# Proof of existance of non-singular matrix C for a positive definite matrix. [closed]

Prove that $$A$$ is a positive definite matrix i.e., $$x^TAx>0$$ for any $$x$$ if and only if there exists a non-singluar matrix $$C$$ such that $$C^TAC = I$$.

I am not able to do prove both the direction.

• Can you do one direction? – Berci Feb 23 at 8:12
• No, I am not able to – Abc1729 Feb 23 at 8:16
• And, maybe you have any ideas, seeing a connection..? Use that $x\mapsto Cx$ is a bijection. Also, do you know the spectral theorem (a symmetric matrix has orthogonal eigenvectors and is diagonalizable)? – Berci Feb 23 at 8:19

First, if $$C^TAC=I$$ with an invertible matrix $$C$$, then for any $$x$$ consider $$y=C^{-1}x$$, then we have $$x=Cy$$ and $$x^TAx=y^TC^TACy=y^TIy=y^Ty=\|y\|^2>0$$ unless $$y=0$$ which is equivalent to $$x=0$$.
Second, if $$A$$ is positive definite, it determines an inner product $$(x,y):=x^TAy$$.
You can use e.g. Gram-Schmidt procedure to obtain a basis $$c_1,\dots,c_n$$ which is orthonormal with respect to the new inner product.
But that just means $$c_i^TAc_j=\delta_{ij}$$ where $$\delta_{ii}=1$$ and $$\delta_{ij}=0$$ if $$i\ne j$$.
Putting these together, it yields $$C^TAC=I$$ where $$C$$ is the matrix whose columns are $$c_i$$.