Anki Deck Changes

Note 1: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Classic
GUID: E.Y/B(ySb^

added

New Note

Front

What is the pseudoinverse in the case where $A \in \mathbb{R}^{n \times m}$ has neither full column nor full row rank?

Back

What is the pseudoinverse in the case where $A \in \mathbb{R}^{n \times m}$ has neither full column nor full row rank?

We have to solve both projecting and finding $||x||^2$ with the smallest norm at once.
We decompose $A = CR'$ where $C$ has full column and $R'$ full row-rank.
Then $A^\dagger = R^\dagger C^\dagger$.

Field-by-field Comparison

Field	Before	After
Front		What is the pseudoinverse in the case where $A \in \mathbb{R}^{n \times m}$ has neither full column nor full row rank?
Back		We have to solve both projecting and finding $\|\|x\|\|^2$ with the smallest norm at once.<br>We decompose $A = CR'$ where $C$ has full column and $R'$ full row-rank.<br>Then $A^\dagger = R^\dagger C^\dagger$.

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::0._Prelude

Note 2: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: FjSbP7PsTQ

added

New Note

Front

$A^\dagger A$ is the projection matrix on $C(A^\top)$

Back

$A^\dagger A$ is the projection matrix on $C(A^\top)$

Field-by-field Comparison

Field	Before	After
Text		$A^\dagger A$ is {{c1::the projection matrix on $C(A^\top)$}}

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::2._Properties

Note 3: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: K9{>JhFX!.

added

New Note

Front

$(A^\top)^\dagger = (A^\dagger)^\top $

Back

$(A^\top)^\dagger = (A^\dagger)^\top $

Field-by-field Comparison

Field	Before	After
Text		$(A^\top)^\dagger = {{c1:: (A^\dagger)^\top }}$

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::2._Properties

Note 4: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: Md-#o7w$ub

added

New Note

Front

For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = m$, the pseudo-inverse $A^\dagger \in \mathbb{R}^{n \times m}$ is a right inverse of $A$: \[ A A^\dagger = I \]Proof Included

Back

For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = m$, the pseudo-inverse $A^\dagger \in \mathbb{R}^{n \times m}$ is a right inverse of $A$: \[ A A^\dagger = I \]Proof Included

Proof Since $A^\top$ has full column rank, $((A^\top)^\top A^\top) = AA^\top$ is invertible: $AA^\dagger = AA^\top(A A^\top)^{-1} = I$.

Field-by-field Comparison

Field	Before	After
Text		For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = m$, the <b>pseudo-inverse</b> $A^\dagger \in \mathbb{R}^{n \times m}$ is {{c1::a right inverse}} of $A$: \[ {{c1:: A A^\dagger = I }}\]<i>Proof Included</i>
Extra		<div><b>Proof</b> Since $A^\top$ has full column rank, $((A^\top)^\top A^\top) = AA^\top$ is invertible: $AA^\dagger = AA^\top(A A^\top)^{-1} = I$.</div>

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::1._Definitions

Note 5: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: Nu?3UH(7q

added

New Note

Front

Given $A \in \mathbb{R}^{m \times n}$ (can have any rank) and a vector $b \in \mathbb{R}^m$, the unique solution to \[ \min_{x \in \mathbb{R}^n} ||x||^2 \] such that $A^\top Ax = A^\top b$ is given by {{c2::$\hat{x} = A^\dagger b$}}.

Back

Given $A \in \mathbb{R}^{m \times n}$ (can have any rank) and a vector $b \in \mathbb{R}^m$, the unique solution to \[ \min_{x \in \mathbb{R}^n} ||x||^2 \] such that $A^\top Ax = A^\top b$ is given by {{c2::$\hat{x} = A^\dagger b$}}.

Field-by-field Comparison

Field	Before	After
Text		Given $A \in \mathbb{R}^{m \times n}$ (can have any rank) and a vector $b \in \mathbb{R}^m$, the {{c1::<b>unique</b>}}<b> </b>solution to \[ \min_{x \in \mathbb{R}^n} \|\|x\|\|^2 \] such that {{c1::$A^\top Ax = A^\top b$}} is given by {{c2::$\hat{x} = A^\dagger b$}}.

