Singular Value Decomposition

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Previously we have seen that for $n\times n$ matrix (say$A$) with $n$ independent eigenvectors factorization is,

$$\begin{matrix} A=S\Lambda S^{-1} \end{matrix}$$

Previously we have seen that if eigenvectors of matrix $A$ are orthogonal then $Q^{-1}=Q^T$, then the factorization is,
(We discussed it in detail HERE)

$$\begin{matrix} A=Q\Lambda Q^{T} \end{matrix}$$

But generally eigenvectors are not orthogonal.

Singular Value Decomposition(SVD) is a factorization of any $m\times n$ matrix.

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$$\begin{matrix} A=U\Sigma V^T \end{matrix}$$

Say that we have a $m\times n$ matrix $A$ and $\text{Rank}(A)=r$,

$$A_{m\times n}=\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end{bmatrix}$$

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IDEA: behind SVD is to map an orthonormal basis in $\mathbb{R}^n$ (Row space + Null Space of matrix $A^TA$) to the orthonormal basis in $\mathbb{R}^m$ (Column space + Left Null Space of matrix $AA^T$).

$\quad$

The row space is in $\mathbb{R}^n$, here our row vectors are,

$$\begin{bmatrix} a_{11} \\ a_{12} \\ \vdots \\ a_{1n} \end{bmatrix}\in\mathbb{R}^n, \begin{bmatrix} a_{21} \\ a_{22} \\ \vdots \\ a_{2n} \end{bmatrix}\in\mathbb{R}^n, \cdots, \begin{bmatrix} a_{m1} \\ a_{m2} \\ \vdots \\ a_{mn} \end{bmatrix}\in\mathbb{R}^n$$

The column space is in $\mathbb{R}^m$, our column vectors are,

$$\begin{bmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{bmatrix}\in\mathbb{R}^m, \begin{bmatrix} a_{12} \\ a_{22} \\ \vdots \\ a_{m2} \end{bmatrix}\in\mathbb{R}^m, \cdots, \begin{bmatrix} a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn} \end{bmatrix}\in\mathbb{R}^m$$

A vector (say) $\vec{v}\in\mathbb{R}^n$ has some portion in Row Space and some in Left Null Space, when we transform that vector $\vec{v}\in\mathbb{R}^n$ using matrix $A$, $A\vec{v}=\vec{u}\in\mathbb{R}^m$ now this vector $\vec{u}\in\mathbb{R}^m$ has some portion in Column Space and some in Null Space
$\vec{u}=\vec{u}_p+\vec{u}_n$ [Reference]

● Row space $(C(A^T))$ (Gather Orthonormal Basis vectors for Row Space)

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Row vectors lives in $n$-dimensional vector space.
$\text{Rank}(A)=r$ so we have $r$ independent row vectors.
So the space spanned by row vectors is $r$-dimensional vector space(Row Space), so to span this whole $r$-dimensional vector space we need $r$ independent basis vectors.
Among those we choose orthonormal basis vectors.

( we can get them using Gram Schmidt method, but here we will use $A^TA$ to get them we will discuss it below)
(say) our orthonormal basis vector for Row Space are $\vec{v}_1,\vec{v}_2,\cdots,\vec{v}_r$.

● Null Space $(N(A))$ (Orthonormal Basis vectors of Null Space)

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Null Space is perpendicular to Row Space(we talked about orthogonal subspaces HERE), so every vector in Null Space is perpendicular to Row Space.
$\text{Rank}(A)=r$ so we have $n-r$ dependent(free) row vectors.
So the space spanned by vectors in Null Space is $(n-r)$-dimensional vector space(Null Space), so to span this whole $(n-r)$-dimensional vector space we need $n-r$ independent basis vectors in Null Space.
(we don't need to find basis vectors for SVD we discuss it below in Column Space$(C(A))$ section )
(say) our orthonormal basis vector for Null Space are $\vec{v}_{r+1},\vec{v}_{r+2},\cdots,\vec{v}_n$.

