# 4 fundamental subspaces

Our\(4\)

fundamental subspaces are,\((C(A))\)

\((N(A))\)

**Row Space**and

**Left Null Space**.

## Row Space

Say we have a\(m\times n\)

matrix \(A\)

.**Row Space**of

\(A\)

is all the **linear combination**of

**rows**of matrix

\(A\)

.We can also say that

**Row space**of

\(A\)

is all the **linear combination**of

**columns**of matrix

\(A^T\)

.SoRow spaceof matrix\(A\)is the Column Space of matrix\(A^T\).Row spaceis\(C(A^T)\)

## Left Null Space

Say we have a\(m\times n\)

matrix \(A\)

.Then

**Left Null Space**is the Null Space of

\(A^T\)

, so, For a matrix\(A\)theLeft Null spaceis the space of all\(\vec{x}\)that solves\(A^T\vec{x}=\vec{0}\)

# Dimensions of fundamental spaces

## Dimensions of Column space

Say we have a\(m\times n\)

matrix \(A\)

and \(\text{Rank}(A)=r\)

.Here we have

\(n\)

columns and each column have \(m\)

components.So everycolumn vectorof matrix\(A\)\(\in\mathbb{R}^m\).

So theColumn spaceof matrix\(A\)lives in\(\mathbb{R}^m\).

\(\text{Rank}(A)=r\)

so we know that we have \(r\)

**independent columns**in matrix

\(A\)

.TheColumn spaceof\(A\)is spanned by these\(r\)independent columns vectorsof matrix\(A\).

So then dimension of thecolumn spaceis\(r\).\[\text{dim }C(A)=r\]

## Dimensions of Null space

Say we have a\(m\times n\)

matrix \(A\)

and \(\text{Rank}(A)=r\)

.For matrix

\(A\)

the **Null space**is the space of all

\(\vec{x}\)

that solves \(A\vec{x}=\vec{0}\)

.We know that\(A\)is a\(m\times n\)matrix, so\(\vec{x}\in\mathbb{R}^n\).

So theNull spaceof matrix\(A\)lives in\(\mathbb{R}^n\).

\(\text{Rank}(A)=r\)

so we know that we have \(r\)

**independent columns**in matrix

\(A\)

.So there are

\(n-r\)

**dependent column vector**(special solution) in matrix

\(A\)

.TheNull spaceis spanned by these\(n-r\)dependent column vectorof matrix\(A\).

So then dimension of thenull spaceof matrix\(A\)is\(n-r\).\[\text{dim }N(A)=n-r\]

## Dimensions of Row space

Say we have a\(m\times n\)

matrix \(A\)

and \(\text{Rank}(A)=r\)

.Here we have

\(m\)

rows and each row have \(n\)

components.So everyrow vectorof matrix\(A\)\(\in\mathbb{R}^n\).

So theRow spaceof matrix\(A\)lives in\(\mathbb{R}^n\).

\(\text{Rank}(A)=\text{Rank}(A^T)\)

[proof]\(\text{Rank}(A)=r\Rightarrow\)

\(\text{Rank}(A^T)=r\)

so we know that we have \(r\)

**independent columns**in matrix

\(A^T\)

.OR say that we have

\(r\)

**independent rows**in matrix

\(A\)

.TheRow spaceof\(A\)is spanned by these\(r\)independent row vectorsof\(A\).ORsay that, theColumn spaceof\(A^T\)is spanned by these\(r\)independent columns vectorsof\(A^T\).

So then dimension of therow spaceof matrix\(A\)is\(r\).\[\text{dim }C(A^T)=r\]

## Dimensions of Left Null space

Say we have a\(m\times n\)

matrix \(A\)

and \(\text{Rank}(A)=r\)

.For a matrix

\(A\)

the **Left Null space**is the space of all

\(\vec{x}\)

that solves \(A^T\vec{x}=\vec{0}\)

.We know that\(A^T\)is a\(n\times m\)matrix, so\(\vec{x}\in\mathbb{R}^m\).

So theLeft Null spaceof matrix\(A\)lives in\(\mathbb{R}^m\).

\(\text{Rank}(A)=\text{Rank}(A^T)\)

[proof]\(\text{Rank}(A)=r\Rightarrow\)

\(\text{Rank}(A^T)=r\)

so we know that we have \(r\)

**independent columns**in matrix

\(A^T\)

.OR say that we have

\(r\)

**independent rows**in matrix

\(A\)

.So there are\(m-r\)dependent column vector(special solution) in matrix\(A^T\).

TheLeft Null spaceis spanned by these\(m-r\)dependent column vectorin matrix\(A\).

So then dimension of theleft null spaceis\(m-r\).\[\text{dim }N(A^T)=m-r\]

# Basis of fundamental spaces

## Basis of Column space

Say we have a\(m\times n\)

matrix \(A\)

and \(\text{Rank}(A)=r\)

.Here we haveWe talked about\(r\)independent columnsin\(A\).

And theseindependent columnsare the basis ofColumn space.