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::1._Definitions

Note 6: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: P+IBIpS^<+

added

New Note

Front

$A^\dagger A$ is symmetric

Back

$A^\dagger A$ is symmetric

Field-by-field Comparison

Field	Before	After
Text		$A^\dagger A$ is {{c1:: symmetric :: property }}

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::2._Properties

Note 7: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Classic
GUID: P/y(FM1x:V

added

New Note

Front

Why is the pseudoinverse (for $A$ with full row-rank) $A^\top (AA^\top)^{-1}$?

Back

Why is the pseudoinverse (for $A$ with full row-rank) $A^\top (AA^\top)^{-1}$?

It uses the multiplication by $A^\top$ to choose an $\hat{x}$ that lies in the row-space, thus minimising the norm.

Field-by-field Comparison

Field	Before	After
Front		Why is the pseudoinverse (for $A$ with full row-rank) $A^\top (AA^\top)^{-1}$?
Back		It uses the multiplication by $A^\top$ to choose an $\hat{x}$ that lies in the row-space, thus minimising the norm.

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::1._Definitions

Note 8: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: PCBMoNL{vn

added

New Note

Front

The nullspace of $A$ $N(A) $ is equal to the nullspace of $N(A^\dagger)$. Proof Included

Back

The nullspace of $A$ $N(A) $ is equal to the nullspace of $N(A^\dagger)$. Proof Included

Intuitively this holds as $A^\dagger$ projects onto the $C(A)$.

Thus anything in $C(A)^\bot = N(A^\top)$ is projected to $0$. In other words, $\forall x \in C(A^\top)^\bot = N(A^\top)$ we have $A^\dagger x = 0$.
We conclude that $N(A) = C(A^\top)^\bot = N(A^\dagger)$.

Field-by-field Comparison

Field	Before	After
Text		The <b>nullspace of </b>$A$ $N(A) $ is equal to {{c1:: the nullspace of $N(A^\dagger)$}}. <i>Proof Included</i>
Extra		Intuitively this holds as $A^\dagger$ projects onto the $C(A)$.<br><br>Thus anything in $C(A)^\bot = N(A^\top)$ is projected to $0$. In other words, $\forall x \in C(A^\top)^\bot = N(A^\top)$ we have $A^\dagger x = 0$.<br>We conclude that $N(A) = C(A^\top)^\bot = N(A^\dagger)$.

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::2._Properties

Note 9: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: QGIRdw&4iR

added

New Note

Front

For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = n$, the pseudoinverse $A^\dagger$ is a left inverse of $A$, meaning \[ A^\dagger A = I \]Proof Included

Back

For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = n$, the pseudoinverse $A^\dagger$ is a left inverse of $A$, meaning \[ A^\dagger A = I \]Proof Included

Proof: Since $A$ has full column rank, $A^\top A$ invertible and then $A^\dagger A = ((A^\top A)^{-1} A^\top)A$ $= (A^\top A)^{-1} (A^\top A) = I$.

Field-by-field Comparison

Field	Before	After
Text		For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = n$, the pseudoinverse $A^\dagger$ is {{c1::a left inverse}} of $A$, meaning \[{{c1:: A^\dagger A = I }}\]<i>Proof Included</i>
Extra		<b>Proof: </b>Since $A$ has full column rank, $A^\top A$ invertible and then $A^\dagger A = ((A^\top A)^{-1} A^\top)A$ $= (A^\top A)^{-1} (A^\top A) = I$.

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::1._Definitions

Note 10: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: b-hsvze>Y9

added

New Note

Front

For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = n$, we define the pseudo-inverse $A^\dagger \in \mathbb{R}^{n \times m}$ as \[ A^\dagger = {{c1::(A^\top A)^{-1} A^\top }}\]

Back

For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = n$, we define the pseudo-inverse $A^\dagger \in \mathbb{R}^{n \times m}$ as \[ A^\dagger = {{c1::(A^\top A)^{-1} A^\top }}\]

Field-by-field Comparison

Field	Before	After
Text		For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = n$, we define the <b>pseudo-inverse</b> $A^\dagger \in \mathbb{R}^{n \times m}$ as \[ A^\dagger = {{c1::(A^\top A)^{-1} A^\top }}\]

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::1._Definitions

Note 11: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Classic
GUID: dle9WT5o|g

added

New Note

Front

What is the pseudoinverse in the case where $A \in \mathbb{R}^{n \times m}$ has independent rows?

Back

What is the pseudoinverse in the case where $A \in \mathbb{R}^{n \times m}$ has independent rows?

