Problem 32 - Differential Equations and Linear Algebra

Question

Let \(V\) be an inner product space with vectors \(\mathbf{v}\) and \(\mathbf{w}\) with \(\|\mathbf{v}\|=3,\|\mathbf{w}\|=4,\) and \(\langle\mathbf{v}, \mathbf{w}\rangle=-2 .\) Compute the following: (a) \(\|\mathbf{v}+\mathbf{w}\|\) (b) \(\langle 3 \mathbf{v}+\mathbf{w},-2 \mathbf{v}+3 \mathbf{w}\rangle\) (c) \(\langle-\mathbf{w}, 5 \mathbf{v}+2 \mathbf{w}\rangle\)

Step-by-Step Solution

Answer

The short answers for the given expressions are: (a) \(\|\mathbf{v}+\mathbf{w}\| = \sqrt{21}\) (b) \(\langle 3 \mathbf{v}+\mathbf{w},-2 \mathbf{v}+3 \mathbf{w}\rangle = -56\) (c) \(\langle-\mathbf{w}, 5 \mathbf{v}+2 \mathbf{w}\rangle = -22\)

Step 1: Apply the norm squared formula

We will use the formula \(\| \mathbf{x} + \mathbf{y} \|^2 = \| \mathbf{x} \|^2 + \| \mathbf{y} \|^2 + 2 \langle \mathbf{x}, \mathbf{y} \rangle\) for vectors \(\mathbf{x}\) and \(\mathbf{y}\). Here, \(\mathbf{x} = \mathbf{v}\) and \(\mathbf{y} = \mathbf{w}\): \[\|\mathbf{v}+\mathbf{w}\|^2 = \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 + 2 \langle\mathbf{v},\mathbf{w}\rangle.\]

Step 2: Substitute given values

We know that \(\|\mathbf{v}\|=3\), \(\|\mathbf{w}\|=4\), and \(\langle\mathbf{v},\mathbf{w}\rangle=-2\). Substituting these values, we get: \[\|\mathbf{v}+\mathbf{w}\|^2 = 3^2 + 4^2 + 2(-2).\]

Step 3: Calculate the norm

Solving the expression, we get: \[\|\mathbf{v}+\mathbf{w}\|^2 = 9 + 16 - 4 = 21,\] and \[\|\mathbf{v}+\mathbf{w}\| = \sqrt{21}.\] So, the magnitude of \(\mathbf{v}+\mathbf{w}\) is \(\sqrt{21}\). (b) \(\langle 3 \mathbf{v}+\mathbf{w},-2 \mathbf{v}+3 \mathbf{w}\rangle\)

Step 1: Use the bilinear properties of inner products

The inner product is bilinear, meaning that it is linear with respect to both arguments. Using this property, we can expand the inner product as follows: \[\langle 3 \mathbf{v}+\mathbf{w},-2 \mathbf{v}+3 \mathbf{w}\rangle = 3 (-2) \langle \mathbf{v}, \mathbf{v} \rangle + 3 (3) \langle \mathbf{w}, \mathbf{v} \rangle - 2 \langle \mathbf{w}, \mathbf{w} \rangle + 3 \langle \mathbf{w}, \mathbf{w} \rangle.\]

Step 2: Substitute given values and norms

We know that \(\|\mathbf{v}\|=3\), \(\|\mathbf{w}\|=4\), and \(\langle\mathbf{v},\mathbf{w}\rangle=-2\). We also know that \(\langle \mathbf{x}, \mathbf{x} \rangle = \| \mathbf{x} \|^2\). Substituting these values, we get: \[ -6 \cdot 3^2 + 9 \cdot (-2) - 2 \cdot 4^2 + 3 \cdot 4^2.\]

Step 3: Calculate the inner product

Solving the expression, we get: \[ -6 \cdot 9 - 9 \cdot 2 - 2 \cdot 16 + 3 \cdot 16 = -54 - 18 - 32 + 48 = -56.\] So, the inner product \(\langle 3 \mathbf{v}+\mathbf{w},-2 \mathbf{v}+3 \mathbf{w}\rangle\) is \(-56\). (c) \(\langle-\mathbf{w}, 5 \mathbf{v}+2 \mathbf{w}\rangle\)