**Basis of Column space**[HERE]

## Basis of Null space

Say we have a\(m\times n\)

matrix \(A\)

and \(\text{Rank}(A)=r\)

.Here we haveWe talked about\(n-r\)special solutions.

And thesespecial solutionsare the basis ofNull space.

**Basis of Null space**[HERE]

## Basis of Row space

Say we have a\(m\times n\)

matrix \(A\)

and \(\text{Rank}(A)=r\)

.\(\text{Rank}(A)=r\Rightarrow\)\(\text{Rank}(A^T)=r\)so here we have\(r\)independent columnsin\(A^T\).

So we can also say that there are\(r\)independent rowsin\(A\).

And theseindependent rowsof\(A\)are the basis ofRow space.

**Example,**

- Say \(A=\begin{bmatrix} 1 & 2 & 3 & 1\\ 1 & 1 & 2 & 1\\ 1 & 2 & 3 & 1\\ \end{bmatrix}\)

It's**reduced row echelon form**is,\(R=\begin{bmatrix} \color{blue}{\fbox{1}} & \color{blue}{0} & \color{red}{1} & \color{red}{1} \\ \color{blue}{0} & \color{blue}{\fbox{1}} & \color{red}{1} & \color{red}{0} \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}\)

Here we can clearly see that rank\((r)\)is\(2\).Our row operation preserve the

**row space**.

It mean our row space is unaffected during row operations while reducing it to reduced row echelon form.

Because we are just taking the**linear combinations**of rows and linear combinations of two vectors in row space remains in row space.

But column space might changed.

Here the basis for**Row space**is first\(r\)row vectors in\(R\).

So our basis are\(\begin{bmatrix} 1\\0\\1\\1\\ \end{bmatrix}\),\(\begin{bmatrix} 0\\1\\1\\0\\ \end{bmatrix}\)

## Basis of Left Null space

Say we have a\(m\times n\)

matrix \(A\)

and \(\text{Rank}(A)=r\)

.For matrix

\(A\)

the **Left Null space**is the space of all

\(\vec{x}\)

that solves \(A^T\vec{x}=\vec{0}\)

.When we reduce

\(A\)

in **reduced row echelon form**, then

**# zero**rows are

\(m-r\)

.So we have

\(m-r\)

**special solutions**for our

**Left null space**.

And these

**special solutions**are the basis of

**Left Null space**.

To find the basis we

**can**take the transpose of

\(A\)

and find the null space of \(A^T\)

.But this is not intuitive for

**row space**perspective.

So what we do is while reducing the matrix

\(A\)

to the **reduced row echelon form**

\((R)\)

, by using Gauss-Jordan elimination technique we can keep the track of our action.And it will result in a matrix say

\(E\)

.\(EA=R\)

then last

\(m-r\)

rows of \(E\)

will be our **basis**for

**Left null space**.

**Example,**

- Say \(A=\begin{bmatrix} 1 & 2 & 3 & 1\\ 1 & 1 & 2 & 1\\ 1 & 2 & 3 & 1\\ \end{bmatrix}\)

Now let's find it's**reduced row echelon form**Gauss Jordan,\(\left[ \begin{array}{cccc|ccc} 1 & 2 & 3 & 1 & 1 & 0 & 0 \\ 1 & 1 & 2 & 1 & 0 & 1 & 0 \\ 1 & 2 & 3 & 1 & 0 & 0 & 1 \\ \end{array} \right] \)\(R_2 \leftarrow R_2 - R_1\)and\(R_3 \leftarrow R_3 - R_1\)\(\left[ \begin{array}{cccc|ccc} \color{blue}{\fbox{1}} & 2 & 3 & 1 & 1 & 0 & 0 \\ 0 & -1 & -1 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 & 1 \\ \end{array} \right] \)\(R_2 \leftarrow -R_2 \)\(\left[ \begin{array}{cccc|ccc} \color{blue}{\fbox{1}} & 2 & 3 & 1 & 1 & 0 & 0 \\ 0 & \color{blue}{\fbox{1}} & 1 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 & 1 \\ \end{array} \right] \)\(R_1 \leftarrow R_1 -R_2\)\(\left[ \begin{array}{cccc|ccc} \color{blue}{\fbox{1}} & \color{blue}{0} & \color{red}{1} & \color{red}{1} & -1 & 2 & 0 \\ \color{blue}{0}& \color{blue}{\fbox{1}} & \color{red}{1} & \color{red}{0} & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & \color{green}{-1} & \color{green}{0} & \color{green}{1} \\ \end{array} \right] \)

What is special in\(\begin{bmatrix} \color{green}{-1}\\\color{green}{0}\\\color{green}{1}\\ \end{bmatrix}\)?

It's suggesting that if we take\(\color{green}{-1}\cdot R_1 + \color{green}{0}\cdot R_2 + \color{green}{1}\cdot R_3\)of\(A\)then we will get\(\vec{0}\).

So our**left null space**is,\[c\cdot \begin{bmatrix} \color{green}{-1}\\\color{green}{0}\\\color{green}{1}\\ \end{bmatrix}; \quad c\in\mathbb{R}\]