Because $rank(A) = r = m$ and thus $n \geq m$

$C(A)$spans $\mathbb{R}^m$ (columns span the space)
$R(A) \subseteq$ $\mathbb{R}^n$

There could be multiple $x \in \mathbb{R}^n$ that map to $T_A(x) = b$. We pick the one with the smallest norm $||x||^2$.
We know $x = x_r + x_n$ for $x_r \in R(A)$ and $x_n \in N(A)$ thus we pick $x = x_r + 0$ to get the smallest norm.

Field-by-field Comparison

Field	Before	After
Front		What is the pseudoinverse in the case where $A \in \mathbb{R}^{n \times m}$ has independent rows?
Back		Because $rank(A) = r = m$ and thus $n \geq m$<ul><li>$C(A)$spans $\mathbb{R}^m$ (columns span the space)</li><li>$R(A) \subseteq$ $\mathbb{R}^n$</li></ul>There could be multiple $x \in \mathbb{R}^n$ that map to $T_A(x) = b$. We pick the one with the smallest norm $\|\|x\|\|^2$.<br>We know $x = x_r + x_n$ for $x_r \in R(A)$ and $x_n \in N(A)$ thus we pick $x = x_r + 0$ to get the smallest norm.<br><br><div>  <img src="paste-4707a6f9abbe720721f1a4ab781ab8c8fda3c76a.jpg"></div>

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::0._Prelude

Note 12: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Classic
GUID: f?EhyxxJ04

added

New Note

Front

Rewrite $A^\dagger = R^\dagger C^\dagger$ in terms of A, R, C:

Back

Rewrite $A^\dagger = R^\dagger C^\dagger$ in terms of A, R, C:

$A^\dagger = R^\top {(RR^\top)}^{-1} {(C^\top C)}^{-1} C^\top =$ $R^\top {(C^\top C R R^\top)}^{-1} C^\top =$ $R^\top {(C^\top A R^\top)}^{-1}C^\top$.

Field-by-field Comparison

Field	Before	After
Front		Rewrite $A^\dagger = R^\dagger C^\dagger$ in terms of A, R, C:
Back		$A^\dagger = R^\top {(RR^\top)}^{-1} {(C^\top C)}^{-1} C^\top =$ $R^\top {(C^\top C R R^\top)}^{-1} C^\top =$ $R^\top {(C^\top A R^\top)}^{-1}C^\top$.

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::1._Definitions

Note 13: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: oX/3/jm{Ku

added

New Note

Front

For a full row rank matrix $A$, the unique solution to\[{{c1:: \min_{x \in \mathbb{R}^n} ||x||^2 \text{ s.t. } Ax = b}}\] is given by the vector $\hat{x} = A^\dagger b$. This $\hat{x}$ is in $C(A^\top)$. Proof Included

Back

For a full row rank matrix $A$, the unique solution to\[{{c1:: \min_{x \in \mathbb{R}^n} ||x||^2 \text{ s.t. } Ax = b}}\] is given by the vector $\hat{x} = A^\dagger b$. This $\hat{x}$ is in $C(A^\top)$. Proof Included

Proof

By Lemma 6.4.5 we only need to show that $\hat{x} = A^\dagger b$ satisfies $A \hat{x} = b$ and that $\hat{x} \in C(A^\top)$.

$A\hat{x} = AA^\dagger b = AA^\top (AA^\top)^{-1}b = b$
$\hat{x} = A^\dagger b = A^\top ((AA^\top)^{-1} b) = A^\top y$ for some $y$ thus $x \in C(A^\top)$.

Field-by-field Comparison

Field	Before	After
Text		<div>For a <b>full row rank</b> matrix $A$, the unique solution to\[{{c1:: \min_{x \in \mathbb{R}^n} \|\|x\|\|^2 \text{ s.t. } Ax = b}}\] is given by the vector $\hat{x} = A^\dagger b$. This $\hat{x}$ is in {{c1:: $C(A^\top)$}}. <i>Proof Included</i></div>
Extra		<div><strong>Proof</strong> </div><div>By Lemma 6.4.5 we only need to show that $\hat{x} = A^\dagger b$ satisfies $A \hat{x} = b$ and that $\hat{x} \in C(A^\top)$.</div><div><ul><li>$A\hat{x} = AA^\dagger b = AA^\top (AA^\top)^{-1}b = b$ </li><li>$\hat{x} = A^\dagger b = A^\top ((AA^\top)^{-1} b) = A^\top y$ for some $y$ thus $x \in C(A^\top)$.</li></ul></div>

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::1._Definitions

Note 14: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: os3Z8p7:x,

added

New Note

Front

We can compute the pseudoinverse from the any full rank (not just CR) factorisation of $A$.

