Prove that $\langle\mathbf{A}, \mathbf{C}\rangle \leq \delta$ equals with $\|\mathbf{A}\|_*\leq\delta$ Given an arbitrary matrix $\mathbf{A}\in R^{n\times n}$  and the basis matrix set $\mathbb{S}=\{\mathbf{C}\in R^{n\times n}: \mathbf{C}^T\mathbf{C}=\mathbf{I}_n\}$. 
 A: We can use the polar decomposition
$$
A=P\sqrt{A^*A}
$$ where $\sqrt{A^*A}$ is positive semi-definite and $P$ is orthogonal, i.e. $PP^T=P^TP=I$. Note that $\text{tr}(AB)=\text{tr}(BA)$ and $\text{tr}(A)=\text{tr}(A^T)$ hold for any $n\times n$ real matrices $A,B$. 
1. We can take $B=P$ and it follows
$$
\|A\|_*=\text{tr}(\sqrt{A^*A})=\text{tr}(B^TP\sqrt{A^*A})=\text{tr}(P\sqrt{A^*A}B^T)=\text{tr}(AB^T)=\langle A,B\rangle \le \delta.
$$
2. We have $$\begin{eqnarray}\langle A,B\rangle &=&\text{tr}(AB^T)\\&=&\text{tr}(P\sqrt{A^*A}B^T)\\&=&\text{tr}(B^TP\sqrt{A^*A})=\text{tr}(\tilde{B}\sqrt{A^*A})\end{eqnarray}$$ where $\tilde{B}=B^TP$ is an orthogonal matrix. Since $(I-\tilde{B})\sqrt{A^*A}(I-\tilde{B}^T)$ is positive semi-definite, we have
$$
\text{tr}\left((I-\tilde{B})\sqrt{A^*A}(I-\tilde{B}^T)\right)\ge 0.
$$ This leads to
$$
\text{tr}\left(\tilde{B}\sqrt{A^*A}\right)+\text{tr}\left(\sqrt{A^*A}\tilde{B}^T\right)\le\text{tr}\left(\sqrt{A^*A}\right)+\text{tr}\left(\tilde{B}\sqrt{A^*A}\tilde{B}^T\right).
$$ Since $$\text{tr}\left(\sqrt{A^*A}\tilde{B}^T\right)=\text{tr}\left((\tilde{B}^T)^T\sqrt{A^*A}^T\right)=\text{tr}\left(\tilde{B}\sqrt{A^*A}\right)$$ and $$\text{tr}\left(\tilde{B}\sqrt{A^*A}\tilde{B}^T\right)=\text{tr}\left(\tilde{B}^T\tilde{B}\sqrt{A^*A}\right)=\text{tr}\left(\sqrt{A^*A}\right),$$
 it follows$$
\langle A,B\rangle=\text{tr}\left(\sqrt{A^*A}\tilde{B}^T\right)\le \text{tr}\left(\sqrt{A^*A}\right)=\|A\|_*\le \delta.
$$
