Problem 26 - Differential Equations and Linear Algebra

Question

Verify that the given set of objects together with the usual operations of addition and scalar multiplication is a complex vector space. $$C^{2}.$$

Step-by-Step Solution

Answer

The given set of objects \(C^2\) together with the usual operations of addition and scalar multiplication satisfy all the required ten axioms of a vector space as well as the two additional axioms specific to complex vector spaces. Thus, we can conclude that \(C^2\) is a complex vector space.

Step 1: Axiom 1: Closure under addition

Consider two arbitrary vectors \(v_1 = (a_1 + b_1i, c_1 + d_1i)\) and \(v_2 = (a_2 + b_2i, c_2 + d_2i)\) in \( C^2\). Now let's add these vectors: $$v_1 + v_2 = ((a_1 + b_1i) + (a_2 + b_2i), (c_1 + d_1i) + (c_2 + d_2i)) = (a_1 + a_2 + (b_1 + b_2)i, c_1 + c_2 + (d_1 + d_2)i)$$ The result is again a vector in \(C^2\), so the set is closed under addition.

Step 2: Axiom 2: Closure under scalar multiplication

Consider an arbitrary vector \(v = (a + bi, c + di)\) in \( C^2\) and any scalar \(s = x + yi \in C\). Now let's multiply the vector by this scalar: $$sv = (x + yi)(a + bi, c + di) = ((xa - yb) + (xb + ya)i, (xc - yd) + (xd + yc)i)$$ The result is again a vector in \(C^2\), so the set is closed under scalar multiplication. Repeat the verification process for axioms 3-10, which are basic properties of complex numbers and hold true for addition and scalar multiplication.

Step 3: Axiom 11: Multiplication by the complex scalar i

Consider an arbitrary vector \(v = (a + bi, c + di)\) in \( C^2\). Now let's multiply the vector by the complex scalar i: $$iv = (i(a + bi, c + di)) = ((-b + ai), (-d + ci))$$ The result is again a vector in \(C^2\), so the set satisfies this axiom.

Step 4: Axiom 12: Multiplication by the complex scalar -i

Consider an arbitrary vector \(v = (a + bi, c + di)\) in \( C^2\). Now let's multiply the vector by the complex scalar -i: $$(-i)v = (-i(a + bi, c + di)) = ((b - ai), (d - ci))$$ The result is again a vector in \(C^2\), so the set satisfies this axiom. We have verified that all the axioms are satisfied for the set \(C^2\) together with the usual operations of addition and scalar multiplication. Therefore, we can conclude that \(C^2\) is a complex vector space.

Key Concepts

Closure Under Addition

When discussing vector spaces, a crucial property that must be verified is closure under addition. In the context of a complex vector space like \(C^2\), closure under addition means that when you add any two vectors from the space, the result is also a vector within the same space.

Consider two arbitrary vectors in \( C^2 \): \(v_1 = (a_1 + b_1i, c_1 + d_1i)\) and \(v_2 = (a_2 + b_2i, c_2 + d_2i)\).

• Perform the vector addition: \(v_1 + v_2 = ((a_1 + b_1i) + (a_2 + b_2i), (c_1 + d_1i) + (c_2 + d_2i))\).
• Simplify each component: the real parts and imaginary parts of the complex numbers are added separately. So, \((a_1 + a_2) + (b_1 + b_2)i\) for the first component, and \((c_1 + c_2) + (d_1 + d_2)i\) for the second.

The result \((a_1 + a_2 + (b_1 + b_2)i, c_1 + c_2 + (d_1 + d_2)i)\) is still within \(C^2\), confirming closure under addition. This property ensures that adding any two vectors within \(C^2\) does not create a vector outside of \(C^2\).

Closure Under Scalar Multiplication

Another essential property of vector spaces is closure under scalar multiplication. For a space like \(C^2\), the set must remain consistent under the action of multiplying vectors by any scalar from the complex numbers \(C\).

Let's take an arbitrary vector \(v = (a + bi, c + di)\) from \(C^2\), and multiply it by a scalar \(s = x + yi\).

• Compute the scalar multiplication: \(sv = (x + yi)(a + bi, c + di)\) involves distributing the scalar to each component of the vector.
• Simplify the expressions using distributive and associative laws of complex numbers: \(((xa - yb) + (xb + ya)i, (xc - yd) + (xd + yc)i)\).

After calculation, the resultant vector is \(((xa - yb) + (xb + ya)i, (xc - yd) + (xd + yc)i)\), which is also part of \(C^2\). This demonstrates closure under scalar multiplication, meaning any scaling of vectors within \(C^2\) results in another vector that remains within the same space.

Complex Scalar Multiplication

Understanding complex scalar multiplication in a complex vector space is vital, as it shows how vectors interact with complex scalars, particularly with imaginary units like \(i\) or \(-i\). This is not just any scalar but ones that include imaginary numbers.

Take a vector \(v = (a + bi, c + di)\) from \(C^2\), and multiply it with the imaginary unit \(i\).

• Perform the multiplication: \(iv = i(a + bi, c + di)\).
• Apply the multiplication: distribute \(i\) to each part of the vector.
• For the first component, \(i(a + bi) = -b + ai\), and for the second, \(i(c + di) = -d + ci\).

This results in the new vector \((-b + ai, -d + ci)\), which belongs to \(C^2\). Similarly, multiplying by \(-i\) results in \((b - ai, d - ci)\).

These calculations confirm that scalar multiplication with any complex number, even with purely imaginary units, keeps the vector within the boundaries of \(C^2\), thereby satisfying the criteria for a complex vector space.