Back

We can compute the pseudoinverse from the any full rank (not just CR) factorisation of $A$.

Note to Lorenz: Leave the "the" in, it's for maximum confusion .

Field-by-field Comparison

Field	Before	After
Text		We can compute the pseudoinverse from the {{c1:: any full rank (not just CR)}} factorisation of $A$.
Extra		<i>Note to Lorenz</i>: Leave the "<i>the</i>" in, it's for maximum confusion .

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::1._Definitions

Note 15: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Classic
GUID: p:TgRlDdKr

added

New Note

Front

What is the pseudoinverse in the case where $A \in \mathbb{R}^{n \times m}$ has independent columns?

Back

What is the pseudoinverse in the case where $A \in \mathbb{R}^{n \times m}$ has independent columns?

Because $rank(A) = r = n$ and thus $m \geq n$

$R(A)$ spans $\mathbb{R}^n$(rows span the space)
$C(A) \subseteq$ $\mathbb{R}^m$ (as $A$ is not necessarily square)

We therefore first project into $b$ into $C(A)$ and then invert, which is Least Squares

Field-by-field Comparison

Field	Before	After
Front		What is the pseudoinverse in the case where $A \in \mathbb{R}^{n \times m}$ has independent columns?
Back		Because $rank(A) = r = n$ and thus $m \geq n$<br><ul><li>$R(A)$ spans $\mathbb{R}^n$(rows span the space)</li><li>$C(A) \subseteq$ $\mathbb{R}^m$ (as $A$ is not necessarily square)</li></ul><div>We therefore first project  into $b$ into $C(A)$ and then invert, which is <b>Least Squares</b></div><br><div>  <img src="paste-455009459e5a5c70fa5574bdbcedcfb838341523.jpg"></div>

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::0._Prelude

Note 16: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: tjub=34aze

added

New Note

Front

$AA^\dagger$ is symmetric

Back

$AA^\dagger$ is symmetric

Field-by-field Comparison

Field	Before	After
Text		$AA^\dagger$ is{{c1:: symmetric :: property }}

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::2._Properties

Note 17: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: ve7Q_/^p1y

added

New Note

Front

For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = m$, we define the pseudo-inverse $A^\dagger \in \mathbb{R}^{n \times m}$ as \[ A^\dagger = {{c1::A^\top (A A^\top)^{-1} }}\]

Back

For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = m$, we define the pseudo-inverse $A^\dagger \in \mathbb{R}^{n \times m}$ as \[ A^\dagger = {{c1::A^\top (A A^\top)^{-1} }}\]

For an $A$ with full column-rank, we basically define, $A^\dagger$ as the transpose of the pseudoinverse of the transpose:

Field-by-field Comparison

Field	Before	After
Text		For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = m$, we define the <b>pseudo-inverse</b> $A^\dagger \in \mathbb{R}^{n \times m}$ as \[ A^\dagger = {{c1::A^\top (A A^\top)^{-1} }}\]
Extra		For an $A$ with full column-rank, we basically define, $A^\dagger$ as the transpose of the pseudoinverse of the transpose:<br><img src="paste-ea3dc98b302c74b79fb2bafc8b144f36da289e16.jpg"><br><br>

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::1._Definitions

Note 18: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: v~twgkx^)U

added

New Note

Front

For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = r$ and CR decomposition $A = CR$, we define the pseudoinverse $A^\dagger$ as \[ A^\dagger = R^\dagger C^\dagger \]

Back

For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = r$ and CR decomposition $A = CR$, we define the pseudoinverse $A^\dagger$ as \[ A^\dagger = R^\dagger C^\dagger \]

We can rewrite this as $A^\dagger = R^\top {(RR^\top)}^{-1} {(C^\top C)}^{-1} C^\top =$ $R^\top {(C^\top C R R^\top)}^{-1} C^\top =$ $R^\top {(C^\top A R^\top)}^{-1}C^\top$.