● Column Space $(C(A))$ (Map the Orthonormal Basis vectors from the Row Space to Column Space)

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After we got our orthonormal basis vector for Row Space $\vec{v}_1,\vec{v}_2,\cdots,\vec{v}_n$, we map them to the Column Space using matrix $A$.
$\text{Rank}(A)=r$ so,

$$\begin{matrix} A\vec{v}_i=\sigma_i\vec{u}_i;\quad\forall i\in\{1,2,\cdots,r\} \end{matrix}$$

For $i\in\{1,2,\cdots,r\}$ $A$ maps $\vec{v}_i$ to $\sigma_i\vec{u}_i$ then we normalize it to get $\vec{u}_i$(unit vector), $\sigma_i$ is the length of $A\vec{v}_i$.
Now we got the orthonormal basis vector for our column space $\vec{u}_1,\vec{u}_2,\cdots,\vec{u}_r$

$$\begin{matrix} A\vec{v}_i=\vec{0};\quad\forall i\in\{r+1,r+2,\cdots,n\} \end{matrix}$$

For $i\in\{r+1,r+2,\cdots,n\}$ $\vec{v}_i$ lives in the Null Space of $A$
So $\vec{u}_{r+1},\vec{u}_{r+2},\cdots,\vec{u}_n$ are all $\vec{0}$

But wait WHY all $\vec{u}_i$'s are perpendicular to each other?
we discussed it below in $A^TA$ section.

● Left Null Space $(N(A^T))$ (Map the Orthonormal Basis vectors from the Null Space to Left Null Space)

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Left Null Space is perpendicular to Column Space(we talked about orthogonal subspaces HERE), so every vector in Left Null Space is perpendicular to Column Space.
$\text{Rank}(A)=r$ so we have $m-r$ dependent(free) column vectors.
So the space spanned by vectors in Left Null Space is $(m-r)$-dimensional vector space(Null Space), so to span this whole $(m-r)$-dimensional vector space we need $m-r$ independent basis vectors in Left Null Space.
(say) our orthonormal basis vector for Left Null Space are $\vec{u}_{r+1},\vec{u}_{r+2},\cdots,\vec{u}_n$.

● Terminologies
Say $\vec{v}_1,\vec{v}_2,\cdots,\vec{v}_r$ are Orthonormal Basis vectors for Row Space $(C(A^T))$,
and $\vec{v}_{r+1},\vec{v}_{r+2},\cdots,\vec{v}_n$ are Orthonormal Basis vectors for Null Space $(N(A^T))$

$$\text{(say)} \mathbf{V}'= \begin{bmatrix} \begin{matrix} \vdots & \vdots & & \vdots \\ \vec{v}_1 & \vec{v}_2 & \cdots & \vec{v}_r \\ \vdots & \vdots & & \vdots \end{matrix} & \begin{matrix} \vdots & \vdots & & \vdots \\ \vec{v}_{r+1} & \vec{v}_{r+2} & \cdots & \vec{v}_n \\ \vdots & \vdots & & \vdots \end{matrix} \end{bmatrix}_{n\times n} \\ \text{(say)} \mathbf{U}'= \begin{bmatrix} \begin{matrix} \vdots & \vdots & & \vdots \\ \vec{u}_1 & \vec{u}_2 & \cdots & \vec{u}_r \\ \vdots & \vdots & & \vdots \end{matrix} & \begin{matrix} \vdots & \vdots & & \vdots \\ \vec{u}_{r+1} & \vec{u}_{r+2} & \cdots & \vec{u}_n \\ \underbrace{\vdots}_{=\vec{0}} & \underbrace{\vdots}_{=\vec{0}} & & \underbrace{\vdots}_{=\vec{0}} \end{matrix} \end{bmatrix}_{m\times n} \\ \text{(say)} \Sigma'= \begin{bmatrix} \begin{bmatrix} \sigma_1 & & & \huge0 \\ & \sigma_2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_r \\ \end {bmatrix}_{r\times r} & \mathbf{0}_{r\times (n-r)} \\ \mathbf{0}_{(n-r)\times r} & \mathbf{0}_{(n-r)\times (n-r)} \end{bmatrix}_{n\times n}$$

