# Rank and Nullity of a Matrix, Nullity of Transpose

## Problem 140

Let $A$ be an $m\times n$ matrix. The nullspace of $A$ is denoted by $\calN(A)$.
The dimension of the nullspace of $A$ is called the nullity of $A$.
Prove the followings.

(a) $\calN(A)=\calN(A^{\trans}A)$.

(b) $\rk(A)=\rk(A^{\trans}A)$.

## Hint.

For part (b), use the rank-nullity theorem and the result from (a).
The rank-nullity theorem says that for a $m\times n$ matrix,
$\text{rank of }A+\text{ nullity of }A=n.$

## Proof.

### (a)$\calN(A)=\calN(A^{\trans}A)$.

#### Show $\calN(A) \subset \calN(A^{\trans}A)$

Consider any $\mathbf{x} \in \calN(A)$. Then we have $A\mathbf{x}=\mathbf{0}$. Multiplying it by $A^{\trans}$ from the left, we obtain
$A^{\trans}A\mathbf{x}=A^{\trans}\mathbf{0}=\mathbf{0}.$ Thus $\mathbf{x} \in \calN(A^{\trans}A)$, and hence $\calN(A) \subset \calN(A^{\trans}A)$.

#### Show $\calN(A) \supset \calN(A^{\trans}A)$

On the other hand, let $\mathbf{x} \in \calN(A^{\trans}A)$. Thus we have
$A^{\trans}A\mathbf{x}=\mathbf{0}.$ Multiplying it by $\mathbf{x}^{\trans}$ from the left, we obtain
$\mathbf{x}^{\trans}A^{\trans}A\mathbf{x}=\mathbf{x}^{\trans}\mathbf{0}=\mathbf{0}.$ This implies that we have
$\mathbf{0}=(A\mathbf{x})^{\trans}(A\mathbf{x})=||A\mathbf{x}||^2$ and the length of the vector $A\mathbf{x}$ is zero, thus the vector $A\mathbf{x}=\mathbf{0}$. Hence $\mathbf{x} \in \calN(A)$, and we obtain $\calN(A) \supset \calN(A^{\trans}A)$.

### (b) $\rk(A)=\rk(A^{\trans}A)$

We use the rank-nullity theorem and obtain
\begin{align*}
\rk(A)=n-\dim(\calN(A))=n-\dim(\calN(A^{\trans}A))=\rk(A^{\trans}A).
\end{align*}
(Note that the size of the matrix $A^{\trans}A$ is $n\times n$.)

Let $A_1, A_2, \dots, A_m$ be $n\times n$ Hermitian matrices. Show that if $A_1^2+A_2^2+\cdots+A_m^2=\calO,$ where $\calO$ is the \$n \times...