Consider $\|A\|_{F} = \sqrt{\langle A, A\rangle)}$ where $\langle X, Y\rangle = \text{Tr}(X^{*}Y)$ where '*' denotes the conjugate transpose of a matrix.
I was asked to prove the following question:
Prove that $\|A\|_{F}^{2} = \sigma_{1}^{2} + \sigma_{2}^{2} + \dots + \sigma_{k}^{2}$
I was provided that the singular value decomposition of $A$ is $A = U\Sigma V^{*}$ where * denotes the conjugate transpose. Where $u_{i}$ and $v_{i}$ are column vectors of $U$ and $V$ respectfully, and where $\sigma_{i}$ are the diagonal entries of $\Sigma$. In previous questions I proved that $A = A_{1} + A_{2} + \dots + A_{k}$ where $A_{j} = \sigma_{j}u_{i}v_{i}^{*}$, $\langle A_{i}, A_{j} \rangle = 0$ if $i \neq j$ and $\|A_{j}\|_{F} = \sigma_{j}$. I have made an attempt to at a proof which is displayed below:
Consider the singular value decomposition $A = U\Sigma V^{*}$. From this we know that:
\begin{align*} & \|A\|_{F}^{2} = (\sqrt{\langle A, A \rangle_{F}})^{2} = \langle A, A \rangle \\ &= \langle A_{1} + \dots + A_{k} , A_{1} + \dots + A_{k} \rangle \\ &= \langle A_{1}, A_{1} + \dots + A_{k} \rangle + \dots + \langle A_{k} , A_{1} + \dots + A_{k} \rangle \\ &= \sum_{i=1}^{k} \langle A_{i}, A \rangle \\ &= \sum_{i=1}^{k} \text{Tr}(A_{i}^{*}A) \\ &= \sum_{i=1}^{k} \text{Tr}(v_{i}u_{i}^{*}\sigma_{i}^{*}(\sigma_{1}u_{1}v_{1}^{*} + \dots + \sigma_{k}u_{k}v_{k}^{*})) \\ &= \sum_{i=1}^{k} \text{Tr}(v_{i}u_{i}^{*}\sigma_{i}^{*}\sigma_{1}u_{1}v_{1}^{*} + \dots + v_{i}u_{i}^{*}\sigma_{i}^{*}\sigma_{k}u_{k}v_{k}^{*}) \\ &= \sum_{i=1}^{k} \text{Tr}(\sigma_{i}^{*}\sigma_{1}v_{i}u_{i}^{*}u_{1}v_{1}^{*} + \dots + \sigma_{i}^{*}\sigma_{k}v_{i}u_{i}^{*}u_{k}v_{k}^{*}) \\ &= \sum_{i=1}^{k} \text{Tr}(\sigma_{i}^{*}\sigma_{i}v_{i}u_{i}^{*}u_{i}v_{i}^{*}) \text{ because $\langle A_{i}, A_{j} \rangle = 0$ is 0 for distinct $i$ and $j$} \\ &= \sum_{i=1}^{k} \sigma_{i}^{*}\sigma_{i} \text{Tr}(v_{i}u_{i}u_{i}^{*}v_{i}^{*}) \\ &= \sum_{i=1}^{k} \sigma_{i}^{*}\sigma_{i} = \sum_{i=1}^{k} \sigma_{i}^{2} = \sigma_{1}^{2} + \dots + \sigma_{k}^{2}. \end{align*} From this we can conclude that $\|A\|_{F}^{2} = \sigma_{1}^{2} + \dots + \sigma_{k}^{2}$.
Is this proof correct? Thanks in advance!
It looks okay to me, but I think you can shorten it quite a bit by using $\langle A_{i}, A_{j} \rangle = 0$ if $i \neq j$, $\|A_{j}\|_{F} = \sqrt{\langle A_{i}, A_i \rangle}=\sigma_{j}$, and the linearity property of the inner product.
Looking at \begin{align*} \langle A_{i}, A \rangle &=\langle A_{i}, A_1+A_2+\cdots+A_n \rangle\\ &=\langle A_{i}, A_1 \rangle +\langle A_i,A_2\rangle+\cdots+\langle A_{i},A_n \rangle\\ &=\langle A_{i}, A_i \rangle \text{ (Since all the rest are 0)} \end{align*}
Then \begin{align*} & \|A\|_{F}^{2} = (\sqrt{\langle A, A \rangle_{F}})^{2} = \langle A, A \rangle \\ &= \langle A_{1} + \dots + A_{k} , A_{1} + \dots + A_{k} \rangle \\ &= \langle A_{1}, A_{1} + \dots + A_{k} \rangle + \dots + \langle A_{k} , A_{1} + \dots + A_{k} \rangle \\ &= \sum_{i=1}^{k} \langle A_{i}, A \rangle \\ &= \sum_{i=1}^{k} \langle A_{i}, A_i \rangle \\ &= \sum_{i=1}^{k} \sigma_i^2 \\ \end{align*}