatorname{tr}(\mathbf{A}) = \sum_{i=1}^n \lambda_i</math>
|cellpadding= 6
|border
|border colour = #0073CF
|background colour=#F5FFFA}}

where {{math|&lambda;<sub>1</sub>, ..., &lambda;<sub>''n''</sub>}} are the [[eigenvalue]]s of {{math|'''A'''}} counted with multiplicity. This holds true even if {{math|'''A'''}} is a real matrix and some (or all) of the eigenvalues are complex numbers. This may be regarded as a consequence of the existence of the [[Jordan canonical form]], together with the similarity-invariance of the trace discussed above.

===Trace of commutator===
When both {{math|'''A'''}} and {{math|'''B'''}} are {{math|''n'' × ''n''}} matrices, the trace of the (ring-theoretic) [[commutator]] of {{math|'''A'''}} and {{math|'''B'''}} vanishes: {{math|1=tr(['''A''', '''B''']) = 0}}, because {{math|1=tr('''AB''') = tr('''BA''')}} and {{math|tr}} is linear. One can state this as "the trace is a map of [[Lie algebras]] {{math|gl<sub>''n''</sub> → ''k''}} 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.

Conversely, any square matrix with zero trace is a linear combination of the commutators of pairs of matrices.<ref group="note">Proof: <math>\mathfrak{sl}_n</math> is a [[semisimple Lie algebra]] and thus every element in it is a linear combination of commutators of some pairs of elements, otherwise the [[derived algebra]] would be a proper ideal.</ref> Moreover, any square matrix with zero trace is [[Unitary representation|unitarily equivalent]] to a square matrix with diagonal consisting of all zeros.

===Traces of special kinds of matrices===
{{bulleted list
| The trace of the {{math|''n'' × ''n''}} [[identity matrix]] is the dimension of the space, namely {{mvar|n}}.
<math display="block">\operatorname{tr}\left(\mathbf{I}_n\right) = n</math>
This leads to [[Dimension (vector space)#Trace|generalizations of dimension using trace]].
| The trace of a [[Hermitian matrix]] is real, because the elements on the diagonal are real.
| The trace of a [[permutation matrix]] is the number of [[Fixed point (mathematics)|fixed points]] of the corresponding permutation, because the diagonal term {{math|''a''<sub>''ii''</sub>}} is 1 if the {{math|''i''}}th point is fixed and 0 otherwise.
| The trace of a [[Projection_(linear_algebra)|projection matrix]] is the dimension of the target space.
<math display="block">\begin{align}
                                  \mathbf{P}_\mathbf{X} &= \mathbf{X}\left(\mathbf{X}^\mathsf{T} \mathbf{X}\right)^{-1} \mathbf{X}^\mathsf{T} \\[3pt]
\Longrightarrow
    \operatorname{tr}\left(\mathbf{P}_\mathbf{X}\right) &= \operatorname{rank}(\mathbf{X}).
\end{align}</math>
The matrix {{math|'''P<sub>X</sub>'''}} is idempotent.
| More generally, the trace of any [[idempotent matrix]], i.e. one with {{math|1='''A'''<sup>2</sup> = '''A'''}}, equals its own [[rank (linear algebra)|rank]].
| The trace of a [[nilpotent matrix]] is zero.
{{pb}}
When the characteristic of the base field is zero, the converse also holds: if {{math|1=tr('''A'''<sup>''k''</sup>) = 0}} for all {{mvar|k}}, then {{math|'''A'''}} is nilpotent.
{{pb}}
When the characteristic {{math|''n'' > 0}} is positive, the identity in {{mvar|n}} dimensions is a counterexample, as <math>\operatorname{tr}\left(\mathbf{I}_n^k\right) = \operatorname{tr}\left(\mathbf{I}_n\right) = n \equiv 0</math>, but the identity is not nilpotent.
}}

=== Relationship to the characteristic polynomial ===

The trace of an <math>n \times n</math> matrix <math>A</math> is the coefficient of <math>t^{n-1}</math> in the [[characteristic polynomial]], possibly changed of sign, according to the convention in the definition of the characteristic polynomial.