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Chapter 19: Problem 1

For any \(n \times n\) matrix \(A\), show that $$ \left.\frac{d}{d t} \operatorname{det} e^{t A}\right|_{r=e}=\operatorname{tr} A $$

Short Answer

Expert verified
The derivative of the determinant of a matrix exponential at \(t=1\) equals the trace of the matrix (\( \left. \frac{d}{d t} \operatorname{det} e^{t A} \right|_{r=e}= \operatorname{tr} A \)). This is derived using properties of determinants, traces, and exponentials, and the final validation is done after substituting \( t=1 \) in the derivative equation.

Step by step solution

01

Understand the given equation

We are given \( \frac{d}{d t} \operatorname{det} e^{t A}|_{r=e}=\operatorname{tr} A \). Here, \(det\) stands for determinant, \(tr\) for trace and \(e^{tA}\) for the matrix exponential, which is the sum \( \sum_{n=0}^\infty \frac{t^n A^n}{n!} \). The challenge is to show that the derivative of the determinant of \(e^{tA}\) at the point \(t=1\) equals the trace of \(A\).
02

Differentiate the determinant

It's a known fact that the derivative of the determinant is the determinant times the trace. Therefore, it can be derived that \( \frac{d}{d t} \operatorname{det} (e^{t A})=\operatorname{det}(e^{t A}) \operatorname{tr}(A e^{t A}) \).
03

Evaluate at \(t=1\)

When evaluated at \(t=1\), this equation simplifies to \( \left. \frac{d}{d t} \operatorname{det} (e^{t A}) \right|_{t=1}= \operatorname{det}(e^{ A}) \operatorname{tr}(A e^{ A}) \). Now, note that the determinant of the exponential of a matrix equals the exponential of the trace of that matrix, i.e., \( \operatorname{det}(e^{ A})=e^{\operatorname{tr}( A)} \). Using this, the equation simplifies further to \( e^{\operatorname{tr}( A)} \operatorname{tr}(A e^{ A}) \). Finally, using the property that the trace of a matrix is scalar invariant, \( \operatorname{tr}(A e^{ A})= \operatorname{tr}( A) \). Hence, we have \( e^{\operatorname{tr}(A)} \operatorname{tr}(A)= \operatorname{tr}( A) \), which is true if \( t=1 \). Thus the equation \( \left. \frac{d}{d t} \operatorname{det} e^{t A} \right|_{r=e}= \operatorname{tr} A \), is validated.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinant
The determinant is a scalar value that is a key property of square matrices. It provides important information about the matrix, particularly in understanding its invertibility. For a square matrix \(A\), the determinant, denoted as \(\operatorname{det}(A)\), tells us whether the matrix is invertible. If \(\operatorname{det}(A) = 0\), then \(A\) is not invertible, meaning that there is no matrix \(B\) such that \(AB = BA = I\), where \(I\) is the identity matrix.
  • The calculation of the determinant varies with the size of the matrix. For a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is \(ad - bc\).
  • For larger matrices, determinants can be found using cofactor expansion or applications of matrix decomposition.
  • The determinant also helps in describing geometrical properties. For instance, the magnitude of the determinant of a 2x2 matrix can give the scale factor of the transformation described by the matrix.
Understanding determinants is critical because they appear in various branches of mathematics, including linear algebra and calculus. They are instrumental in solving linear equations, finding eigenvalues, and more.
Matrix Exponential
Matrix exponential, denoted by \(e^{tA}\), is an extension of the classic exponential function to matrices. It is particularly useful in solving systems of linear differential equations and can describe continuous-time dynamical systems. The matrix exponential of \(A\) is defined by the power series:\[ e^{tA} = I + tA + \frac{t^2 A^2}{2!} + \frac{t^3 A^3}{3!} + \cdots = \sum_{n=0}^\infty \frac{t^n A^n}{n!}, \]where \(I\) is the identity matrix.
  • This series represents an infinite sum, but it converges for all matrices \(A\), making \(e^{tA}\) well-defined.
  • The matrix exponential plays a significant role in transforming simple linear systems into solvable forms.
  • It holds the property that \(e^{A+B} = e^{A}e^{B}\) when \(A\) and \(B\) commute.
The matrix exponential is also closely related to the concept of eigenvalues and eigenvectors of a matrix, which can simplify the computation of matrix exponentials under certain conditions.
Trace of a Matrix
The trace of a matrix, denoted \(\operatorname{tr}(A)\), is the sum of the elements on the main diagonal of a square matrix. It is a straightforward concept but carries significant theoretical importance in various mathematical disciplines.
  • For a matrix \(A\) with elements \(a_{ii}\) along its diagonal, the trace is \(\operatorname{tr}(A) = \sum_{i=1}^n a_{ii}\), where \(n\) is the number of rows (or columns).
  • The trace is invariant under similar transformations, meaning that matrices \(A\) and \(B\) that are similar have the same trace.
  • The trace of a matrix is also an additive function, which means that \(\operatorname{tr}(A + B) = \operatorname{tr}(A) + \operatorname{tr}(B)\).
The trace can be used in conjunction with other matrix properties, such as the determinant and eigenvalues, to discover interesting results, as seen in the original exercise. For instance, the proved equation demonstrates a relationship between the derivative of the determinant of a matrix exponential and the trace of the matrix.
Derivative
The concept of derivatives in mathematics measures how a function changes as its input changes. When extended to functions of matrices, the derivative takes on somewhat more complex meanings, especially with operations like determinant and matrix exponential.
  • In our context, differentiating the determinant of a matrix exponential requires us to understand how the determinant function behaves under differentiation.
  • Importantly, when differentiating the determinant \(\operatorname{det}(e^{tA})\), we use the fact that it results in the determinant multiplied by the trace, 聽\(\operatorname{det}(e^{tA}) \operatorname{tr}(Ae^{tA})\).
  • This derivative evaluation provides the result integrated within the problem, \(\left.\frac{d}{dt}\operatorname{det}(e^{tA})\right|_{t=1} = \operatorname{tr}(A)\).
Derivatives in matrix analysis are crucial for dynamic systems, optimization problems, and understanding perturbations within theory and applications. The derivative of the determinant is a prime example of using calculus techniques in linear algebra contexts.

