Question

Prove that to every \(A \in L\left(R^{n}, R^{1}\right)\) corresponds a unique \(y \in R^{n}\) such that \(A \mathrm{x}=\mathbf{x} \cdot \mathrm{y}\). Prove also that \(\|A\|=|\mathbf{y}|\). Hint: Under certain conditions, equality holds in the Schwarz inequality.

Short answer

Question: Prove that there is a unique \(y\) in \(R^n\) such that \(A \boldsymbol{x} = \boldsymbol{x} \cdot \boldsymbol{y}\) for all \(A \in L(R^n, R^1)\) and that the norm of the linear operator \(A\) is equal to the absolute value of the vector \(y\). Answer: To prove the uniqueness of \(y\) and that the norm of the linear operator \(A\) is equal to the absolute value of the vector \(y\), we followed these steps: 1. We showed that there exists a unique \(y\) satisfying \(A \boldsymbol{x} = \boldsymbol{x} \cdot \boldsymbol{y}\) by expressing a linear transformation as a matrix and finding the unique vector \(y\). 2. We proved that \(\|A\| = |\boldsymbol{y}|\) using the Schwarz Inequality, noting that equality holds when \(\boldsymbol{x}\) and \(\boldsymbol{y}\) are linearly dependent. Therefore, we concluded that the unique \(y\) exists and the norm of the linear operator \(A\) is equal to the absolute value of the vector \(y\).

Step-by-step solution

Show that there exists a unique \(y\) satisfying \(A \boldsymbol{x} = \boldsymbol{x} \cdot \boldsymbol{y}\)

Let's denote the standard basis in \(R^n\) as \(\boldsymbol{e_1}, \boldsymbol{e_2}, \dots, \boldsymbol{e_n}\). We have: $$A \boldsymbol{e_i} = A(\delta_{1i},\delta_{2i},\dots,\delta_{ni}) = (a_{1i},a_{2i},\dots,a_{ni})$$ Now, let's define a vector \(\boldsymbol{y} \in R^n\) such that \(\boldsymbol{y} = (y_1, y_2, \dots, y_n)\) and try to find the unique vector \(\boldsymbol{y}\) that satisfies \(A \boldsymbol{x} = \boldsymbol{x} \cdot \boldsymbol{y}\). We can do this by using the idea that a linear transformation can be expressed as a matrix: $$\boldsymbol{x} \cdot \boldsymbol{y} = x_1 y_1 + x_2 y_2 + \dots + x_n y_n = \left[\boldsymbol{x} \cdot A(\boldsymbol{e_1}), \boldsymbol{x} \cdot A(\boldsymbol{e_2}), \dots, \boldsymbol{x} \cdot A(\boldsymbol{e_n})\right]$$ As the equation \(A \boldsymbol{x} = \boldsymbol{x} \cdot \boldsymbol{y}\) should hold for all \(\boldsymbol{x}\), we can take the dot product of \(\boldsymbol{e_i}\) with both sides: $$\boldsymbol{e_i} \cdot A(\boldsymbol{x}) = A(\boldsymbol{x})\cdot\boldsymbol{e_i}=\boldsymbol{x} \cdot A(\boldsymbol{e_i}) = \boldsymbol{x} \cdot (a_{1i},a_{2i},\dots,a_{ni})$$ Now, compare this with the definition of scalar product: $$\boldsymbol{x} \cdot \boldsymbol{y} = x_1 y_1 + x_2 y_2 + \dots + x_n y_n$$ We can see that if we let \(\boldsymbol{y} = \sum_{i=1}^{n} A(\boldsymbol{e_i}) = (a_{11},a_{21},\dots,a_{n1}) + (a_{12},a_{22},\dots,a_{n2}) + \dots + (a_{1n},a_{2n},\dots,a_{nn})\), we can find the unique vector \(\boldsymbol{y}\) satisfying the given equation.

