Chapter 4 Matrix Exponential

The exponential of a matrix plays a crucial role in the theory of Lie groups. The exponential enters into the definition of the Lie algebra of a matrix Lie group and is the mechanism for passing information from the Lie algebra to the Lie group.

(#def:mat_exp) (Matrix Exponential) If \(X\) is an \(n \times n\) matrix, we define the exponential of \(X\), denoted \(e^X\) or \(\exp(X)\), by the usual power series \[\begin{equation} e^X = \sum_{m=0}^{\infty} \frac{X^m}{m!},\tag{4.1} \end{equation}\] where \(X^0=I\), and \(X^m\) denotes the repeated matrix product of \(X\) with itself.

Proposition 4.1 The series (4.1) converges for all \(X \in M_n(\mathbb{C})\), and \(e^X\) is a continuous function of \(X\).

Theorem 4.1 Let \(X,Y \in M_n(\mathbb{C})\). Then the following properties hold:

  1. \(e^{0} = I\)
  2. \((e^{X})^{\mathrm{T}} = e^{X^{\mathrm{T}}}\qquad\text{and}\qquad(e^{X})^{*} = e^{X^{*}}\)
  3. If \(A\) is an invertible \(n \times n\) matrix, then
    \(e^{AXA^{-1}} = A\, e^{X}\, A^{-1}\)
  4. \(\det(e^{X}) = e^{\operatorname{trace}(X)}\)
  5. If \(XY = YX\), then \(e^{X+Y} = e^{X} e^{Y}\)
  6. \(e^{X} \text{ is invertible and } (e^{X})^{-1} = e^{-X}\)
  7. Even if \(XY \neq YX\), we have \(e^{X+Y} = \lim_{m\to\infty} \left(e^{X/m} e^{Y/m}\right)^{m}\) (This is a special case of the Trotter Product Formula.)

Proof.

  1. It is trivial from the defnition @ref(def:mat_exp).
  2. This is follows by taking term-by-term adjoints in the series for \(e^X\).
  3. This is also easily verified using term-by-term computation.
  4. On schur decomposition, Over \(\mathbb{C}\), any matrix \(X\) is similar to an upper triangular matrix \(T\). That is, there exists an invertible matrix \(S\) such that, (X=STS^{-1}), where \(T\) is upper triangular. By proptery 3, \[ e^X = e^{STS^{-1}}= Se^TS^{-1} \] Then,\[\det(e^X)=\det(Se^TS^{-1})=\det(e^T).\] For the trace, \[\operatorname{tr}(X)=\operatorname{tr}(STS^{-1})=\operatorname{tr}(T).\] Furthur we can calualate

\[ \det(e^{T}) = \prod_{i=1}^{n} e^{\lambda_i} = e^{\lambda_1} e^{\lambda_2} \cdots e^{\lambda_n} = e^{\lambda_1 + \lambda_2 + \cdots + \lambda_n} = e^{\operatorname{tr}(T)}. \]

By some computaiaonal,\(e^{T}\) is upper triangular with diagonal entries \(e^{\lambda_1}, \dots, e^{\lambda_n}\). Thus, \(\det(e^X)=\det(e^T)\).

Now combining all the results, \[ \det(e^{X}) = \det(e^{T}) = e^{\operatorname{tr}(T)} = e^{\operatorname{tr}(X)}. \]

  1. First, note that both series both series \(e^X\) and \(e^Y\) converge absolutely. So, we can multiply them term by term, \[\begin{equation} e^X e^Y = \sum_{m=0}^{\infty} \sum_{k=0}^{m} \frac{X^k}{k!} \frac{Y^{\,m-k}}{(m-k)!} = \sum_{m=0}^{\infty} \frac{1}{m!} \sum_{k=0}^{m} \binom{m}{k} X^k Y^{\,m-k}. \tag{4.2} \end{equation}\] If \(X\) and \(Y\) commute, then \[(X+Y)^m = \sum_{k=0}^{m} \binom{m}{k} X^k Y^{\,m-k},\] so (4.2) becomes \[e^X e^Y = \sum_{m=0}^{\infty} \frac{(X+Y)^m}{m!} = e^{X+Y}.\]
  2. This is foolowed from Property 5 by taking Y = −X and applying Property.
  3. Proof is ommited. (Becasue it is quite long.)

4.1 Computing Exponential

If \(X\) is diagonalizable, then there exist invertible \(A\) and \(\lambda_1,...,\lambda_n\) such that

\[X = A \begin{bmatrix} {\lambda_1} & 0 & \cdots & 0 \\ 0 & {\lambda_2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & {\lambda_n} \end{bmatrix} A^{-1}\]

Then by Theorem 4.1 fourth property, we get the following:

\[e^X = A \begin{bmatrix} e^{\lambda_1} & 0 & \cdots & 0 \\ 0 & e^{\lambda_2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & e^{\lambda_n} \end{bmatrix}A^{-1}\]

If \(X\) is nilpotent then the series that defines \(e^X\) terminates.

Example 4.1 Consider the matrices

\[X_1 = \begin{bmatrix} 0 & -a \\ a & 0 \end{bmatrix}, \quad X_2 = \begin{bmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{bmatrix}, \quad X_3 = \begin{bmatrix} a & b \\ 0 & a \end{bmatrix}.\]

Then

\[e^{X_1} = \begin{bmatrix} \cos a & -\sin a \\ \sin a & \cos a \end{bmatrix}, \text{ and } e^{X_2} = \begin{bmatrix} 1 & a & b + \tfrac{ac}{2} \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}, \text{ and } e^{X_3} = \begin{bmatrix} e^a & e^a b \\ 0 & e^a \end{bmatrix}.\]

Proof. \[e^{X_1} = \begin{bmatrix} 1 & i \\ i & 1 \end{bmatrix} \begin{bmatrix} e^{-ia} & 0 \\ 0 & e^{ia} \end{bmatrix} \frac{1}{2} \begin{bmatrix} 1 & -i \\ -i & 1 \end{bmatrix}.\]