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Chapter 6: Problem 5

If \(Q\) is a real symmetric matrix satisfying \(Q^{k}=1\) for \(k \geq 1\), prove that \(Q^{2}=1\).

Short Answer

Expert verified
Since the eigenvalues of \(Q\) are \(1\) or \(-1\), \(Q^2 = I\).

Step by step solution

01

Understanding the Given Information

We are given that \(Q\) is a real symmetric matrix and \(Q^k = I\) for some integer \(k \geq 1\), where \(I\) is the identity matrix. This condition tells us that \(Q\) is an involutory matrix when \(k = 2\). We need to prove that \(Q^2 = I\).
02

Using Properties of Real Symmetric Matrices

Recall that if \(Q\) is a real symmetric matrix, it is diagonalizable with real eigenvalues. Let \(\lambda\) be any eigenvalue of \(Q\) with corresponding eigenvector \(v\), so \(Qv = \lambda v\).
03

Analyzing the Eigenvalue Condition

By the given condition \(Q^k = I\), eigenvalues of \(Q\) must satisfy \(\lambda^k = 1\). These values are the \(k^{th}\) roots of unity, which are typically complex numbers.
04

Considering Real Solutions for Eigenvalues

Since \(Q\) is real and symmetric, its eigenvalues are real. The real \(k^{th}\) roots of unity that are also real can only be \(1\) and \(-1\).
05

Proving \(Q^2 = I\) Using Eigenvalue Properties

If the eigenvalues of \(Q\) are only \(1\) and \(-1\), then \(\lambda^2 = 1^2 = 1\) or \((-1)^2 = 1\) for any eigenvalue \(\lambda\). Thus, the matrix \(Q^2\) will have all eigenvalues equal to \(1\), meaning \(Q^2 = I\).
06

Conclusion

Since the eigenvalues are real and \(Q^k = I\) implies that all eigenvalues raised to any power \(k\) remain \(1\), we conclude that the matrix \(Q^2 = I\) must hold. Thus proving the statement.

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Key Concepts

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

Eigenvalues
Eigenvalues are special numbers associated with a matrix. When you multiply a matrix by some vector, the result is usually very different from a simple scaling of that vector. But in the case of eigenvectors, multiplying by the matrix is equivalent to multiplying by a scalar, which is the eigenvalue. Mathematically, if a matrix \(A\) has an eigenvalue \(\lambda\) with an eigenvector \(v\), it's expressed as:
\[ Av = \lambda v. \]
To find eigenvalues, you must solve the characteristic equation:
  • Compute the determinant of \(A - \lambda I = 0\), where \(I\) is the identity matrix.
  • Find solutions for \(\lambda\) by solving this polynomial equation.
For real symmetric matrices, eigenvalues are always real, which means they can be safely used in real-world applications without encountering imaginary numbers.
Involutory Matrix
An involutory matrix is a matrix that, when multiplied by itself, returns the identity matrix. Mathematically, this is expressed as \(A^2 = I\), where \(I\) is the identity matrix. This property makes involutory matrices unique in linear algebra due to:
  • Stability: applying the same transformation twice returns the system to its original state.
  • Eigenvalues: such matrices have eigenvalues of exactly \(1\) or \(-1\).
In the context of the given problem with a matrix \(Q\), the involutory property suggests that even after transformations (represented by powers), \(Q^2\) will still be the identity matrix if \(Q\) also satisfies \(Q^k = I\). This important property is pivotal in proving statements about matrices in symmetrical systems.
Diagonalizable Matrix
A matrix is diagonalizable if it can be transformed into a diagonal matrix. This means the matrix can be rewritten with its eigenvalues on the diagonal and zeros elsewhere. If a matrix is real and symmetric, like in our problem, it's always diagonalizable due to the spectral theorem. Key points about diagonalizable matrices include:
  • A matrix \(A\) is similar to a diagonal matrix \(D\), such that \(A = PDP^{-1}\).
  • Here, \(P\) is a matrix whose columns are the eigenvectors of \(A\), and \(D\) consists of eigenvalues.
  • Diagonalization simplifies exponentiation, making problems involving powers easier to compute.
Since our given matrix is symmetric, it essentially guarantees that it's diagonalizable, which is a crucial step in concluding properties like involutory behavior.
k-th Roots of Unity
The k-th roots of unity are complex numbers that satisfy the equation \(z^k = 1\). In simpler terms, these are the numbers that become 1 when raised to a power of \(k\). For a real symmetric matrix, which can only have real eigenvalues, the solutions that meet both criteria are \(1\) and \(-1\). Key properties include:
  • The k-th roots of unity form a regular polygon on the complex plane.
  • The only real numbers among them are \(1\) and \(-1\).
  • These roots relate to the characteristic polynomial of the matrix.
In our context, with \(Q^k = I\) implying its eigenvalues are among the k-th roots of unity, the restriction to real numbers explains why they must be \(1\) or \(-1\), leading to the conclusion that \(Q^2 = I\). This unique property assists in solving equations for power matrices effectively.

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Most popular questions from this chapter

Prove that if \(A \geq 0\) then \(A\) has a unique nonnegative square root.

(a) Prove that the matrix $$ \left(\begin{array}{rrr} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 1 & 1 & 0 \end{array}\right) $$ is nilpotent, and find its invariants and Jordan form. (b) Prove that the matrix in part (a) is not similar to $$ \left(\begin{array}{rrr} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 1 & 0 & 0 \end{array}\right) . $$

If \(F\) is the field of integers modulo 2 and if \(V\) is two-dimensional over \(F\), compute the group of regular elements in \(A(V)\) and prove that this group is isomorphic to \(S_{3}\), the 'symmetric group of degree 3 .

Prove that, given the matrix $$ A=\left(\begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6 \end{array}\right) \quad \in F_{3} $$ (where the characteristic of \(F\) is not 2 ), then: (a) \(A^{3}-6 A^{2}+11 A-6=0\). (b) There exists a matrix \(C \in F_{3}\) such that $$ C A C^{-1}=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) $$

If \(T \in A(V)\), let \(V_{0}=\left\\{v \in V \mid v T^{k}=0\right.\) for some \(\left.k\right\\}\). Prove that \(V_{0}\) is a subspace and that if \(v T^{m} \in V_{0}\), then \(v \in V_{0}\)
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