# Rank of matrices

Consider a\(m\times n\)

matrix \(A\)

of rank \(r\)

.So we have

\(n\)

column vectors in \(\mathbb{R}^m\)

, and there are \((n-r)\)

column vectors which we can get by the linear combination of other \(r\)

column vectors.We describe rank as number of pivots of a matrix.

#pivots can't be\(\gt\)#rows (\(m\)), so\(r\leq m\)#pivots can't be\(\gt\)#columns (\(n\)), so\(r\leq n\)

## ▶ Full column rank matrix

- Consider a \(m\times n\)matrix\(A\)of rank\(r\)and\(m\gt n\).

Here\(\text{Rank}(A)=r=n\)it means that all of the column vectors are independent,**#**pivots\(=n\)\(\Rightarrow\)**#**free column vectors\(=0\).

We have no free vectors here so**Can we always get a solution, can we always find**\(x\)such that\(A\vec{x}=\vec{b}\)for every\(\vec{b}\)?

Here in\(A\)we have\(n\)independent vectors in\(\mathbb{R}^m\)and\(m\gt n\), to fill a\(m\)dimensional space we alteast need\(m\)independent vectors, but we had\(n\)independent vectors and\(n\lt m\)so we don't have enough vectors to fill\(\mathbb{R}^m\).

Linear combination of\(n\)independent vectors can't give us a space in\(\mathbb{R}^m\)

So we can't get every vector\(\vec{b}\)by the linear combinations of those independent vectors.

So answer is**No**we can't get every\(\vec{b}\).**Null Space**of\(A\)is just\(\vec{0}\), so\(A\vec{x}=\vec{0}\)has only one solution that is\(\vec{x}=\vec{0}\)

Complete solution of\(A\vec{x}=\vec{b}\)is\(\vec{x}=\vec{x}_p+\vec{x}_n=\vec{x}_p+\vec{0}=\vec{x}_p\)So complete solution is just a single vector

\(\vec{x}_p\)**if it exists**.**#**solution =\(\left\{\begin{matrix} 1 & \text{ if there is a solution} \\ 0 & \text{ otherwise} \\ \end{matrix}\right.\)**Example:**\(A = \begin{bmatrix} 1 & 4\\ 2 & 3\\ 3 & 2\\ 4 & 1\\ \end{bmatrix}\)

row reduced echelon form is**Can we always get a solution, can we always find**\(x\)such that\(A\vec{x}=\vec{b}\)for every\(\vec{b}\)?

Here in\(A\)we have\(2\)vectors in\(\mathbb{R}^4\).

and linear combination of\(2\)vector can give us a space in\(\mathbb{R}^2\)so we can only get those\(\vec{b}\)which are in the linear combination(plane) of those two vectors.

possible\(\vec{b}\)is\(\vec{b}= \alpha\begin{bmatrix} 1\\2\\3\\4\\ \end{bmatrix} +\beta\begin{bmatrix} 4\\3\\2\\1\\ \end{bmatrix};\quad\)\(\alpha,\beta\in\mathbb{R}\)

So answer is**No**we can't get every\(\vec{b}\)\(\begin{bmatrix} \fbox{1} & 0 \\ 0 & \fbox{1} \\ 0 & 0 \\ 0 & 0 \\ \end{bmatrix} = \begin{bmatrix} I_{2\times2}\\0 \end{bmatrix} \)

So**#**of pivots are\(2\)

## ▶ Full row rank matrix

- Consider a \(m\times n\)matrix\(A\)of rank\(r\)and\(m\lt n\).

Here\(\text{Rank}(A)=r=m\)now all of the column vectors are not independent,**#**pivots\(=m\)\(\Rightarrow\)**#**free column vectors\(=n-r\).

Here we have**Can we always get a solution, can we always find**\(x\)such that\(A\vec{x}=\vec{b}\)for every\(\vec{b}\)?

