Question

If $\mathbf{x}=\left(\begin{array}{c}2\\ 3i\\ 1-i\end{array}\right)$ and $\mathbf{y}=\left(\begin{array}{c}-1+i\\ 2\\ 3-i\end{array}\right),$ find (a) ${\mathbf{x}}^T\mathbf{y}$ (b) ${\mathbf{y}}^T\mathbf{y}$ (c) $(\mathbf{x},\mathbf{y})$ (d) $(\mathbf{y},\mathbf{y})$

Short answer

In summary, we have found the following: (a) ${\mathbf{x}}^T\mathbf{y}=-1+5i$ (b) ${\mathbf{y}}^T\mathbf{y}=13-8i$ (c) $(\mathbf{x},\mathbf{y})=-5-6i$ (d) $(\mathbf{y},\mathbf{y})=14+8i$

Step-by-step solution

(a) Find ${\mathbf{x}}^T\mathbf{y}$

To find ${\mathbf{x}}^T\mathbf{y}$, perform the standard dot product of $\mathbf{x}$ and $\mathbf{y}$: ${\mathbf{x}}^T\mathbf{y}=(2)(-1+i)+(3i)(2)+(1-i)(3-i)$ Calculate the result: ${\mathbf{x}}^T\mathbf{y}=-2+2i+6i+2-3i+3+1-i$ Therefore, ${\mathbf{x}}^T\mathbf{y}=-1+5i.$

(b) Find ${\mathbf{y}}^T\mathbf{y}$

To find ${\mathbf{y}}^T\mathbf{y}$, perform the standard dot product of $\mathbf{y}$ and itself: ${\mathbf{y}}^T\mathbf{y}=(-1+i)(-1+i)+(2)(2)+(3-i)(3-i)$ Calculate the result: ${\mathbf{y}}^T\mathbf{y}=1-2i+i^2+4+9-6i+i^2$ Recall that $i^2=-1$: ${\mathbf{y}}^T\mathbf{y}=1-2i-1+4+9-6i-1$ Therefore, ${\mathbf{y}}^T\mathbf{y}=13-8i.$

(c) Find $(\mathbf{x},\mathbf{y})$

To find $(\mathbf{x},\mathbf{y})$, perform the inner product of $\mathbf{x}$ and $\mathbf{y}$, which means conjugating the first vector in the product: $(\mathbf{x},\mathbf{y})=(2{)}^{\ast }(-1+i)+(3i{)}^{\ast }(2)+(1-i{)}^{\ast }(3-i)$ Calculate the conjugates and the result: $(\mathbf{x},\mathbf{y})=2(-1+i)-3i(2)+(-1+i)(3-i)$ Therefore, $(\mathbf{x},\mathbf{y})=-2+2i-6i-2-3i+3-1+i$ So, $(\mathbf{x},\mathbf{y})=-5-6i.$

(d) Find $(\mathbf{y},\mathbf{y})$

To find $(\mathbf{y},\mathbf{y})$, perform the inner product of $\mathbf{y}$ and itself, which means conjugating the first vector in the product: $(\mathbf{y},\mathbf{y})=(-1+i{)}^{\ast }(-1+i)+(2{)}^{\ast }(2)+(3-i{)}^{\ast }(3-i)$ Calculate the conjugates and the result: $(\mathbf{y},\mathbf{y})=(-1-i)(-1+i)+2(2)+(3+i)(3-i)$ Therefore, $(\mathbf{y},\mathbf{y})=1+2i+i^2+4+9+6i-i^2$ Recall that $i^2=-1$: $(\mathbf{y},\mathbf{y})=1+2i-1+4+9+6i+1$ So, $(\mathbf{y},\mathbf{y})=14+8i.$

Key concepts

Dot Product

When working with vectors in both real and complex spaces, one of the most fundamental operations is the dot product. The dot product is essentially a way to multiply two vectors, resulting in a scalar quantity. In mathematical terms, to compute the dot product, we multiply corresponding components of the vectors and then sum them up.

In the context of complex spaces, the dot product formula changes slightly due to the nature of complex numbers. With complex vectors, instead of just multiplying elements directly, it is essential to work carefully to ensure calculations are correct. To find the dot product, simply multiply each component of vector $\mathbf{x}$ with its corresponding component in vector $\mathbf{y}$, and add these products together. This operation is typically denoted as ${\mathbf{x}}^T\mathbf{y}$.

In an example like that of the vectors $\mathbf{x}$ and $\mathbf{y}$ given in the exercise, the operation is straightforward but requires attention to detail especially with imaginary components. Making careful computations ensures that when we multiply and sum values from vectors, the answer reflects both the real and imaginary nature of the components.

Inner Product

The inner product is a closely related concept to the dot product, but it includes a crucial difference: complex conjugation. In complex spaces, the inner product alters one of the vectors by applying this conjugation process, typically before performing multiplication and summing.

Unlike the usual dot product, when calculating the inner product, you must first take the complex conjugate of each element in the first vector. For example, if the element is $a+bi$, its complex conjugate is $a-bi$. This operation is denoted as ${\mathbf{x}}^{\ast }$ when performed on vector $\mathbf{x}$ before computing the inner product $(\mathbf{x},\mathbf{y})$.

This concept underlies the calculation to find $(\mathbf{x},\mathbf{y})$ where one must execute these procedural steps methodically: compute the conjugates of components, multiply their results with the components of the second vector, then complete by summing all of these products. It is a little more involved, but crucial for precisely manipulating vectors in complex spaces.

Complex Conjugate

One of the basic yet extremely useful operations in complex numbers is finding the complex conjugate. This operation reflects a complex number across the real axis in the complex plane, changing the sign of the imaginary part while keeping the real part unchanged. It acts as a foundational tool for operations like the inner product.

To find the complex conjugate of a number represented as $a+bi$, simply switch the sign of the imaginary part, resulting in $a-bi$. This procedure doesn't change the magnitude of the number, only its direction in the complex plane.

Using complex conjugates becomes crucial when dealing with various mathematical operations in complex spaces, such as division and computing the inner product. The conjugate serves to manage imaginary numbers, often aiding in simplifying equations and even in computing the modulus of complex numbers.

Understanding how and when to use complex conjugates can greatly enhance your capability to navigate through complex problems involving vector operations, like those seen in exercises involving inner products.