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.

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 $n\times n$.)

Hyperplane in $n$-Dimensional Space Through Origin is a Subspace
A hyperplane in $n$-dimensional vector space $\R^n$ is defined to be the set of vectors
\[\begin{bmatrix}
x_1 \\
x_2 \\
\vdots \\
x_n
\end{bmatrix}\in \R^n\]
satisfying the linear equation of the form
\[a_1x_1+a_2x_2+\cdots+a_nx_n=b,\]
[…]

Idempotent Matrices are Diagonalizable
Let $A$ be an $n\times n$ idempotent matrix, that is, $A^2=A$. Then prove that $A$ is diagonalizable.
We give three proofs of this problem. The first one proves that $\R^n$ is a direct sum of eigenspaces of $A$, hence $A$ is diagonalizable.
The second proof proves […]

Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$
Let $A$ be an $m \times n$ matrix and $B$ be an $n \times l$ matrix. Then prove the followings.
(a) $\rk(AB) \leq \rk(A)$.
(b) If the matrix $B$ is nonsingular, then $\rk(AB)=\rk(A)$.
Hint.
The rank of an $m \times n$ matrix $M$ is the dimension of the range […]

Column Rank = Row Rank. (The Rank of a Matrix is the Same as the Rank of its Transpose)
Let $A$ be an $m\times n$ matrix. Prove that the rank of $A$ is the same as the rank of the transpose matrix $A^{\trans}$.
Hint.
Recall that the rank of a matrix $A$ is the dimension of the range of $A$.
The range of $A$ is spanned by the column vectors of the matrix […]

Given a Spanning Set of the Null Space of a Matrix, Find the Rank
Let $A$ be a real $7\times 3$ matrix such that its null space is spanned by the vectors
\[\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, \begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}, \text{ and } \begin{bmatrix}
1 \\
-1 \\
0
[…]

A Matrix Representation of a Linear Transformation and Related Subspaces
Let $T:\R^4 \to \R^3$ be a linear transformation defined by
\[ T\left (\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \,\right) = \begin{bmatrix}
x_1+2x_2+3x_3-x_4 \\
3x_1+5x_2+8x_3-2x_4 \\
x_1+x_2+2x_3
\end{bmatrix}.\]
(a) Find a matrix $A$ such that […]