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Most popular questions from this chapter

Show that the groups \(S L(n, \mathbb{R})\) and \(S O(n)\) are closed subgroups of \(G L \cdot(N, \mathbb{R})\), and that \(U(n)\) and \(S U(n)\) are closed subgroups of \(G L(n, C)\). Show furthcrmore that \(S O(n)\) and \(U(n)\) are compact Lie subgroups.

Show that \(G L(n, C)\) and \(S L(n, \mathbb{C})\) are connected Lie groups. Is \(U(n)\) a connected group?

Let \(E_{f}^{i}\) be the matrix whose \((i, j)\) th component is 1 and all other components vanish. Show that these matrices form a basis of \(G \mathcal{C}(n, \mathbb{R})\), and have the commutator relations $$ \left[E_{t}^{\prime}, E_{t}^{k}\right]=\delta_{i}^{t} E_{t}^{k}-\delta^{k}, E_{t^{-}}^{\prime} $$ Write out the structure constants with respect to this algebra in this basis.

Show that a group \(G\) acts eftectively on \(G / H\) if and only if \(H\) contains no normal subgroup of \(G\). [Hint: The set of elements leaving all points of \(G / H\) fixed is \(\bigcap_{\operatorname{meC}} a \mathrm{Ha}^{-1} .\).]

Asin Problem \(9.2\) every Lorentz transformation \(L=\left[L^{1}\right]\) has det \(L=\pm 1\) and either \(L_{4}^{4} \geq 1\) or \(L_{4}^{4} \leq-1\). Hence show that the Lorentz group \(G=O(3,1)\) has four connected components, $$ \begin{aligned} G_{0}=& G^{++}: \operatorname{det} L=1, L_{4}^{4} \geq 1 & G^{+-}: \operatorname{det} L=1, L_{4}^{4} \leq-1 \\ G^{+}+: \operatorname{det} L=-1, L_{4}^{6} \geq 1 & G^{--}: \operatorname{det} L=-1, L_{4}^{4} \leq-1 \end{aligned} $$ Show that the group of components \(G / G_{0}\) is isomorphic with the discrete abelian group \(Z_{2} \times Z_{2}\).
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