Proving the Positive Semidefiniteness of AA^T | Mathematical Analysis of Eigenvalues and Determinants.

. For all n × n matrices A, det(AA^T ) ≥ 0. (T/F)

The statement is true

The statement is true.

To prove this, let’s begin by defining some terms.

Given an n × n matrix A, the transpose of A, denoted by A^T, is the matrix obtained by interchanging the rows and columns of A. More formally, if A = [a_{ij}] is an n × n matrix, then A^T = [b_{ij}], where b_{ij} = a_{ji} for all 1 ≤ i, j ≤ n.

Now, let’s consider the product of the matrix A with its transpose, denoted as AA^T. The (i,j)-entry of AA^T is given by the dot product of the ith row of A with the jth column of A^T. Mathematically, it can be expressed as:

(AA^T)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + … + a_{in}b_{nj}

Since the transpose of A is obtained by interchanging rows and columns, we have b_{kj} = a_{jk}, so we can rewrite the above expression as:

(AA^T)_{ij} = a_{i1}a_{1j} + a_{i2}a_{2j} + … + a_{in}a_{nj}

Now, consider the determinant of the matrix AA^T, denoted as det(AA^T). The determinant is a scalar value calculated using the entries of the matrix and has several properties. One important property is that the determinant of a matrix is non-negative if and only if the matrix is positive semidefinite.

To prove that det(AA^T) ≥ 0, we need to show that all the eigenvalues of AA^T are non-negative. If a matrix has all eigenvalues greater than or equal to zero, it is positive semidefinite.

Let λ be an eigenvalue of AA^T and v be the corresponding eigenvector. We can express this relationship as:

AA^T v = λv

Multiplying both sides of the equation by v^T (transpose of v), we get:

v^T AA^T v = v^T λv

Since v^T AA^T v = (AA^T v)^T v, we can rewrite the equation as:

(A^T v)^T (A^T v) = (A^T v)^T λv

Expanding, we have:

v^T A (A^T v) = λ (v^T v)

Taking the determinant on both sides:

det(v^T A (A^T v)) = λ^n det(v^T v)

Since the determinant of a product is the product of determinants, we have:

det(v^T A) det(A^T v) = λ^n (det(v)^2)

Note that det(v^T A) = det(A^T v) because the two are transposes of each other. Therefore, we can simplify the equation to:

(det(A^T v))^2 = λ^n (det(v)^2)

Since the determinant of a vector is a scalar value, det(v)^2 is non-negative. Hence, the left-hand side of the equation is non-negative as well.

In order for the equation to hold, λ^n must also be non-negative. And since n is a positive integer, λ^n is non-negative for any real number λ.

Therefore, all eigenvalues of AA^T are non-negative, implying that det(AA^T) ≥ 0. Hence, the statement is true.

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