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Chapter 6: Problem 10

Let \(A\) be a singular \(n \times n\) matrix. Show that \(A^{T} A\) is positive semidefinite, but not positive definite.

Short Answer

Expert verified
In conclusion, given a singular \(n \times n\) matrix \(A\), \(A^{T} A\) is positive semidefinite because the quadratic form \(x^{T}A^{T}Ax = y^T y \geq 0\). However, \(A^{T} A\) is not positive definite because there exists a non-zero vector \(x\) with \(Ax = 0\) such that \(x^{T}A^{T}Ax = 0\).

Step by step solution

01

Show that \(A^{T} A\) is positive semidefinite

Given a singular \(n \times n\) matrix \(A\), we need to show that \(A^{T} A\) is positive semidefinite. To do this, we will examine the quadratic form \(x^{T}A^{T}Ax\) for any non-zero vector \(x\): \[x^{T}A^{T}Ax\] Notice that if we define a new vector \(y = Ax\), then we can rewrite the quadratic form as: \[x^{T}A^{T}Ax = x^{T}y^Ty = y^{T}y\] Now we know that \(y^Ty\) is the sum of squares of the elements of \(y\): \[ y^T y = y_1^2 + y_2^2 + \dots + y_n^2 \] Since the sum of squares is always non-negative, it follows that: \[ x^{T}A^{T}Ax = y^T y \geq 0\] Thus, \(A^{T} A\) is positive semidefinite.
02

Show that \(A^{T} A\) is not positive definite

As we know that \(A\) is a singular matrix, this implies that its determinant is 0: \[ \det(A) = 0 \] This means that \(A\) has at least one non-trivial (non-zero) solution in its null space. So, there exists a non-zero vector \(x\) such that: \[ Ax = 0 \] Now, let's examine the quadratic form for this specific vector \(x\): \[ x^{T}A^{T}Ax \] As \(Ax = 0\), we can replace \(Ax\) by \(0\): \[ x^{T}A^{T}Ax = x^{T}0 = 0 \] So, we have found a non-zero vector \(x\) for which the quadratic form \(x^{T}A^{T}Ax = 0\). Hence, \(A^{T} A\) is not positive definite. In conclusion, \(A^{T} A\) is positive semidefinite, but not positive definite for a singular \(n \times n\) matrix \(A\).

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