Prove Matrix X Non-Singular When X ≠ 0

Hey everyone! Let's dive into an interesting problem from Munkres' "Analysis on Manifolds." We're going to explore a different way to prove that a certain matrix ${X}$ is non-singular (i.e., invertible) when the vector ${x}$ is not the zero vector. This is a common topic in linear algebra and analysis, and understanding different proofs can give you a more robust grasp of the underlying concepts.

Problem Statement

Given a vector ${x = (x_1, \dots, x_n)}$, we define a matrix ${X}$ as follows:

${ X = \begin{pmatrix} 2x_1^2 + ||x||^2 & 2x_1x_2 & \cdots & 2x_1x_n \\ 2x_2x_1 & 2x_2^2 + ||x||^2 & \cdots & 2x_2x_n \\ \vdots & \vdots & \ddots & \vdots \\ 2x_nx_1 & 2x_nx_2 & \cdots & 2x_n^2 + ||x||^2 \end{pmatrix} }$

where ${||x||^2 = x_1^2 + x_2^2 + \cdots + x_n^2}$ is the squared Euclidean norm of ${x}$.

We want to prove that if ${x \neq 0}$, then the matrix ${X}$ is non-singular. In other words, we want to show that ${X}$ is invertible, or equivalently, that the determinant of ${X}$ is non-zero.

Why is this important?

Understanding the conditions under which a matrix is non-singular is crucial in many areas of mathematics, including:

• Solving Linear Systems: A non-singular matrix guarantees a unique solution to a system of linear equations.
• Eigenvalue Problems: Non-singularity relates to the eigenvalues of a matrix.
• Differential Equations: Analyzing the stability of solutions often involves checking the non-singularity of certain matrices.
• Manifold Theory: As seen from the original textbook, the properties of matrices are important for the analysis on manifold. For example, to check the smoothness of a map.

Proof Strategy

Instead of directly computing the determinant (which can be messy), we'll use a clever trick involving linear transformations. We'll show that ${Xv = 0}$ implies ${v = 0}$. This demonstrates that the null space of ${X}$ contains only the zero vector, meaning that ${X}$ has full rank and is therefore invertible. This approach avoids complicated determinant calculations and provides a more conceptual understanding.

The Proof

Suppose that ${Xv = 0}$ for some vector ${v = (v_1, v_2, \dots, v_n)}$. We want to show that this implies ${v = 0}$.

The equation ${Xv = 0}$ can be written as:

${ \sum_{j=1}^n X_{ij}v_j = 0 \quad \text{for all } i = 1, 2, \dots, n }$

Substituting the expression for ${X_{ij}}$, we get:

${ \sum_{j=1}^n (2x_ix_j + ||x||^2 \delta_{ij})v_j = 0 \quad \text{for all } i = 1, 2, \dots, n }$

Here, ${\delta_{ij}}$ is the Kronecker delta, which is 1 if ${i = j}$ and 0 otherwise. Expanding the sum, we have:

${ 2x_i \sum_{j=1}^n x_jv_j + ||x||^2 v_i = 0 \quad \text{for all } i = 1, 2, \dots, n }$

Let's denote the sum ${\sum_{j=1}^n x_jv_j}$ as ${c}$. Then the equation becomes:

${ 2x_i c + ||x||^2 v_i = 0 \quad \text{for all } i = 1, 2, \dots, n }$

Solving for ${v_i}$, we get:

${ v_i = -\frac{2cx_i}{||x||^2} \quad \text{for all } i = 1, 2, \dots, n }$

Now, let's substitute this expression for ${v_i}$ back into the definition of ${c}$:

${ c = \sum_{j=1}^n x_jv_j = \sum_{j=1}^n x_j \left(-\frac{2cx_j}{||x||^2}\right) = -\frac{2c}{||x||^2} \sum_{j=1}^n x_j^2 = -\frac{2c}{||x||^2} ||x||^2 = -2c }$

So we have ${c = -2c}$, which implies ${3c = 0}$, and therefore ${c = 0}$.

Now, going back to the equation for ${v_i}$:

${ v_i = -\frac{2cx_i}{||x||^2} = -\frac{2(0)x_i}{||x||^2} = 0 \quad \text{for all } i = 1, 2, \dots, n }$

Thus, ${v_i = 0}$ for all ${i}$, which means ${v = 0}$.

Since ${Xv = 0}$ implies ${v = 0}$, the null space of ${X}$ is trivial, and therefore ${X}$ is non-singular (invertible).

Alternative Perspectives and Implications

Geometric Interpretation

Consider the vector ${u = \frac{x}{||x||}}$, which is a unit vector in the direction of ${x}$. The matrix ${X}$ can be related to projections and reflections involving this vector. Specifically, ${X}$ is related to a transformation that amplifies the component of any vector ${v}$ along the direction of ${x}$ and leaves the orthogonal component unchanged. Since ${x \neq 0}$, this transformation is invertible.

Connection to Eigenvalues

The eigenvalues of ${X}$ can also provide insight into its non-singularity. It can be shown that the eigenvalues of ${X}$ are ${||x||^2}$ (with multiplicity ${n-1}$) and ${3||x||^2}$ (with multiplicity 1). Since ${x \neq 0}$, all eigenvalues are non-zero, implying that ${X}$ is invertible.

Generalizations

This result can be generalized to other types of matrices with similar structures. For example, matrices of the form ${A = B + uv^T}$, where ${B}$ is a non-singular matrix and ${u}$ and ${v}$ are vectors, can be analyzed using similar techniques to determine their invertibility. The Sherman-Morrison formula is a useful tool in such cases.

Conclusion

So, there you have it! We've shown that the matrix ${X}$ is non-singular when ${x \neq 0}$ using a method that focuses on linear transformations and the null space of the matrix. This approach not only proves the result but also gives us a deeper understanding of why it holds. Remember, exploring different proofs and perspectives can really solidify your understanding of mathematical concepts. Keep exploring, guys!

This exploration provides a solid understanding and some techniques that helps tackle similar problems, making it a valuable concept in linear algebra. Understanding the relationships between vectors, matrices, and their properties is fundamental to solving more complex problems in various fields.