Mapping from $\mathbb{R}^n$ to $\mathbb{R}^m$ is,

danger

$$\begin{matrix} A\vec{v}_i=\sigma_i\vec{u}_i;\quad\forall i\in\{1,2,\cdots,n\} \end{matrix}$$

We can Write it as,

$$A\underbrace{\begin{bmatrix} \vdots & \vdots & & \vdots \\ \vec{v}_1 & \vec{v}_2 & \cdots & \vec{v}_n \\ \vdots & \vdots & & \vdots \end{bmatrix}}_{ V'_{_{n\times n}} } = \underbrace{\begin{bmatrix} \vdots & \vdots & & \vdots \\ \vec{u}_1 & \vec{u}_2 & \cdots & \vec{u}_n \\ \vdots & \vdots & & \vdots \end{bmatrix}}_{ U'_{m\times n} } \underbrace{\begin{bmatrix} \begin{bmatrix} \sigma_1 & & & \huge0 \\ & \sigma_2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_r \\ \end {bmatrix}_{r\times r} & \mathbf{0}_{r\times (n-r)} \\ \mathbf{0}_{(n-r)\times r} & \mathbf{0}_{(n-r)\times (n-r)} \end {bmatrix}}_{ \Sigma'_{n\times n} }$$

So,

$$AV'_{n\times n}=U'_{m\times n}\Sigma'_{n\times n}$$

Full SVD

Say that all columns of our matrix $A$ are independent $\Rightarrow \text{Rank}(A)=m$, so

• Dimension of Row Space is: $m$

• Dimension of Null Space is: $n-m$

• Dimension of Column Space is: $m$

• Dimension of Left Null Space is: $m-m=0$ Our equation was,

  $$A\underbrace{\begin{bmatrix} \vdots & \vdots & & \vdots \\ \vec{v}_1 & \vec{v}_2 & \cdots & \vec{v}_n \\ \vdots & \vdots & & \vdots \end{bmatrix}}_{ V'_{_{n\times n}} } = \underbrace{\begin{bmatrix} \vdots & \vdots & & \vdots \\ \vec{u}_1 & \vec{u}_2 & \cdots & \vec{u}_n \\ \vdots & \vdots & & \vdots \end{bmatrix}}_{ U'_{m\times n} } \underbrace{\begin{bmatrix} \begin{bmatrix} \sigma_1 & & & \huge0 \\ & \sigma_2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_m \\ \end {bmatrix}_{m\times m} & \mathbf{0}_{m\times (n-m)} \\ \mathbf{0}_{(n-m)\times m} & \mathbf{0}_{(n-m)\times (n-m)} \end {bmatrix}}_{ \Sigma'_{n\times n} }$$

  So,

  $$AV'_{n\times n}=U'_{m\times n}\Sigma'_{n\times n}$$

  And as we discussed above that $\vec{u}_{r+1},\vec{u}_{r+2},\cdots,\vec{u}_n$ are all $\vec{0}$, so we can discard those vectors from our equation so now our equation will become,

  $$A\underbrace{\begin{bmatrix} \vdots & \vdots & & \vdots \\ \vec{v}_1 & \vec{v}_2 & \cdots & \vec{v}_n \\ \vdots & \vdots & & \vdots \end{bmatrix}}_{ V_{_{n\times n}} } = \underbrace{\begin{bmatrix} \vdots & \vdots & & \vdots \\ \vec{u}_1 & \vec{u}_2 & \cdots & \vec{u}_m \\ \vdots & \vdots & & \vdots \end{bmatrix}}_{ U_{m\times m} } \underbrace{\begin{bmatrix} \begin{bmatrix} \sigma_1 & & & \huge0 \\ & \sigma_2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_m \\ \end {bmatrix}_{m\times m} & \mathbf{0}_{m\times (n-m)} \end {bmatrix}}_{ \Sigma_{m\times n} }$$

  So,

  $$\begin{matrix} A_{m\times n}V_{n\times n}=U_{m\times m}\Sigma_{m\times n} \end{matrix}$$

