Can someone please help me to understand the following statement?
Let $Q \in \mathbb{R}^{d \times d}$ be a positive definite matrix with eigenvalues $\lambda_{\max} = \lambda_{1} \geq \dots \geq \lambda_{d} = \lambda_{\min} > 0$. Then the following optimization problem $$\min_{x \in \mathbb{R}^{d}} \dfrac{1}{2} x^{T}Qx$$ is equivalent to $$\min_{x \in \mathbb{R}^{d}} \dfrac{1}{2} x^{T}Dx$$ where $Q = UDU^{T}$ is an eigendecomposition of $Q$ and $D$ is the diagonal matrix with the eigenvalues along the diagonal.