Math Problem Statement

I don't understand how you expanded the quadratic expression?

Solution

Certainly! Let's carefully go through the expansion of the quadratic expression.

Given Expression:

We start with the expression given in the problem: (tx+y)(tx+y)0.(t\vec{x} + \vec{y})(t\vec{x} + \vec{y})^\top \geq 0. Here, tt is a scalar, and x\vec{x} and y\vec{y} are row vectors in Rn\mathbb{R}^n. The notation (tx+y)(t\vec{x} + \vec{y})^\top represents the transpose of the row vector (tx+y)(t\vec{x} + \vec{y}), making it a column vector.

Expansion of the Expression:

Let's expand (tx+y)(tx+y)(t\vec{x} + \vec{y})(t\vec{x} + \vec{y})^\top step by step.

  1. First Term: tx(tx)=txtx=t2(xx).t\vec{x}(t\vec{x})^\top = t\vec{x} \cdot t\vec{x}^\top = t^2 (\vec{x}\vec{x}^\top). Here, xx\vec{x}\vec{x}^\top is a scalar (since it's a row vector multiplied by its transpose).

  2. Second Term: txy.t\vec{x} \cdot \vec{y}^\top. This is a scalar since it is the dot product of two vectors.

  3. Third Term: y(tx)=t(yx).\vec{y} \cdot (t\vec{x})^\top = t(\vec{y} \cdot \vec{x}^\top). Again, this is a scalar.

  4. Fourth Term: yy.\vec{y}\vec{y}^\top. This term is also a scalar.

Full Expansion:

Now, let's put it all together:

(tx+y)(tx+y)=t2(xx)+t(xy)+t(yx)+yy.(t\vec{x} + \vec{y})(t\vec{x} + \vec{y})^\top = t^2 (\vec{x}\vec{x}^\top) + t (\vec{x}\vec{y}^\top) + t (\vec{y}\vec{x}^\top) + \vec{y}\vec{y}^\top.

But note that xy\vec{x}\vec{y}^\top and yx\vec{y}\vec{x}^\top are actually the same scalar, so we can combine these terms:

=t2(xx)+2t(xy)+yy.= t^2 (\vec{x}\vec{x}^\top) + 2t (\vec{x}\vec{y}^\top) + \vec{y}\vec{y}^\top.

This is a quadratic expression in tt, with:

  • a=xxa = \vec{x}\vec{x}^\top (coefficient of t2t^2),
  • b=2xyb = 2\vec{x}\vec{y}^\top (coefficient of tt),
  • c=yyc = \vec{y}\vec{y}^\top (constant term).

Summary:

Thus, the expression (tx+y)(tx+y)(t\vec{x} + \vec{y})(t\vec{x} + \vec{y})^\top expands into the quadratic form: t2(xx)+2t(xy)+yy0.t^2 (\vec{x}\vec{x}^\top) + 2t (\vec{x}\vec{y}^\top) + \vec{y}\vec{y}^\top \geq 0. This is the basis for using the discriminant to establish the inequality.

Would you like to explore further or clarify any other steps?

Related Questions:

  1. Why does the quadratic expression need to be non-negative?
  2. How do the vectors x\vec{x} and y\vec{y} interact geometrically in this inequality?
  3. What role does the discriminant play in determining the nature of quadratic equations?
  4. Can this proof technique be applied to other forms of inequalities?
  5. What is the significance of the Cauchy-Schwarz inequality in inner product spaces?

Tip:

When expanding expressions like this, carefully track each term and how they combine. Visualizing vector multiplication can help you understand the nature of each term.

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Math Problem Analysis

Mathematical Concepts

Quadratic Expressions
Vector Algebra
Matrix Algebra

Formulas

Quadratic expression expansion

Theorems

Cauchy-Schwarz inequality

Suitable Grade Level

Advanced High School