  This equation is what we call Full SVD

Reduced SVD

If some of the columns of our matrix $A$ are dependent and say $\text{Rank}(A)=r$, so

• Dimension of Row Space is: $r$

• Dimension of Null Space is: $n-r$

• Dimension of Column Space is: $r$

• Dimension of Left Null Space is: $m-r$ Our equation was,

  $$A\underbrace{\begin{bmatrix} \vdots & \vdots & & \vdots \\ \vec{v}_1 & \vec{v}_2 & \cdots & \vec{v}_n \\ \vdots & \vdots & & \vdots \end{bmatrix}}_{ V'_{_{n\times n}} } = \underbrace{\begin{bmatrix} \vdots & \vdots & & \vdots \\ \vec{u}_1 & \vec{u}_2 & \cdots & \vec{u}_n \\ \vdots & \vdots & & \vdots \end{bmatrix}}_{ U'_{m\times n} } \underbrace{\begin{bmatrix} \begin{bmatrix} \sigma_1 & & & \huge0 \\ & \sigma_2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_m \\ \end {bmatrix}_{m\times m} & \mathbf{0}_{m\times (n-m)} \\ \mathbf{0}_{(n-m)\times m} & \mathbf{0}_{(n-m)\times (n-m)} \end {bmatrix}}_{ \Sigma'_{n\times n} }$$

  So,

  $$AV'_{n\times n}=U'_{m\times n}\Sigma'_{n\times n}$$

  And as we discussed above that $\vec{v}_{r+1},\vec{u}_{r+2},\cdots,\vec{u}_n$ are all $\vec{0}$, so these vectors are redundant we can discard those vectors from our equation so now our equation will become,

  $$A\underbrace{\begin{bmatrix} \vdots & \vdots & & \vdots \\ \vec{v}_1 & \vec{v}_2 & \cdots & \vec{v}_r \\ \vdots & \vdots & & \vdots \end{bmatrix}}_{ V_{_{n\times r}} } = \underbrace{\begin{bmatrix} \vdots & \vdots & & \vdots \\ \vec{u}_1 & \vec{u}_2 & \cdots & \vec{u}_r \\ \vdots & \vdots & & \vdots \end{bmatrix}}_{ U_{m\times r} } \underbrace{\begin{bmatrix} \sigma_1 & & & \huge0 \\ & \sigma_2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_r \\ \end {bmatrix}}_{ \Sigma_{r\times r} }$$

  So,

  $$\begin{matrix} A_{m\times n}V_{n\times r}=U_{m\times r}\Sigma_{r\times r} \end{matrix}$$

  This equation is what we call Reduced SVD

We have seen that we can write any $m\times n$ matrix $A$ as $AV=U\Sigma$ where $V$ and $U$ are Orthonormal matrix.
Now the real challange is to find these $V$ and $U$ Orthonormal matrix.
Once we got those $V$ and $U$ Orthonormal matrix then we can find $A$ as,
$AV=U\Sigma$
$A=U\Sigma V^{-1}$
And because $V$ is an orthogonal matrix so we know that $V^{-1}=V^T$
To achieve it we need $V$ to be square matrix so we have to use Full SVD method, then we get

danger

$$\begin{matrix} A_{m\times n}=U_{m\times m}\Sigma_{m\times n}V_{n\times n}^{T} \end{matrix}$$

Everything looks good but how to find these $V$ and $U$ Orthonormal matrix.

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IDEA: to find the $V$ and $U$ are the Matrices $A^TA$ and $AA^T$

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• Row space of $A$ is same as Row space of $A^TA$.

• It is easy to find the basis for Row space of $A^TA$, so rather than finding basis for Row space of $A$ we will find basis for Row space of $A^TA$*

Explanation:

$A_{m\times n}=\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end{bmatrix}$
Let's rewrite $A^TA$
$A^TA=A^T\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end{bmatrix}$

Here we can see that we are mapping the row vectors of $A$ onto the Row Space of $A^T$.

And if $\text{Rank}(A)=r$ then $\Rightarrow \text{Rank}(A^T)=r$ so $A$ and $A^T$ both are $r$-dimensions Row Space.

