Let $A$ be a real $7\times 3$ matrix such that its null space is spanned by the vectors
\[\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, \begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}, \text{ and } \begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix}.\]
Then find the rank of the matrix $A$.

(Purdue University, Linear Algebra Final Exam Problem)

We first determine the nullity of $A$ and deduce the rank of $A$ by the rank-nullity theorem.
The null space $\calN(A)$ of the matrix $A$ is spanned by
\[\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, \begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}, \text{ and } \begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix}.\]

Let us find a basis of the null space $\calN(A)$ among these vectors.
We use the “leading 1 method”.
Form the matrix whose column vectors are these three vectors and we apply elementary row operations as follows.
\begin{align*}
\begin{bmatrix}
1 & 2 & 1 \\
2 &1 &-1 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_2-2R_1}
\begin{bmatrix}
1 & 2 & 1 \\
0 &-3 & -3 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{-\frac{1}{3}R_2}\\[10pt]
\begin{bmatrix}
1 & 2 & 1 \\
0 &1 & 1 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_1-2R_2}
\begin{bmatrix}
1 & 0 & -1 \\
0 &1 & 1 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form and the first and second column contain the leading 1’s.

Therefore, the first two vectors
\[\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, \begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}\]
form a basis of the null space $\calN(A)$.
Hence the nullity, which is the dimension of $\calN(A)$, is $2$.

Since the size of the matrix $A$ is $7 \times 3$, the rank-nullity theorem gives
\[3=\text{nullity of } A + \text{rank of } A.\]
Thus, the rank of $A$ is $1$.

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The dimension of the nullspace of $A$ is called the nullity of $A$.
Prove the followings.
(a) $\calN(A)=\calN(A^{\trans}A)$.
(b) $\rk(A)=\rk(A^{\trans}A)$.
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For part (b), […]

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(b) Find a basis for the null space of $A$.
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Find a basis for the range $\calR(A)$ of $A$ that consists of columns of $A$.
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A hyperplane in $n$-dimensional vector space $\R^n$ is defined to be the set of vectors
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x_2 \\
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x_n
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0 &1 &0
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(b) Find the rank of $A$.
(c) Find an orthonormal basis of the row space of $A$.
(The Ohio State University, Linear Algebra Exam […]

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x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \,\right) = \begin{bmatrix}
x_1+2x_2+3x_3-x_4 \\
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x_1+x_2+2x_3
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3 &6 &4
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-3 \\
1 \\
1
\end{bmatrix}, \qquad \mathbf{b}=\begin{bmatrix}
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1 \\
0
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