Is this proof correct? Exercise/theorem: Let B be an invertible matrix. Show that A=B^(T)*B is positive definite

Question

Exercise 5.5.29, Linear algebra: a modern introduction

• Because A has a quadratic form, A can be written as A=QDQ^(T), where D is an orthogonal matrix with orthonormal eigenvectors as columns and D is a diagonal matrix with the eigenvalues of A on the diagonal.

• Consider A=B^(T)*B. Because B is invertible, we can write A=B^(T)BI=B^(T)BB^(-1)*B=B^(T)IB. If we let Q=B^(T) and D=I, then A=QDQ^(T)=B^(T)IB.

• Because I has the eigenvalue 1, it follows that A has eigenvalue 1. Hence A is positive definite.

One thing im not really sure about, is if i can let Q=B^(T), because B^(T) might not be orthogonal.

Answer 1

You're right, this approach may not work, since there's no guarantee that $B^T$ is an orthogonal matrix.

Simply observe that $A$ is symmetric, and that $$\langle x, Ax\rangle =x^TB^TBx=\langle Bx, Bx\rangle=\|Bx\|^2$$ which is always $\ge 0$, and since $B$ is injective, it is zero iff $x=0$.

Answer 2

Let $Bx=y$. Since $B$ is invertible $y=0$ if and only if $x=0$. Now suppose that $x \neq 0$ we have

$$ x^T A x= x^T B^T B x = (Bx)^TBx=y^Ty=y_1^2 + \cdots +y_n^2>0$$

since $x \neq 0$ implies $y \neq 0$. Thus $x^T A x >0$ when $x \neq 0$ showing $A$ is positive definite.