Optional further context — Matrix exponentials, determinants, and Lie algebras
This contents of this section are taken from the video Matrix exponentials, determinants, and Lie algebras by Michael Penn. Although this section is not needed in the context of the FLT proof, it demonstrates a nice application of many of the concepts dicussed in this chapter on linear algebra nonetheless.
🎯 Goal
Suppose that \(A\) is an \(n\times n\) matrix. Then
\[ \boxed{e^{\text{tr}(A)} = \det(e^A)} \]
where
\[ e^A = I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \ldots \]
✍️ Example
\[ A=\begin{pmatrix}4 && 2 \\ 1 && 5\end{pmatrix} \]
Find the trace
Let’s first calculate the left-hand side of the equation.
\[ \boxed{\text{tr}(A) = 9 \implies e^{\text{tr}(A)} = e^9} \]
Find the determinant
Eigenvalues and eigenvectors
Next, let’s calculate the right-hand side of the equation. We do this by finding the eigenvalues and eigenvectors first, as it allows us to diagonalize A and carry out the exponentiation much easier.
The characteristic polynomial for this matrix \(A\) is:
\[ \begin{array}{ll} \chi_A(x) &=& \det(A- xI)\\ &=& \det \begin{pmatrix} 4 - x && 1 \\ 2 && 5 - x \end{pmatrix} \\ &=& (4-x)(5-x) -2 \\ &=& x^2 -9x + 18 \\ &=& (x-3)(x-6) \end{array} \]
So our eigenvalues are: \(\boxed{\lambda_1=3, \lambda_2=6}\)
Next, let’s find our eigenvectors. This is equivalent to determining the kernel of \((A - \lambda I)\) for each of the two eigenvalues as:
\[ Av=\lambda v \Rightarrow (A - \lambda I)v = 0 \Rightarrow v\in \ker(A - \lambda I) \]
So for \(\lambda_1\) we have
\[ \begin{array}{ll} \ker(A - \lambda_1 I) &=& \ker \begin{pmatrix} 4 - 3 && 1 \\ 2 && 5 - 3 \end{pmatrix}\\ &=& \ker \begin{pmatrix} 1 && 1 \\ 2 && 2 \end{pmatrix} \\ &=& \ker \begin{pmatrix} 1 && 1 \\ 0 && 0 \end{pmatrix} \Rightarrow \begin{pmatrix} 1 && 1 \\ 0 && 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{array} \]
This implies that \(x = -y\), so we have \(\boxed{\overrightarrow{e_1} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}}\).
Analogously, we find that \(\boxed{\overrightarrow{e_2} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}}\).
Conseqently, the diagonalizing matrix \(P\) becomes
\[ P = (\overrightarrow{e_1}\ \overrightarrow{e_2}) = \begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix} \Rightarrow P^{-1} =\frac{1}{3} \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix} \]
So
\[ PAP^{-1} = \begin{pmatrix} 3 & 0 \\ 0 & 6 \end{pmatrix} \Rightarrow A = P\begin{pmatrix} 3 & 0 \\ 0 & 6 \end{pmatrix}P^{-1} \Rightarrow A^n = P\begin{pmatrix} 3 & 0 \\ 0 & 6 \end{pmatrix}^n P^{-1} \]
Now we can move on to calculate the exponential:
\[ \begin{array}{ll} e^A &=& \sum_{n=0}^{\infty} \frac{1}{n!}A^n \\ &=& P\bigg[\sum_{n=0}^{\infty}\begin{pmatrix} 3^n & 0 \\ 0 & 6^n \end{pmatrix}\bigg] P^{-1} \\ &=& P\begin{pmatrix} e^3 & 0 \\ 0 & e^6 \end{pmatrix} P^{-1} \\ &=& \begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} e^3 & 0 \\ 0 & e^6 \end{pmatrix}\frac{1}{3} \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix} \\ &=& \frac{1}{3}\begin{pmatrix} e^3 & e^6 \\ -e^3 & 2e^6 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix} \\ &=& \frac{1}{3}\begin{pmatrix} 2e^3 + e^6 & -e^3+e^6 \\ -2e^3+2e^6 & e^3+2e^6 \end{pmatrix} \end{array} \]
Finally, we can caculate the determinant:
\[ \begin{array}{ll} \det(e^A) &=& \frac{1}{9}\big((2e^3 + e^6)(e^3+2e^6)+(e^3-e^6)(-2e^3+2e^6) \big) \\ &=& \frac{1}{9}(2e^6+4e^9 +e^9+2e^{12} -2e^6+2e^9+2e^9-2e^{12} ) \\ &=& e^9 = e^{\text{tr}(A)}\ ꪜ \end{array} \]
📌 What is going on?
👉 \(\text{tr}(A)\) is the sum of eigenvalues of \(A\).
👉 \(\det(A)\) is the product of eigenvalues of \(A\).
👉 \(\lambda\) is an eigenvalue of \(A \Longleftrightarrow e^\lambda\) is an eigenvector of \(e^A\).
From these it quickly follows that \(e^{\text{tr}(A)} = \det(e^A)\).
The proof these statements in general is a bit tricky, so we limit ourselves to the diagonalizable case here, with the following lemma (that can easily be proven using the definition of matrix multiplication):
\[ \text{tr}(xy)=\text{tr}(yx) \]
Let’s assume that \(A\) is diagonalizable:
\[ PAP^{-1} = \begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & \lambda_n\end{pmatrix} \]
Then we can write
\[ \text{tr}(A) = \text{tr}\left (P\begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & \lambda_n\end{pmatrix}P^{-1}\right ) \]
Using the above lemma, we can commute the \(P\) and \(P^{-1}\), so \(\text{tr}(A) = \lambda_1+\ldots+\lambda_n\). This gives us
\[ e^{\text{tr}(A)}=e^{\lambda_1+\ldots+\lambda_n}=\boxed{e^{\lambda_1} \ldots e^{\lambda_n}} \]
At the same time,
\[ e^A = P\begin{pmatrix} e^{\lambda_1} & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & e^{\lambda_n}\end{pmatrix}P^{-1} \]
from which follows
\[ \det(e^A)=\det(P)\left [e^{\lambda_1}\ldots e^{\lambda_n}\right ]\det(P^{-1}) = \boxed{e^{\lambda_1}\ldots e^{\lambda_n}} \]
📌 Lie groups and Lie algebras
\[ \begin{array}{l} SL_n(\mathbb{C})=\{X\ |\ \det(X) = 1\} \\ sl_n(\mathbb{C})=\{x\ |\ \text{tr}(x)=0 \},\ \text{e.g.}\ \begin{pmatrix} -2 & 3 \\ 4 & 2 \end{pmatrix} \in sl_2 \end{array} \]
Note that if \(x\in sl_n(\mathbb{C})\), set \(X=e^x\). Then \(\det(X) = \det(e^x) = e^{\text{tr}(x)} = 1\). That is exactly the requirement that we have for \(X\) to be an element of \(SL_2(\mathbb{C})\)!
🧠 Concluding, \(\boxed{e^{\text{tr}(A)} = \det(e^A)}\) gives us a mechanism to go from the Lie group to the Lie algebra and vice versa.