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单词 MatrixExponential
释义

matrix exponential


The exponential of a real valued square matrixMathworldPlanetmath A, denotedby eA, is defined as

eA=k=01k!Ak
=I+A+12A2+

Let us check that eA is a real valued square matrix.Suppose M is a real number such |Aij|<M for allentries Aij of A.Then |(A2)ij|<nM2 for all entries in A2,where n is the order of A. (Alternatively, one could argue using matrix normsMathworldPlanetmath: We have ||eA||e||A|| for the 2-norm, and hence the entries of eA are bounded by M=||eA||.) Thus,in general, we have |(Ak)i,j|<nkMk+1. Sincek=0nkk!Mk+1 converges, we see thateA converges to real valued n×n matrix.

Example 1. Suppose A is nilpotent, i.e., Ar=0 for some naturalnumberMathworldPlanetmath r. Then

eA=I+A+12!A2++1(r-1)!Ar-1.

Example 2. If A is diagonalizable, i.e., of the formA=LDL-1, where D is a diagonal matrixMathworldPlanetmath, then

eA=k=01k!(LDL-1)k
=k=01k!LDkL-1
=LeDL-1.

Further, ifD=diag{a1,,an}, thenDk=diag{a1k,,ank} whence

eA=Ldiag{ea1,,ean}L-1.

For diagonalizable matrix A, it follows thatdeteA=etrA.However, this formulaMathworldPlanetmathPlanetmath is, in fact, valid for all A.


Let A be a square n×n real valued matrix.Then the matrix exponentialMathworldPlanetmath satisfies the following properties

  1. 1.

    For the n×n zero matrixMathworldPlanetmath O, eO=I, where I is then×n identity matrixMathworldPlanetmath.

  2. 2.

    If A=Ldiag{a1,,an}L-1 for an invertiblePlanetmathPlanetmathPlanetmath n×nmatrix L, then

    eA=Ldiag{ea1,,ean}L-1.
  3. 3.

    If A and B commute,then eA+B=eAeB.

  4. 4.

    The trace of A and the determinantMathworldPlanetmath of eA are related by the formula

    deteA=etrA.

    In effect, eA is always invertible. The inversePlanetmathPlanetmathPlanetmathPlanetmath is given by

    (eA)-1=e-A.
  5. 5.

    If eA is a rotational matrix, then trA=0.

A relevant example on property 3.
We report an interesting example where the cited property is valid. In the field of complex numbersMathworldPlanetmathPlanetmath consider the complex matrix

C=A+iB,(1)

being C hermitian, i.e. C=C¯ (here ”” and overline ”-” stand for tranposition and conjugation, respectively) and orthogonalMathworldPlanetmathPlanetmath, i.e C-1=C . From (1),

C=A+iB.

Since C is orthogonal, from the complex equation CC=I (I is the identity matrix), we have

CC=(A+iB)(A+iB)=(AA-BB)+i(BA+AB)=I,

whence the imaginary part leads to the equation

BA+AB=0.(2)

But C is also hermitian, so that

C=A+iB=C¯=A-iB,

therefore A=A is symmetricPlanetmathPlanetmathPlanetmathPlanetmath, and B=-B is skew-symmetric. From these and (2), BA=AB, and this implies that exp(A)exp(B)=exp(A+B). So that, the real and imaginary parts of an orthogonal and hermitian matrix verifies the property. Likewise, it is easy to show that if the complex matrix is symmetric and unitary, its real an imaginaryPlanetmathPlanetmath componentsPlanetmathPlanetmath also verify this property.

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更新时间:2025/5/4 4:46:20