Let A\in \mathbb{R}^{m\times n} (the results also holds for \mathbb{C}^{m\times n}). Then, the **column rank**/**row rank** of A is defined as the dimension of the column/row space of A, i.e. the dimension of the vector space spanned by the columns/rows of A; this is then equivalent to the number of linearly independent columns/rows (column/rows vector) of A.

**Theorem:** column rank of A = row rank of A.

**Definition:** the rank of a matrix, **rank(****A****)**, is the dimension of either the column or row space of A; simply the number of linearly independent columns or rows of A.

**Definition:** for a linear map f: \mathbb{R}^n\longrightarrow \mathbb{R}^m, the rank of the linear map is defined as the dimension of the image of f. This definition is equivalent to the definition of the matrix rank as every linear map f:\mathbb{R}^n\longrightarrow \mathbb{R}^m has a matrix A\in \mathbb{R}^{m\times n} by which it can be written as f(x)=Ax.

**Proposition**: \text{rank}(A)\le \min(m,n). This leads to these definitions: A matrix A is said to be **full rank** iff \text{rank}(A)= \min(m,n), i.e. the largest possible rank, and it is said to be **rank deficient** iff \text{rank}(A)\lt \min(m,n), i.e. not having full rank.

## Properties of rank

For A\in \mathbb{R}^{m\times n}:

**1-** only a zero matrix has rank zero.

**2-** If B\in \mathbb{R}^{n\times k}, then \text{rank}(AB)\le \min( \text{rank}(A) , \text{rank}(B))

**3-** \text{rank}(A+B)\le \text{rank}(A)+ \text{rank}(B)

**4-** \text{rank}(A)=\text{rank}(A^{\text T})

**5-** \text{rank}(A)= \text{rank}(A^{\text T}) =\text{rank}(AA^{\text T})= \text{rank}(A^{\text T}A)

**6-** If v\in \mathbb{R}^{r\times 1}, then for V=vv^\text T \in \mathbb{R}^{r\times r}, \text{rank}(V)=1. In addition, for v_i \in \mathbb{R}^{r\times 1} , \text{rank}(S=\sum_{i=1}^{n} v_iv_i^\text T)\le n, i.e. S has at most rank *n*.

**7-** A square matrix C\in \mathbb{R}^{n\times n} can be decomposed as C=U\Gamma U^{-1} where \Gamma is a diagonal matrix containing the eigenvalues of C. Then, \text{rank}(C)=\text{rank}(\Gamma), i.e. the number of non-zero eigenvalues of C.

**8-** For a square matrix C\in \mathbb{R}^{n\times n} , then equivalently C is full rank, is invertible, has non-zero determinant, and has *n* non-zero eigenvalues.

## Proofs

**P6:**

where v_i\in \mathbb{R} are coordinates of the vector v. This indicates that each column of V is a scalar multiple of any other columns of V; therefore, the column space is one dimensional. Hence, \text{rank}(V)=1.

For the second part, property 3 proves the statement.