Rank and Nullity

Definition

Let $A$ : Matrix

Then

• $n=\dim [R(A)]=\dim [C(A)]$
• We call $n$ the rank of $A$
• We denote $n=\mathrm{rank}(A)$
• We call $\dim [N(A)]$ the nullity of $A$

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Theorems

Let $A$ : $m\times n$ matrix

Rank-Nullity Theorem

$$\mathrm{rank}(A)+\mathrm{nullity}(A)=n$$

Row Echelon Form Theorem

Let $R$ : REF (or RREF) of $A$

Then

• $\mathrm{rank}(A)=\text{number of nonzero rows in }R$
• $\mathrm{rank}(A)$ equals the number of pivot positions in $A$.

Transpose Rank Theorem

$$\mathrm{rank}(A)=\mathrm{rank}(A^T)$$

Note

This follows from $\dim [R(A)]=\dim [C(A)]$.

Tranpose product rank theorem

$\mathrm{rank}(A^{T}A)=\mathrm{rank}(A)$

Proof

If $A\mathbf{x}=0$, then $A^{T}A\mathbf{x}=A^T(0)=0$

If $A^{T}A\mathbf{x}=0$, then ${\mathbf{x}}^T{A}^{T}A\mathbf{x}=(A\mathbf{x}{)}^T(A\mathbf{x})=|A\mathbf{x}{|}^2=0$, so $A\mathbf{x}=0$

Since $N(A^{T}A)=N(A)$, we have $\mathrm{nullity}(A^{T}A)=\mathrm{nullity}(A)$.

Both $A^{T}A$ and $A$ have $n$ columns, so by Rank-Nullity Theorem:

$\mathrm{rank}(A^{T}A)=n-\mathrm{nullity}(A^{T}A)=n-\mathrm{nullity}(A)=\mathrm{rank}(A)$

Product Rank Theorem

Let $B$ : $n\times p$ matrix

Then

$$\mathrm{rank}(AB)\le \min \{\mathrm{rank}(A),\mathrm{rank}(B)\}$$

Invertible Matrix Rank Theorem

Suppose $m=n$, i.e., $A$ be a square matrix of order $n$

Then $A$ is invertible if and only if $\mathrm{rank}(A)=n$.

Equivalently, $A$ is invertible if and only if $\mathrm{nullity}(A)=0$.

Full Rank Theorem

Let $A$ be an $m\times n$ matrix.

Then

• $A$ has full column rank if $\mathrm{rank}(A)=n$ (number of columns)
  • This occurs if and only if the columns of $A$ are linearly independent
  • This occurs if and only if $\mathrm{nullity}(A)=0$
• $A$ has full row rank if $\mathrm{rank}(A)=m$ (number of rows)
  • This occurs if and only if the rows of $A$ are linearly independent

Equality of Row and Column Space Dimensions Theorem

Let $A$ : Matrix

Then $\dim [R(A)]=\dim [C(A)]$

Theorems: Rank with eigenvalues and eigenvectors

Number of Nonzero Eigenvalues Theorem

If $A$ is diagonalizable

Then $\mathrm{rank}(A)=\text{number of nonzero eigenvalues (counting multiplicity)}$

Zero Eigenvalue and Rank Deficiency Theorem

$\lambda =0$ is an eigenvalue of $A$

If and only if $A$ is singular (rank-deficient)

Geometric Multiplicity and Nullity

The geometric multiplicity of $\lambda =0$ (dimension of its eigenspace) equals $\mathrm{nullity}(A)$.

Since the eigenspace for $\lambda =0$ is $\{\mathbf{x}:A\mathbf{x}=0\mathbf{x}\}=N(A)$:

$\dim (\text{eigenspace of }\lambda =0)=\dim [N(A)]=\mathrm{nullity}(A)$

Rank via Rank-Nullity Theorem

For an $n\times n$ matrix:

$\mathrm{rank}(A)=n-\mathrm{nullity}(A)=n-\text{geometric multiplicity of }\lambda =0$

Footnotes

1. Column space, Row space, Null space, Dimension ↩