Nuclear norm of a symmetric matrix

For any symmetric matrix $\mathbf{Z} \in \mathbb{S}^{d\times d}$, the nuclear norm is defined as:

\[\begin{align*} || \mathbf{Z} ||_* &= \text{Tr}\Big((\mathbf{Z}^2)^{\frac{1}{2}}\Big) \\ &= \text{Tr}\Big((\mathbf{P\Lambda P}^\top\mathbf{P\Lambda P}^\top)^{\tfrac{1}{2}}\Big) && \text{Eigendecomposition of } \mathbf{Z} \\ &= \text{Tr}\Big((\mathbf{P\Lambda}^2 \mathbf{P}^\top)^{\tfrac{1}{2}}\Big) && \text{Orthogonality of } \mathbf{P} \\ &= \text{Tr}\Big(\mathbf{P}(\mathbf{\Lambda}^2)^{\tfrac{1}{2}}\mathbf{P}^\top \Big) && \text{square root of an eigendecomposition} \\ &= \text{Tr}\Big((\mathbf{\Lambda}^2)^{\tfrac{1}{2}} \Big) && \text{Note2 below} \\ &= \text{Tr}\Big( |\mathbf{\Lambda}| \Big) \\ &= ||\mathbf{\lambda}||_1 && \mathbf{\Lambda}=\text{diag}(\lambda) \end{align*}\]

Note1: square root of an eigendecomposition

Given an eigendecomposition $ \mathbf{Z} = \mathbf{P\Lambda P}^\top $, then the following holds:

\[\Big(\mathbf{P\Lambda P}^\top \Big)^\frac{1}{2} = \mathbf{P\Lambda}^{\frac{1}{2}} \mathbf{P}^\top\]

Indeed:

\[\begin{align*} \Big( \mathbf{P\Lambda}^{\frac{1}{2}} \mathbf{P}^\top \Big)^2 &= \mathbf{P\Lambda}^{\frac{1}{2}} \mathbf{P}^\top \mathbf{P\Lambda}^{\frac{1}{2}} \mathbf{P}^\top \\ &= \mathbf{P\Lambda}^{\frac{1}{2}} \mathbf{\Lambda}^{\frac{1}{2}} \mathbf{P}^\top \\ &= \mathbf{P\Lambda P}^\top \end{align*}\]

Note2: trace properties

The matrices in a trace of a product can be switched without changing the result. If $\mathbf{A}$ is an $m \times n$ matrix and $\mathbf{B}$ is an $n \times m$ matrix, then the following holds:

\[\text{Tr}(\mathbf{AB}) = \text{Tr}(\mathbf{BA})\]

Then:

\[\text{Tr}\Big(\mathbf{P}(\mathbf{\Lambda}^2)^{\tfrac{1}{2}}\mathbf{P}^\top \Big) = \text{Tr}\Big(\mathbf{P}\big((\mathbf{\Lambda}^2)^{\tfrac{1}{2}}\mathbf{P}^\top \big) \Big) = \text{Tr}\Big(\big((\mathbf{\Lambda}^2)^{\tfrac{1}{2}}\mathbf{P}^\top \big) \mathbf{P} \Big) = \text{Tr}\Big((\mathbf{\Lambda}^2)^{\tfrac{1}{2}} \Big)\]

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eigendecomposition