其中U是 $AA^T $ 之eigenvectors;V是 $A^TA$ 之eigenvectors
4. For any matrix A, $AA^T$ and $A^TA$ are both symmetric.
sol. $(AA^T)^T=AA^T$; $(A^TA)^T = A^T A$
5. For arbitrary matrices A, B, AB and BA have same nonzero eigenvalues. (if AB, BA both exist)
sol. $AB\vec{u} = \lambda \vec{u} \rightarrow since \lambda, \vec{u} \neq 0, B\vec{u}\neq 0$ and
$BA(B\vec{u})_{(\rightarrow set=\vec{y})} = B(AB\vec{u}) = \lambda B \vec{u} = \lambda \vec{y}$
\begin{cases}
AB\vec{u} = \lambda\vec{u} \\
BA\vec{y} = \lambda\vec{y} \end{cases}
\begin{cases}
AB\vec{u} = \lambda\vec{u} \\
BA\vec{y} = \lambda\vec{y} \end{cases}
From 4. and 5., we can write $A^TA$ and $AA^T$ as:
$AA^T = UDU^T$; $A^TA = V \tilde{D}V$ and D, $\tilde{D}$ have the same nonzero eigenvalues.
6. Eigenvalues of $AA^T$ are all $\geq$ 0
sol. Let $\vec{v}$ be an arbitrary vector with shape (A.shape[0], 1).
$\vec{v}^T_{1\times m}A_{m\times n}A^T_{n\times m} \vec{v}_{m\times 1} = \vec{y}^T\vec{y}\geq 0$ (set $y=\vec{v}^TA$)
Now we set $\vec{v} =$ eigenvector of $AA^T$ $(\vec{u})$
$\vec{u}^TAA^T\vec{u} = \vec{u}^T\lambda\vec{u} = \lambda\vec{u}^T\vec{u}$
Since $\vec{u}^T\vec{u}\geq 0 $ and $\vec{u}^TAA^T\vec{u}\geq 0$, $\lambda$ always $\geq 0$
And from 6. we can rewrite A as $USV^T$ such that
$AA^T = USV^TVS^TU^T = UDU^T$ and $A^TA = VS^TU^TUSV^T = V\tilde{D}V^T$
$S_{i,i}$的值 =
\begin{cases}
\sqrt{\lambda_i}必存在 \text{ (for nonzero eigenvalue }\lambda_i\text{)}\\
0 \end{cases}
\sqrt{\lambda_i}必存在 \text{ (for nonzero eigenvalue }\lambda_i\text{)}\\
0 \end{cases}
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