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The inner product of \(\mathbf{u}=\operatorname{col}\left(u_{1}, \ldots, u_{n}\right)\) and \(\mathbf{v}=\operatorname{col}\left(v_{1}, \ldots, v_{n}\right)\) is defined in Section \(1.14\) as \(u_{1} v_{1}+\cdots+u_{n} v_{n}\) and hence equals \(\mathbf{u}^{\prime} \mathbf{v}=\mathbf{v}^{\prime} \mathbf{u}\). Also \(|\mathbf{u}|\) is defined as \(\left(\mathbf{u}^{\prime} \mathbf{u}\right)^{1 / 2} ; \mathbf{u}\) and \(\mathbf{v}\) are termed orthogonal if \(\mathbf{u}^{\prime} \mathbf{v}=0\). Now let \(A\) be a real symmetric \(n \times n\) matrix, so that \(A=A^{\prime}\). Let \(\lambda_{1}, \lambda_{2}\) be eigenvalues of \(A\) which are unequal and let \(\mathbf{v}_{1}, \mathbf{v}_{2}\) be corresponding eigenvectors. Show that \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are orthogonal. [Hint: First show that \(\mathbf{x}^{\prime} A \mathbf{y}=\mathbf{y}^{\prime} A \mathbf{x}\) for all \(\mathbf{x}, \mathbf{y}\). Then from \(A \mathbf{v}_{1}=\lambda_{1} \mathbf{v}_{1}, A \mathbf{v}_{2}=\lambda_{2} \mathbf{v}_{2}\) deduce that \(\lambda_{1} \mathbf{v}_{2}^{\prime} \mathbf{v}_{1}=\lambda_{2} \mathbf{v}_{1}^{\prime} \mathbf{v}_{2}\) and hence that \(\mathbf{v}_{1}^{\prime} \mathbf{v}_{2}=0\).]

Step by step solution

01

Show that x' * A * y = y' * A * x for any vectors x and y

Since A is symmetric, we have A = A'. Using this property, we can write the following equation: x' * A * y = x' * A' * y = (x' * A' * y)' = (y' * (A * x))' = y' * A * x Thus, x' * A * y = y' * A * x for all vectors x and y.

02

Obtain equations from given eigenvalue-eigenvector pairs

We are given two eigenvalue-eigenvector pairs: (位鈧�, v鈧�) and (位鈧�, v鈧�). By definition of eigenvalue-eigenvector pairs, we have the following equations: A * v鈧� = 位鈧� * v鈧� (1) A * v鈧� = 位鈧� * v鈧� (2)

03

Use the results from Step 1 on v鈧� and v鈧�

Since the result derived in Step 1 holds for any vectors x and y, we can use it for vector pairs (v鈧�, v鈧�) and (v鈧�, v鈧�). This gives us: v鈧�' * A * v鈧� = v鈧�' * A * v鈧� (3)

04

Substitute the eigenvalue-eigenvector equations in Step 3

Now, we will substitute the equations obtained in Step 2 into equation (3). Replacing A * v鈧� with 位鈧� * v鈧� and A * v鈧� with 位鈧� * v鈧�, we have: v鈧�' * (位鈧� * v鈧�) = v鈧�' * (位鈧� * v鈧�) Which simplifies to: 位鈧� * v鈧�' * v鈧� = 位鈧� * v鈧�' * v鈧�

05

Show that v鈧� and v鈧� are orthogonal

We have shown that 位鈧� * v鈧�' * v鈧� = 位鈧� * v鈧�' * v鈧�. Since 位鈧� 鈮� 位鈧�, this equation implies that v鈧�' * v鈧� = 0 and v鈧�' * v鈧� = 0. This means that vectors v鈧� and v鈧� are orthogonal, as their inner product is zero.

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Key Concepts

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

Inner Product

Imagine you have two vectors, each representing a list of numbers. The inner product between these two vectors is a way to measure how much they 'align' with each other, almost like having a conversation where the vectors 'agree' on certain terms. By multiplying their corresponding entries and summing these products, you end up with a single number, the inner product.

To get an intuitive grasp on this concept, consider \( \mathbf{u} \) and \( \mathbf{v} \) as two strings of numbers. If you were to pair each number from \( \mathbf{u} \) with its counterpart in \( \mathbf{v} \) and multiply them together, then add up all those results, you'd have calculated the inner product, symbolically represented as \( \mathbf{u}^\prime \mathbf{v} = u_1v_1 + u_2v_2 + \cdots + u_nv_n \).

Now, let's talk about orthogonality. If the result of this inner product operation is zero, it means the vectors are orthogonal - they're at a right angle to each other in their multi-dimensional space. It's like two people in a conversation who have absolutely nothing in common; their dialogue (the inner product) adds up to zero.

Eigenvalues and Eigenvectors

Picture a complex shape that can undergo transformations like scaling or rotating without changing its essence. Eigenvectors are direction lines that remain undisturbed during such transformations, while eigenvalues tell us how much they're stretched or compressed.

Consider a matrix \( A \) as a function that transforms vectors, and we're on the lookout for special vectors that, when multiplied by \( A \), simply scale up or down, without changing their direction. These special vectors are the eigenvectors, and the scale factors are the eigenvalues. Written mathematically, for matrix \( A \) and eigenvector \( \mathbf{v} \) corresponding to eigenvalue \( \lambda \), the relation is \( A\mathbf{v} = \lambda\mathbf{v} \).

An enlightening property is that when you have two different eigenvalues, their corresponding eigenvectors are guaranteed to be orthogonal in the case of a real symmetric matrix. That鈥檚 like saying two different stretching forces applied to our shape in their respective directions will never intermingle or twist the shape around.

Symmetric Matrices

Imagine symmetric matrices as mirrors with numerical reflections. A matrix is symmetric if it's equal to its transpose, meaning if you were to flip it over its diagonal, the numbers would match perfectly; this is notated as \( A = A^\prime \).

Symmetric matrices are fascinating because they behave nicely in various mathematical contexts. They have real eigenvalues and their eigenvectors are real and orthogonal when the eigenvalues are distinct. Returning to our visualization of a transforming shape, if that transformation is described by a symmetric matrix, any resulting pushes and pulls along eigenvectors do not twist the shape around; instead, they maintain the shape's 'posture' while expanding or contracting it.