Prove that \(\|A\| = |\boldsymbol{y}|\) using the Schwarz Inequality

Now, let's prove that \(\|A\| = |\boldsymbol{y}|\). Recall the Schwarz Inequality: $$|\boldsymbol{x} \cdot \boldsymbol{y}| \leq |\boldsymbol{x}| |\boldsymbol{y}|$$ From Step 1, we know \(A \boldsymbol{x} = \boldsymbol{x} \cdot \boldsymbol{y}\). Using the Schwarz Inequality, we get: $$|A \boldsymbol{x}| = |\boldsymbol{x} \cdot \boldsymbol{y}| \leq |\boldsymbol{x}| |\boldsymbol{y}|$$ Dividing both sides by \(|\boldsymbol{x}|\), we get: $$\frac{|A \boldsymbol{x}|}{|\boldsymbol{x}|} \leq |\boldsymbol{y}|$$ Now, taking the supremum over all non-zero \(\boldsymbol{x}\), we have: $$\|A\| = \sup_{\boldsymbol{x} \neq 0} \frac{|A \boldsymbol{x}|}{|\boldsymbol{x}|} \leq |\boldsymbol{y}|$$ However, as noted in the hint, equality holds under certain conditions in the Schwarz Inequality. Specifically, equality holds when \(\boldsymbol{x}\) and \(\boldsymbol{y}\) are linearly dependent, i.e., \(\boldsymbol{x} = \lambda \boldsymbol{y}\) for some scalar \(\lambda\). So, if we take \(\boldsymbol{x} = \boldsymbol{y}\), we have: $$|A \boldsymbol{y}| = |\boldsymbol{y} \cdot \boldsymbol{y}| = |\boldsymbol{y}|^2$$ Dividing both sides by \(|\boldsymbol{y}|\), we have: $$\frac{|A \boldsymbol{y}|}{|\boldsymbol{y}|} \geq \|A\|$$ So, we conclude that \(\|A\| = |\boldsymbol{y}|\).

Key concepts

Schwarz Inequality

Imagine you are balancing two weights on a seesaw in perfect harmony. The Schwarz Inequality is like the principle ensuring both sides remain balanced. In mathematics, it relates to vectors and confirms that the absolute value of the dot product of two vectors is always less than or equal to the product of their magnitudes. Written out, it looks like this:

• \(|\boldsymbol{x} \cdot \boldsymbol{y}| \leq |\boldsymbol{x}| |\boldsymbol{y}|\)

This inequality is a cornerstone in vector analysis and measures how closely two vectors align. It finds applications in various fields including physics and engineering.
An intuitive way to understand this is by considering when the inequality becomes an equality, meaning a perfect alignment. This occurs when the vectors are linearly dependent, essentially meaning one vector is a scaled version of the other. As such, the Schwarz Inequality not only compares vector magnitudes but also provides insights into their directional similarities.

Dot Product

The dot product, sometimes called a scalar product, is a key operation involving two vectors. It multiplies corresponding components of the vectors and sums up these products. In an equation, it is represented as:

• \(\boldsymbol{x} \cdot \boldsymbol{y} = x_1 y_1 + x_2 y_2 + \dots + x_n y_n\)

The result of a dot product is a scalar, which is a single real number. This makes the dot product special as it converts vector information into a single value.
The mathematical and geometric significance is notable. It can determine the angle between two vectors, aiding in concepts like orthogonal (perpendicular) vectors where the dot product equals zero. If you've ever wondered why a right angle is significant in geometry, the dot product helps explain why it carries mathematical weight.
Additionally, practical applications abound. From graphics programming to physics, understanding the dot product allows for calculations involving projections and component forces along a direction.

Scalar Product

The term "scalar product" is synonymous with the dot product, with the word scalar highlighting the type of result derived from this operation. It is called "scalar" because the result, unlike other vector operations, is not another vector but a magnitude – a single number.
This operation plays a vital role in many vector space applications. For example, whenever you encounter expressions needing simplification or calculations of spatial relationships, the scalar product comes into play.

• It is defined as \(\boldsymbol{x} \cdot \boldsymbol{y} = x_1 y_1 + x_2 y_2 + \dots + x_n y_n\), summarizing vector influence along common paths.
• Used in defining the norm or length of a vector, influencing how we interpret vector quantity: \(|\boldsymbol{x}| = \sqrt{x_1^2 + x_2^2 + \ldots + x_n^2}\).

Being able to convert vector directional information into scalar form is essential in disciplines like computer science, where it supports algorithms in graphics and physics simulations, enhancing realism and providing accuracy in directional calculations.