Field-by-field Comparison

Field	Before	After
Text		For $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = r$ and CR decomposition $A = CR$, we define the pseudoinverse $A^\dagger$ as \[ A^\dagger = {{c1::R^\dagger C^\dagger }}\]<br>
Extra		We can rewrite this as $A^\dagger = R^\top {(RR^\top)}^{-1} {(C^\top C)}^{-1} C^\top =$ $R^\top {(C^\top C R R^\top)}^{-1} C^\top =$ $R^\top {(C^\top A R^\top)}^{-1}C^\top$.

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::1._Definitions

Note 19: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: x|?8Eu84o6

added

New Note

Front

$AA^\dagger$ is the projection matrix on $C(A)$.

Back

$AA^\dagger$ is the projection matrix on $C(A)$.

Field-by-field Comparison

Field	Before	After
Text		$AA^\dagger$ is {{c1:: the projection matrix on $C(A)$}}.

Tags: ETH::1._Semester::LinAlg::6._Applications_of_orthogonality_and_projections::4._The_Pseudoinverse,_also_known_as_Moore-Penrose_Inverse::2._Properties

Field	Before	After
Text		For \(A \in \mathbb{R}^{m \times n}\) with \(\text{rank}(A) = m\), the <b>pseudo-inverse</b> \(A^\dagger \in \mathbb{R}^{n \times m}\) is {{c1::a right inverse}} of \(A\): \[ {{c1:: A A^\dagger = I }}\]<i>Proof Included</i>
Extra		<div><b>Proof</b> Since \(A^\top\) has full column rank, \(((A^\top)^\top A^\top) = AA^\top\) is invertible: \(AA^\dagger = AA^\top(A A^\top)^{-1} = I\).</div>

Field	Before	After
Front		Why is the pseudoinverse (for \(A\) with full row-rank) \(A^\top (AA^\top)^{-1}\)?
Back		It uses the multiplication by \(A^\top\) to choose an \(\hat{x}\) that lies in the row-space, thus minimising the norm.

Field	Before	After
Text		The <b>nullspace of </b>\(A\) \(N(A) \) is equal to {{c1:: the nullspace of \(N(A^\dagger)\)}}. <i>Proof Included</i>
Extra		Intuitively this holds as \(A^\dagger\) projects onto the \(C(A)\).<br><br>Thus anything in \(C(A)^\bot = N(A^\top)\) is projected to \(0\). In other words, \(\forall x \in C(A^\top)^\bot = N(A^\top)\) we have \(A^\dagger x = 0\).<br>We conclude that \(N(A) = C(A^\top)^\bot = N(A^\dagger)\).

Field	Before	After
Front		Rewrite \(A^\dagger = R^\dagger C^\dagger\) in terms of A, R, C:
Back		\(A^\dagger = R^\top {(RR^\top)}^{-1} {(C^\top C)}^{-1} C^\top =\) \(R^\top {(C^\top C R R^\top)}^{-1} C^\top =\) \(R^\top {(C^\top A R^\top)}^{-1}C^\top\).

Field	Before	After
Text		<div>For a <b>full row rank</b> matrix \(A\), the unique solution to\[{{c1:: \min_{x \in \mathbb{R}^n} \|\|x\|\|^2 \text{ s.t. } Ax = b}}\] is given by the vector \(\hat{x} = A^\dagger b\). This \(\hat{x}\) is in {{c1:: \(C(A^\top)\)}}. <i>Proof Included</i></div>
Extra		<div><strong>Proof</strong> </div><div>By Lemma 6.4.5 we only need to show that \(\hat{x} = A^\dagger b\) satisfies \(A \hat{x} = b\) and that \(\hat{x} \in C(A^\top)\).</div><div><ul><li>\(A\hat{x} = AA^\dagger b = AA^\top (AA^\top)^{-1}b = b\) </li><li>\(\hat{x} = A^\dagger b = A^\top ((AA^\top)^{-1} b) = A^\top y\) for some \(y\) thus \(x \in C(A^\top)\).</li></ul></div>

Note 1: ETH::LinAlg

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Note 2: ETH::LinAlg

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Note 3: ETH::LinAlg

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Note 6: ETH::LinAlg

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Note 7: ETH::LinAlg

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Note 8: ETH::LinAlg

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Note 9: ETH::LinAlg

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Note 10: ETH::LinAlg

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Note 11: ETH::LinAlg

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Note 12: ETH::LinAlg

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Note 13: ETH::LinAlg

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Note 14: ETH::LinAlg

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Note 15: ETH::LinAlg

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Note 16: ETH::LinAlg

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