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Chapter 8: Problem 11

Let \(B\) be an \(n \times n\) matrix, and let \(V_{1}, V_{2}, \ldots, V_{k}\) be vectors in \(\mathbb{R}^{n}\) such that \(V_{1} \neq 0\), \(B V_{1}=0, B V_{2}=V_{1}, \ldots, B V_{k}=V_{k-1}\). Use induction on \(k\) to prove that \(\left\\{V_{1}, V_{2}, \ldots, V_{k}\right\\}\) is linearly independent. How large can \(k\) be?

Short Answer

Expert verified
The set is linearly independent by induction. Maximum \(k = n\).

Step by step solution

01

Base Case for Induction

Let's start with the base case where \(k = 1\). We are given that \(B V_1 = 0\) and \(V_1 eq 0\). A single non-zero vector is always linearly independent by definition. Thus, the set \(\{V_1\}\) is linearly independent. This proves our base case.
02

Inductive Hypothesis

Assume that for some positive integer \(m\), the set \(\{V_1, V_2, \ldots, V_m\}\) is linearly independent. This is our inductive hypothesis.
03

Inductive Step

We need to show that \(\{V_1, V_2, \ldots, V_m, V_{m+1}\}\) is also linearly independent. Suppose that there is a linear combination such that \(c_1 V_1 + c_2 V_2 + \cdots + c_m V_m + c_{m+1} V_{m+1} = 0\). Apply matrix \(B\) to both sides.
04

Apply Matrix B to Induction Equation

When we apply \(B\), we have \(B(c_1 V_1 + c_2 V_2 + \cdots + c_m V_m + c_{m+1} V_{m+1}) = B 0 = 0\). Using the properties \(B V_{m+1} = V_m, B V_m = V_{m-1}, \ldots, B V_2 = V_1, B V_1 = 0\), the equation becomes \(c_2 V_1 + c_3 V_2 + \cdots + c_{m+1} V_m = 0\).
05

Conclusion from Inductive Step

From the equation \(c_2 V_1 + c_3 V_2 + \cdots + c_{m+1} V_m = 0\), and since \(\{V_1, V_2, \ldots, V_m\}\) is linearly independent by the inductive hypothesis, we conclude that \(c_2 = c_3 = \cdots = c_{m+1} = 0\). In the original equation, this means \(c_1 = 0\) as well, proving \(\{V_1, V_2, \ldots, V_{m+1}\}\) is linearly independent. This completes the inductive step.
06

Determine Maximum Size k

The vectors result from applying \(B\) repeatedly. Since \(B\) is \(n \times n\), it can be reduced to its minimal polynomial of degree \(\leq n\). Therefore, \(k\) can be at most \(n\) since beyond that, the vectors would not be independent due to the dimensions of the null space.

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Key Concepts

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

Matrix Theory and Linearity
In matrix theory, understanding the behavior of linear transformations is pivotal. An \(n \times n\) matrix \(B\) acts on vectors in \(\mathbb{R}^n\), providing a way to map these vectors into new spaces.
Here, each operation of \(B\) on vector \(V_i\) changes or translates the vector to another, based on the given conditions. This direct application enhances our grasp of linear transformations.
  • Zero Transformation: The condition \(BV_1 = 0\) reflects the kernel or null space of \(B\).
  • Transformation Chain: Each transformation \(BV_{i+1} = V_i\) illustrates a linkage, establishing a sequence of dependencies, yet exhibiting potential linear independence within the system.
Familiarizing oneself with these transformations simplifies deeper understandings in matrix theory, such as exploring eigenvalues, eigenvectors, and systems solutions.
Induction and Proof Techniques
Mathematical induction is a powerful tool for proving statements about integers. The given problem uses induction to establish the linear independence of vectors. This process requires two core steps: the base case and the inductive step.
  • Base Case: Starting small, with \(k=1\), it directly proves the initial condition with the simplest scenario鈥攕ingle vector non-zero鈥攖hat is independent by nature.
  • Inductive Hypothesis: This assumption expects that for a set \(\{V_1, V_2, \ldots, V_m\}\), the linear independence holds.
  • Inductive Step: Crucially, this step extends the assumption to \(\{V_1, V_2, \ldots, V_{m+1}\}\), demonstrating through transformation and algebra that independence persists as the set grows.
By systematically tackling the problem with induction, the proof becomes both elegant and rigorous, delivering insights into the hierarchical nature of linear transformations applied progressively.
Vector Spaces and Linear Independence
Vector spaces are fundamental in understanding linear systems. The concept of linear independence within a vector space asks if sets of vectors can be expressed as combinations of others in the set.
  • Linear Independence: The vectors \(V_1, V_2, \ldots, V_k\) are linearly independent if no vector in the set can be written as a combination of others. This intrinsic characteristic makes solving systems and decomposing spaces coherent and reliable.
  • Dimensions and Maximum Extent: The problem hints at the maximum dimension \(k\) can have for \(\{V_1, V_2, \ldots, V_k\}\) being independent, essentially bounded by the space itself (\(n\)) due to its structural capacity, symbolized by its rank or nullity.
Understanding these geometries of vector spaces not only aids in grasping linear transformations but also grounds potential theoretical advancements in linear algebraic exploration.

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Most popular questions from this chapter

Find the general solution of the system $$ \left\\{\begin{array}{l} x^{\prime}=3 x-4 y+e^{t} \\ y^{\prime}=x-y-e^{t} \end{array}\right. $$ Hint: Try functions of the form \(e^{t}, t e^{t}\), and \(t^{2} e^{t}\).

Solve the boundary-value problem $$ \left\\{\begin{array}{ll} x^{\prime \prime}=x^{2} & \\ x(0)=2 / 3 & x(1)=3 / 8 \end{array}\right. $$

Solve the boundary-value problem $$ \left\\{\begin{array}{ll} x^{\prime \prime}=x^{3} & \\ x(0)=1 & x(1)=2-\sqrt{2} \end{array}\right. $$

Consider the two-point boundary-value problem $$ \left\\{\begin{array}{l} x^{\prime \prime}=f(t, x) \\ x(0)=x(1)=0 \end{array}\right. $$ For which function(s) \(f\) can we be sure that a unique solution exists? i. \(f(t, x)=t^{2} x^{3}\) ii. \(f(t, x)=t\left(\tan ^{-1} x\right)\) iii. \(f(t, x)=t x^{4 / 3}\) iv. \(f(t, x)=\log \left(1+x^{2}\right)\) v. \(f(t, x)=|t x|\)

Write this system of higher-order ordinary differential equations as a system of first-order equations: $$ \left\\{\begin{aligned} y_{1}^{(n)} &=f_{1}\left(t, y_{1}, y_{1}^{\prime}, \ldots, y_{1}^{(n-1)}\right) \\ y_{2}^{(n)}=& f_{2}\left(t, y_{2}, y_{2}^{\prime}, \ldots, y_{2}^{(n-1)}\right) \\ & \vdots \\ y_{m}^{(n)} &=f_{m}\left(t, y_{m}, y_{m}^{\prime}, \ldots, y_{m}^{(n-1)}\right) \end{aligned}\right. $$
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