Find the transformation matrix $A$ of a random vector $X $such that $Y = XA^T$ has uncorrelated components.

Question

I am trying to solve this problem. I need to find matrix A such that transformed vector's components are uncorrelated.

Answer 1

Changing the notation (for simplicity) to column vectors $X_{2\times1}$ and $Y_{2\times1}$, the equation is written as $Y=AX$.

Eigenvalue decomposition of $C_n$ is $$C_n=QDQ^{-1}=\begin{bmatrix} -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix}\begin{bmatrix} 4 & 0\\ 0 & 2 \end{bmatrix}\begin{bmatrix} -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix}$$ where $D$ is diagonal and $Q^T=Q^{-1}$.

To decorrelate $Y$ elements, we need to have $$\mathsf{E}(YY^T)=\Lambda$$ where $\Lambda$ is a diagonal matrix.

Choosing $A=Q^T$ yields $Y=Q^TX$ and

$$\mathsf{E}(YY^T)=\mathsf{E}(Q^TXX^TQ)=Q^T\mathsf{E}(XX^T)Q=Q^{-1}C_nQ=Q^{-1}(QDQ^{-1})Q=D$$ Hence, $\Lambda=D$, and the decorrelating transform is $$Y=\begin{bmatrix} -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix}X$$