# X^t(AA^t)X ≥ 0 for all x∈R^n?

• GenAlg
In summary, to show that the quadratic form h(x) = X^t(AA^t)X is greater than or equal to 0 for all x in R^n, it can be rewritten as X^t(AA^t)X = (Y^tY)^2, where Y = A^tX. Alternatively, one can prove that all eigenvalues of AA^t are positive by considering an eigenvector y and using the fact that (AA^t y) \cdot y = (AA^t)^t y^t y = (AA^t y)^t y = \lambda y^t y.

## Homework Statement

Show that the quadratic form $$h(x) = X^t(AA^t)X$$ is greater than or equal to 0 for all $$x$$ in $$R^n$$.

## The Attempt at a Solution

Since $$(AA^t)^t = (A^t)^t A^t = AA^t$$, $$AA^t$$ can (according to the spectral theorem) be diagonalized in an orthonormal eigenvector basis. Assuming $$X = TX'$$ to be the relation between the bases, it follows that (since T is orthogonal):

$$h(x) = X^t(AA^t)X = (TX')^t(AA^t)(TX') = (X'^t T^t) (AA^t)(TX') = X'^t (T^{-1} AA^t T)X' = X'^t D X' = \mu_1 x_1'^2 + \mu_2 x_2'^2 + ... \mu_n x_n'^2$$

The problem is thus equivalent to showing that $$\mu_1 \ge 0, \mu_2 \ge 0, ... , \mu_n \ge 0$$. I haven't gotten anywhere with that approach.

I attempted to work directly on $$h(x) = X^t(AA^t)X$$ as well by trying to complete the square for 2x2, 3x3, ... matrices $$AA^t$$ (i.e. by using induction), and the proof would, if some relevant transformation is valid and all eigenvalues are ≥0, follow from Sylvester's law of inertia. I didn't get anywhere with that approach either.

Ideas?

I don't know most of the methods you used in your solution attempt, however...

Rewrite $$h(x)$$ in terms of $$Y = A^tX$$

Hint: Try rewriting $$X^t (AA^t) X$$ as an inner product of something with itself.

Alternatively, if you want to explicitly prove that all eigenvalues are positive, consider an eigenvector $$y$$, with $$AA^t y = \lambda y$$. What's $$(AA^t y) \cdot y$$? Now apply a basic result about the transpose to get what you want.

## 1. What does the equation X^t(AA^t)X ≥ 0 for all x∈R^n mean?

This equation is known as a positive semi-definite quadratic form and it represents a type of mathematical inequality. It states that the inner product of the vector X and the matrix product (AA^t)X must be greater than or equal to zero for all possible values of X in the real n-dimensional space.

## 2. What is the significance of X^t(AA^t)X ≥ 0 for all x∈R^n in mathematics?

This equation is commonly used in areas of mathematics such as linear algebra, optimization, and statistics. It can be used to solve various problems, including finding optimal solutions and determining the minimum or maximum values of a function.

## 3. How is X^t(AA^t)X ≥ 0 for all x∈R^n related to positive definiteness?

If the matrix (AA^t) is positive definite, meaning all of its eigenvalues are positive, then X^t(AA^t)X ≥ 0 for all x∈R^n is a strict inequality. This means that the inner product of X and (AA^t)X will always be greater than zero, unless X is the zero vector.

## 4. What is the intuition behind X^t(AA^t)X ≥ 0 for all x∈R^n?

The equation represents a constraint on the values of X in the real n-dimensional space. It ensures that the inner product of X and (AA^t)X will always be a non-negative value. This can be thought of as a way to ensure that the solution to a problem is always feasible and does not violate any constraints.

## 5. Can X^t(AA^t)X ≥ 0 for all x∈R^n be extended to higher dimensions?

Yes, the equation can be extended to higher dimensions, such as the complex n-dimensional space. However, the concept of positive semi-definite quadratic forms may differ in higher dimensions and may require different mathematical techniques for solving problems.