So when we map row vector of $A$ onto the Row Space of $A^T$ then the resulting Row Space also has $r$-dimensions, so we can say that $\text{Rank}(A^TA)=r$

Now we can say that Row Space of $A$ is same as Row Space of $A^TA$.

$(A^TA)^T=A^TA \Rightarrow A^TA$ is symmetric matrix, so
Row Space of $A^TA=$ Column Space of $A^TA=$ Row Space of $A$

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• Column Space of $A$ is same as Column Space of $AA^T$

• It is easy to find the basis for Column Space of $AA^T$, so rather than finding basis for Column Space of $A$ we will find basis for Column Space of $AA^T$*

Explanation:
$A^T_{m\times n}=\begin{bmatrix} a_{11} & a_{21} & \cdots & a_{m1} \\ a_{12} & a_{22} & \cdots & a_{m2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & \cdots & a_{mn} \\ \end{bmatrix}$
Let's rewrite $AA^T$
$AA^T=A\begin{bmatrix} a_{11} & a_{21} & \cdots & a_{m1} \\ a_{12} & a_{22} & \cdots & a_{m2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & \cdots & a_{mn} \\ \end{bmatrix}$

Here we can see that we are mapping the column vectors of $A^T$ onto the Column Space of $A$.

And if $\text{Rank}(A)=r$ then $\Rightarrow \text{Rank}(A^T)=r$ so $A$ and $A^T$ both are $r$-dimensions Column Space.

So when we map column vector of $A^T$ onto the Column Space of $A$ then the resulting Column Space also has $r$-dimensions, so we can say that $\text{Rank}(AA^T)=r$

Now we can say that Column Space of $A$ is same as Column Space of $AA^T$.

$(AA^T)^T=AA^T \Rightarrow AA^T$ is symmetric matrix, so
Column Space of $AA^T=$ Row Space of $AA^T=$ Column Space of $A$

Finding orthonormal Row vectors$(V)$ for matrix $A$

We discussed that Row space of $A$ is same as Row space of $A^TA$.
So we need to find orthonormal Row vectors for matrix $A^TA$.

$A_{m\times n}=U_{m\times m}\Sigma_{m\times n}V_{n\times n}^{T}$
$(\Sigma^T_{m\times n}=\Sigma_{n\times m})$
$\Rightarrow (A^T)_{n\times m} = V_{n\times n}\Sigma_{n\times m}U_{m\times m}^{T}$

$$\Rightarrow (A^T)_{n\times m}A_{m\times n} = V_{n\times n}\Sigma_{n\times m} \underbrace{U_{m\times m}^{T} U_{m\times m}}_{=\mathcal{I}_m} \Sigma_{m\times n}V_{n\times n}^{T}$$

$$\Sigma_{n\times m}\Sigma_{m\times n}= \underbrace{\begin{bmatrix} \begin{bmatrix} \sigma_1 & & & \huge0 \\ & \sigma_2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_m \\ \end {bmatrix}_{m\times m} \\ \mathbf{0}_{(n-m)\times m} \end {bmatrix}}_{ \Sigma_{n\times m} } \underbrace{\begin{bmatrix} \begin{bmatrix} \sigma_1 & & & \huge0 \\ & \sigma_2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_m \\ \end {bmatrix}_{m\times m} & \mathbf{0}_{m\times (n-m)} \end {bmatrix}}_{ \Sigma_{m\times n} }$$

$$\Sigma_{n\times m}\Sigma_{m\times n}= \underbrace{\begin{bmatrix} \begin{bmatrix} \sigma_1^2 & & & \huge0 \\ & \sigma_2^2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_m^2 \\ \end {bmatrix}_{m\times m} & \mathbf{0}_{m\times (n-m)} \\ \mathbf{0}_{(n-m)\times m} & \mathbf{0}_{(n-m)\times (n-m)} \\ \end{bmatrix}}_{ \Sigma^2_{n\times n}} = \Sigma^2_{n\times n}$$

$$\Sigma^2_{n\times n}= \begin{bmatrix} \begin{bmatrix} \sigma_1^2 & & & \huge0 \\ & \sigma_2^2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_m^2 \\ \end {bmatrix}_{m\times m} & \mathbf{0}_{m\times (n-m)} \\ \mathbf{0}_{(n-m)\times m} & \mathbf{0}_{(n-m)\times (n-m)} \\ \end{bmatrix}_{n\times n}$$

