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.

  • 1
    $\begingroup$ Can you do one direction? $\endgroup$ – Berci Feb 23 at 8:12
  • $\begingroup$ No, I am not able to $\endgroup$ – Abc1729 Feb 23 at 8:16
  • $\begingroup$ 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)? $\endgroup$ – 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 $c_i$.


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