Here in\(A\)we have\(m\)independent vectors in\(\mathbb{R}^m\), so we have enough vectors to fill\(\mathbb{R}^m\).

Linear combination of\(m\)independent vectors can give us a space in\(\mathbb{R}^m\)

So we can get every vector\(\vec{b}\)by the linear combinations of those independent vectors.

So answer is**Yes**we can get every\(\vec{b}\).\(n-r\)free vectors so**Null Space**of\(A\)is in\(\mathbb{R}^{n-r}\).**#**solutions are\(\infty\)**Example:**\(A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1 \\ \end{bmatrix}\)

row reduced echelon form is\(x\)such that\(A\vec{x}=\vec{b}\)for every\(\vec{b}\)?

Here in\(A\)we have\(4\)vectors in\(\mathbb{R}^2\)and out of those\(2\)vectors are independent.

And linear combination of\(2\)independent vector can give us a space in\(\mathbb{R}^2\). So we covered the whole\(2\)-dimensional space.

So answer is**Yes**we can get every\(\vec{b}\).\(\begin{bmatrix} \fbox{1} & 0 & -1 & -2 \\ 0 & \fbox{1} & 2 & 3 \\ \end{bmatrix} = \begin{bmatrix} I_{2\times2} & F_{2\times2} \end{bmatrix} \)

So**#**of pivots are\(2\)

## ▶ Full rank matrix

- Consider a \(m\times n\)matrix\(A\)of rank\(r\)and\(m = n\).

Here\(\text{Rank}(A)=r=n\)it means that all of the column vectors are independent,**#**pivots\(=n\)\(\Rightarrow\)**#**free column vectors\(=0\).

We have no free vectors here so\(x\)such that\(A\vec{x}=\vec{b}\)for every\(\vec{b}\)?

Here in\(A\)we have\(n\)independent vectors in\(\mathbb{R}^n\), so we have enough vectors to fill\(\mathbb{R}^n\).

Linear combination of\(n\)independent vectors can give us a space in\(\mathbb{R}^n\)

So we can get every vector\(\vec{b}\)by the linear combinations of those independent vectors.

So answer is**Yes**we can get every\(\vec{b}\).**Null Space**of\(A\)is just\(\vec{0}\), so\(A\vec{x}=\vec{0}\)has only one solution that is\(\vec{x}=\vec{0}\)**#**solutions =\(1\)

row reduced echelon form of\(A\)is\(I_{n\times n}\)

# Dimensions and Basis

Say we have a matrix\(A_{m\times n}\in\mathbb{R}^{m\times n}\)

## ▶ Column Space

- We know that the column space of matrix \(A\)lives in\(R^{m}\).

But what are the**basis**for that**column space?**

So**Basis**are those**(minimum)**vectors whose linear combination gives us the desired space.

for example, for space of\(R^2\), linear combinations of any\(2\)**non-parallel(independent)**vectors will give us\(R^2\).The

What is the**basis**for**column space**are those**(independent)pivot columns****Dimension**of**column space?**Number of dimensions are defined by it's

**basis vectors**so, Number of dimensions\(=\)number of pivot columns\(=\)rank\((r)\)so,**# Dimensions**\(= r\)

## ▶ Null Space

- Null space of matrix \(A\)also lives in\(R^{m}\).
Recall those

What are the**free variables**of matrix\(A\),**free variables**are the dependent column vectors of matrix\(A\), these free variables gives us the**special solutions**and linear combination of those**special solutions**gives us our**Null space.****basis**for that**Null space?**We discussed that

What is the**Null space**is the linear combination of those**special solutions**so,

The**basis**for**Null space**are those**special solutions.****Dimension**of**Null space?**Number of dimensions are defined by it's

**basis vectors**so, Number of dimensions\(=\)number of**special solution**and we discussed in**Null Space**section that, Number of**special solution**are\(n-r\).**# Dimensions**\(= n-r\)