$\Rightarrow (A^T)_{n\times m}A_{m\times n} = V_{n\times n}\Sigma^2_{n\times n}V_{n\times n}^{T}$
Those $3$ $\mathbf{0}$ matrices are interesting,

• $\mathbf{0}_{m\times (n-m)}$
• $\mathbf{0}_{(n-m)\times m}$
• $\mathbf{0}_{(n-m)\times (n-m)}$ Remember that in Full SVD we said that $\text{Rank}(A)=m$ so there must be $n-m$ dependent row vectors. and these $\mathbf{0}$ matrices are negating those dependent vectors.
  

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Remember $[A=Q\Lambda Q^T]$ where $A$ is symmetric matrix.

$A^TA$ is a symmetric matrix, so

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Columns of $V$ are eigenvectors of matrix $A^TA$, and $\sigma_i^2$ are the $i^{th}$ eigenvalue of matrix $A^TA$

Finding orthonormal Column Vectors$(U)$ for matrix $A$

We discussed that Column Space of $A$ is same as Column Space of $AA^T$.
So we need to find orthonormal Column vectors for matrix $AA^T$.

$A_{m\times n}=U_{m\times m}\Sigma_{m\times n}V_{n\times n}^{T}$
$(\Sigma^T_{m\times n}=\Sigma_{n\times m})$
$\Rightarrow (A^T)_{n\times m} = V_{n\times n}\Sigma_{n\times m}U_{m\times m}^{T}$
$\Rightarrow A_{m\times n}(A^T)_{n\times m} = U_{m\times m}\Sigma_{m\times n} \underbrace{V_{n\times n}^{T} V_{n\times n}}_{=\mathcal{I}_n} \Sigma_{n\times m}U_{m\times m}^{T}$

$$\Sigma_{m\times n}\Sigma_{n\times m}= \underbrace{\begin{bmatrix} \begin{bmatrix} \sigma_1 & & & \huge0 \\ & \sigma_2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_m \\ \end {bmatrix}_{m\times m} & \mathbf{0}_{m\times (n-m)} \end {bmatrix}}_{ \Sigma_{m\times n} } \underbrace{\begin{bmatrix} \begin{bmatrix} \sigma_1 & & & \huge0 \\ & \sigma_2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_m \\ \end {bmatrix}_{m\times m} \\ \mathbf{0}_{(n-m)\times m} \end {bmatrix}}_{ \Sigma_{n\times m} }$$

$$\Sigma_{n\times m}\Sigma_{m\times n}= \underbrace{\begin{bmatrix} \sigma_1^2 & & & \huge0 \\ & \sigma_2^2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_m^2 \\ \end{bmatrix}}_{ \Sigma^2_{m\times m}} = \Sigma^2_{m\times m}$$

$$\Sigma^2_{m\times m}= \begin{bmatrix} \sigma_1^2 & & & \huge0 \\ & \sigma_2^2 & & \\ & & \ddots & \\ \huge0 & & & \sigma_m^2 \\ \end{bmatrix}_{m\times m}$$

$\Rightarrow A_{m\times n}(A^T)_{n\times m} = U_{m\times m}\Sigma^2_{m\times m}U_{m\times m}^{T}$
This time there isn't any $\mathbf{0}$ matrix, because as we discussed, in Full SVD $\text{Rank}(A)=m$ so all columns vectors and row vectors of $U$ are independent.

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Remember $[A=Q\Lambda Q^T]$ where $A$ is symmetric matrix.

$AA^T$ is a symmetric matrix, so

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Columns of $U$ are eigenvectors of matrix $AA^T$, and $\sigma_i^2$ are the $i^{th}$ eigenvalue of matrix $AA^T$