Problem 18 - Differential Equations and Linear Algebra

Question

Consider the vector space \(\mathbb{R}^{2}\). Define the mapping \(\langle,\rangle \mathrm{by}\) $$\langle\mathbf{v}, \mathbf{w}\rangle=2 v_{1} w_{1}+v_{1} w_{2}+v_{2} w_{1}+2 v_{2} w_{2}$$ for all vectors \(\mathbf{v}=\left(v_{1}, v_{2}\right)\) and \(\mathbf{w}=\left(w_{1}, w_{2}\right)\) in \(\mathbb{R}^{2}\) Verify that Equation ( 5.1 .14 ) defines an inner product on \(\mathbb{R}^{2}\).

Step-by-Step Solution

Answer

In summary, the given equation defines an inner product on \(\mathbb{R}^{2}\) as it satisfies all the axioms of an inner product: Conjugate Symmetry, Linearity in the first argument, and Positive-definiteness.

Step 1: Axiom 1: Conjugate Symmetry

To show conjugate symmetry, we need to demonstrate that \(\langle\mathbf{v},\mathbf{w}\rangle=\langle\mathbf{w},\mathbf{v}\rangle\). For the given equation: \(\langle\mathbf{v},\mathbf{w}\rangle = 2v_1w_1+v_1w_2+v_2w_1+2v_2w_2\) \(\langle\mathbf{w},\mathbf{v}\rangle = 2w_1v_1+w_1v_2+w_2v_1+2w_2v_2\) As the terms in both expressions are the same, we conclude that the conjugate symmetry is satisfied.

Step 2: Axiom 2: Linearity in the first argument

To show linearity in the first argument, we need to prove that \(\langle a\mathbf{v} + b\mathbf{u},\mathbf{w}\rangle = a\langle\mathbf{v},\mathbf{w}\rangle+b\langle\mathbf{u},\mathbf{w}\rangle\), for all scalars \(a,b\). Let \(\mathbf{v} = (v_1, v_2)\), \(\mathbf{u} = (u_1, u_2)\), and \(\mathbf{w} = (w_1, w_2)\): \(\langle a\mathbf{v} + b\mathbf{u},\mathbf{w}\rangle = \langle (av_1 + bu_1, av_2+bu_2), (w_1, w_2)\rangle\) \(= 2(av_1+bu_1)w_1 + (av_1+bu_1)w_2 + (av_2+bu_2)w_1 + 2(av_2+bu_2)w_2\) Now, distribute the terms: \(= 2av_1w_1 + 2bu_1w_1 + av_1w_2 + bu_1w_2 + av_2w_1 + bu_2w_1 + 2av_2w_2 + 2bu_2w_2\) Now regroup the terms by the scalars \(a\) and \(b\): \(= (2v_1w_1 + v_1w_2 + v_2w_1 + 2v_2w_2)a + (2u_1w_1 + u_1w_2 + u_2w_1 + 2u_2w_2)b\) As this expression is equal to \(a\langle\mathbf{v},\mathbf{w}\rangle+b\langle\mathbf{u},\mathbf{w}\rangle\), we conclude that linearity in the first argument is satisfied.

Step 3: Axiom 3: Positive-definiteness

To verify if the given equation satisfies positive-definiteness, we need to check if \(\langle\mathbf{v},\mathbf{v}\rangle\geq0\), and \(\langle\mathbf{v},\mathbf{v}\rangle=0\) if and only if \(\mathbf{v}=0\). Consider the expression \(\langle\mathbf{v},\mathbf{v}\rangle\): \(\langle\mathbf{v},\mathbf{v}\rangle = 2v_1^2+v_1v_2+v_2v_1+2v_2^2 = 2v_1^2 + 2v_1v_2 + 2v_2^2\) As this expression is a sum of squares, it is always non-negative (\(\geq 0\)). Also, this expression will only be equal to zero if all the terms are zero, which implies that both \(v_1\) and \(v_2\) are zero, so \(\mathbf{v} = 0\). Hence, the positive-definiteness axiom is satisfied. As all three axioms are verified to be satisfied by the given equation, we conclude that it defines an inner product on \(\mathbb{R}^{2}\).

Key Concepts

Vector Spaces

Vector spaces are an essential concept in linear algebra. They provide a framework where vectors can be added together and multiplied by scalars. In the case of the vector space \(\mathbb{R}^{2}\), you have vectors with two components. Each component is a real number. This means any vector in \(\mathbb{R}^{2}\) can be represented as \((v_1, v_2)\).

One of the key properties of vector spaces is that they allow linear combinations of vectors. This includes operations such as vector addition and scalar multiplication, which must satisfy specific rules. For instance, vector addition in \(\mathbb{R}^{2}\) is defined as \((v_1, v_2) + (w_1, w_2) = (v_1 + w_1, v_2 + w_2)\). Scalar multiplication is defined as multiplying each component of a vector by a scalar.

Positive Definiteness

Positive definiteness is a crucial property required for a mapping to be considered an inner product. This property ensures that the result of applying the inner product to a vector with itself is always non-negative.

In mathematical terms, for a vector \(\mathbf{v}\), it must hold that \(\langle \mathbf{v}, \mathbf{v} \rangle \geq 0\), with equality if and only if \(\mathbf{v} = \mathbf{0}\). For the exercise, the expression \(\langle \mathbf{v}, \mathbf{v} \rangle = 2v_1^2 + 2v_1v_2 + 2v_2^2\) demonstrates positive definiteness because it's a sum of squares, making it non-negative. It becomes zero only when both \(v_1\) and \(v_2\) are zero.

Conjugate Symmetry

Conjugate symmetry ensures that the inner product of two vectors \(\mathbf{v}\) and \(\mathbf{w}\) is equal to the inner product of \(\mathbf{w}\) and \(\mathbf{v}\). This symmetry is crucial for the inner product's consistency.

In this exercise, it can be shown that \(\langle \mathbf{v}, \mathbf{w} \rangle = 2v_1w_1 + v_1w_2 + v_2w_1 + 2v_2w_2 \) is equal to \(\langle \mathbf{w}, \mathbf{v} \rangle = 2w_1v_1 + w_1v_2 + w_2v_1 + 2w_2v_2 \). Since these expressions are identical, conjugate symmetry is satisfied. This property is often more straightforward in real vector spaces, as complex conjugation isn't required.

Linearity

Linearity is a critical aspect of inner products, especially regarding linearity in the first argument. This means that the inner product distributes over vector addition and is compatible with scalar multiplication.

For any vectors \(\mathbf{v}\), \(\mathbf{u}\), and \(\mathbf{w}\), and scalars \(a\) and \(b\), the linearity condition states \(\langle a\mathbf{v} + b\mathbf{u}, \mathbf{w} \rangle = a\langle \mathbf{v}, \mathbf{w} \rangle + b\langle \mathbf{u}, \mathbf{w} \rangle\). Using the given mapping, this property is validated through distribution and regrouping, showing it holds true. Thus, it confirms linearity in the first argument, which is essential for defining an inner product.