Find a Basis of the Subspace of All Vectors that are Perpendicular to the Columns of the Matrix

Math exam problems and solutions at Harvard University

Problem 40

Find a basis for the subspace $W$ of all vectors in $\R^4$ which are perpendicular to the columns of the matrix
\[A=\begin{bmatrix}
11 & 12 & 13 & 14 \\
21 &22 & 23 & 24 \\
31 & 32 & 33 & 34 \\
41 & 42 & 43 & 44
\end{bmatrix}.\]

(Harvard University exam)

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Hint.

  1. Show that $W=\ker(A^{\trans})$.
  2. Find a basis of $\ker(A^{\trans})$ by reducing the matrix $A^{\trans}$.

The kernel of $A$ is also called the null space of $A$ and it is denoted by $\calN(A)$.
So $\ker(A)=\calN(A)$.

Solution.

Let us write $A=[A_1 \, A_2 \, A_3 \, A_4]$, where $A_i$ is the $i$-th column vector of $A$ for $i=1,2,3,4$.
First we claim that a vector $x\in \R^4$ is perpendicular to all column vectors $A_i$ if and only if $x\in \ker(A^{\trans})$.
To see this, we compute
\begin{align*}
A^{\trans}x=\begin{bmatrix}
A_1^{\trans} \\
A_2^{\trans} \\
A_3^{\trans} \\
A_4^{\trans}
\end{bmatrix}x
=\begin{bmatrix}
A_1^{\trans}x \\
A_2^{\trans} x\\
A_3^{\trans} x\\
A_4^{\trans} x
\end{bmatrix}.
\end{align*}
From this equality the claim follows immediately.

So we proved that $\ker(A^{\trans}) =W$. From this, we see that $W$ is actually a subspace in $\R^4$.
Thus, we need to find a basis for the kernel of the transpose $A^{\trans}$.

We apply elementary row operations to $A^{\trans}$ and obtain a reduced row echelon form
\[A^{\trans} \to \begin{bmatrix}
1 & 0 & -1 & -2 \\
0 &1 & 2 & 3 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}.\] The last two columns correspond to two free variables. Let $s$ and $t$ be free variable.
Then $x=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \in \ker(A^{\trans})$ if and only if $x$ satisfies
\begin{align*}
x_1 &=s+2t \\
x_2 &=-2s-3t\\
x_3 &=s\\
x_4 &=t,
\end{align*}
equivalently
\begin{align*}
x=s\begin{bmatrix}
1 \\
-2 \\
1 \\
0
\end{bmatrix}
+t\begin{bmatrix}
2 \\
-3 \\
0 \\
1
\end{bmatrix}.
\end{align*}
Therefore a basis of $W=\ker(A^{\trans})$ is
\[ \begin{bmatrix}
1 \\
-2 \\
1 \\
0
\end{bmatrix} \text{ and } \begin{bmatrix}
2 \\
-3 \\
0 \\
1
\end{bmatrix}.\]


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