Step 1: Use the bilinear properties of inner products

Using the bilinear properties, we can expand the inner product as follows: \[\langle-\mathbf{w}, 5 \mathbf{v}+2 \mathbf{w}\rangle = -5 \langle \mathbf{w}, \mathbf{v} \rangle - 2 \langle \mathbf{w}, \mathbf{w} \rangle.\]

Step 2: Substitute given values and norms

We know that \(\|\mathbf{v}\|=3\), \(\|\mathbf{w}\|=4\), and \(\langle\mathbf{v},\mathbf{w}\rangle=-2\). We also know that \(\langle \mathbf{x}, \mathbf{x} \rangle = \| \mathbf{x} \|^2\). Substituting these values, we get: \[ -5 (-2) - 2 \cdot 4^2.\]

Step 3: Calculate the inner product

Solving the expression, we get: \[ 10 - 2 \cdot 16 = 10 - 32 = -22.\] So, the inner product \(\langle-\mathbf{w}, 5 \mathbf{v}+2 \mathbf{w}\rangle\) is \(-22\).

Key Concepts

Norm of a Vector

Understanding the norm of a vector is pivotal when studying linear algebra and vector spaces. Simply put, the norm refers to the length or magnitude of a vector. It's a numerical value that gives us a sense of how 'long' the vector is when we think of it in either a geometric context (like an arrow in a plane or space) or an abstract numerical space.

To calculate the norm of a vector, often denoted as \(\|\mathbf{v}\|\), we use the concept of an inner product to find its length. The inner product space allows us to define this for vectors with real or complex components. For real components, the norm is defined by \(\|\mathbf{v}\|^2 = \langle\mathbf{v}, \mathbf{v}\rangle\), resulting in the square of the distance from the origin to the point represented by the vector.

In practice, to compute the norm of a vector, we simply take the square root of the sum of the squares of its components. For a vector \(\mathbf{v}\) in a 2-dimensional space with components \(x\) and \(y\), the norm is \(\|\mathbf{v}\| = \sqrt{x^2 + y^2}\). This is akin to the Pythagorean theorem in elementary geometry. For the problem at hand, the norm was given, and it became the foundation for more complex calculations.

Bilinear Properties of Inner Products

The bilinear properties of inner products are foundational in understanding how to manipulate and work with inner products in vector spaces. Bilinearity means that the inner product operation respects the structure of linear combination—which is the addition of vectors and the multiplication of vectors by scalars.

More formally, an inner product is bilinear if it satisfies two key properties for all vectors \(\mathbf{u}\), \(\mathbf{v}\), \(\mathbf{w}\) and scalars \(\alpha\) and \(\beta\):

• Additivity in the first argument: \(\langle\alpha\mathbf{u} + \beta\mathbf{v}, \mathbf{w}\rangle = \alpha\langle\mathbf{u},\mathbf{w}\rangle + \beta\langle\mathbf{v},\mathbf{w}\rangle\)
• Homogeneity in the second argument: \(\langle\mathbf{v}, \alpha\mathbf{u} + \beta\mathbf{w}\rangle = \alpha\langle\mathbf{v},\mathbf{u}\rangle + \beta\langle\mathbf{v},\mathbf{w}\rangle\).

These properties allow us to break down and simplify computations of inner products involving linear combinations of vectors, just as we saw in the step-by-step solution above. The additivity and homogeneity cut down complex inner product scenarios into smaller, more manageable pieces.

Computing Inner Products

When computing inner products, there are specific tactics we can employ to simplify the process, which directly stem from the bilinear properties previously discussed. The inner product of two vectors, \(\langle\mathbf{v}, \mathbf{w}\rangle\), provides a measure of the vectors' interaction, and it is a central concept in defining both angles and lengths within an inner product space.

To compute an inner product, we use the fact that it is linear in both arguments and distribute over addition and scalar multiplication. This means we can take a complex combination of vectors and break it down into a sum of products of individual vector components and their respective scalar multipliers.

For instance, if you're given a problem involving the inner product of scaled vectors, like \(\langle\alpha\mathbf{v}, \beta\mathbf{w}\rangle\), the bilinear property allows you to simplify it to \(\alpha\beta\langle\mathbf{v}, \mathbf{w}\rangle\). This greatly simplifies calculations, as seen in the example where we computed \(\langle 3\mathbf{v}+\mathbf{w},-2\mathbf{v}+3\mathbf{w}\rangle\) by expanding it into a combination of products involving the given norms and inner products between \(\mathbf{v}\) and \(\mathbf{w}\).