How to prove: a quadratic form with a matrix $ B = CC^T $ is positive defined?

Question

Let a matrix $ C \in \Bbb K^{n \mathtt x n} : det(c) \ne 0 $ (K is any field - C or R) $ \Rightarrow $ a quadratic form with a matrix $ B = CC^T $ is positive defined one. How to prove it?

Answer 1

Here $x^TBx=x^TCC^Tx=(C^Tx)^T(C^Tx)\ge 0$, and we have equality iff $x=0$.

Answer 2

Let $$ C=\begin{pmatrix} c_{11} & c_{12} & \cdots &c_{1n}\\ c_{21} & c_{22} & \cdots & c_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ c_{n1} & c_{n2} & \cdots & c_{nn} \end{pmatrix}. $$ Then, $$ x'C= \begin{pmatrix}\sum_{i=1}^nx_ic_{i1}& \sum_{i=1}^nx_ic_{i2}&\cdots &\sum_{i=1}^nx_ic_{in}\end{pmatrix} $$ so it follows that $$ x'CC'x= \begin{pmatrix}\sum_{i=1}^nx_ic_{i1}& \sum_{i=1}^nx_ic_{i2}&\cdots &\sum_{i=1}^nx_ic_{in}\end{pmatrix}\begin{pmatrix}\sum_{i=1}^nx_ic_{i1}\\ \sum_{i=1}^nx_ic_{i2}\\\sum_{i=1}^nx_ic_{in}\end{pmatrix}\\=\left(\sum_{i=1}^nx_ic_{i1}\right)^2+ \left(\sum_{i=1}^nx_ic_{i2}\right)^2+\cdots +\left(\sum_{i=1}^nx_ic_{in}\right)^2\\=\sum_{j=1}^n\left(\sum_{i=1}^n x_ic_{ij}\right)^2>0. $$ The inequality is quite obvious from the fact that we have a sum of squares in conjunction with the non-zero determinant, as this rules out the positive semi-definite case (allowing a zero determinant would mean it'd be possible for $C$ to be a zero matrix). In other words, since $x$ is by definition non-zero the sum of squares would be equal to zero if and only if $C$ was a zero matrix, but since its determinant is non-zero we know that at least $n$ elements are non-zero. Hence, their squares are positive.