If Two Vectors Satisfy $A\mathbf{x}=0$ then Find Another Solution

Stanford University Linear Algebra Exam Problems and Solutions

Problem 395

Suppose that the vectors
\[\mathbf{v}_1=\begin{bmatrix}
-2 \\
1 \\
0 \\
0 \\
0
\end{bmatrix}, \qquad \mathbf{v}_2=\begin{bmatrix}
-4 \\
0 \\
-3 \\
-2 \\
1
\end{bmatrix}\] are a basis vectors for the null space of a $4\times 5$ matrix $A$. Find a vector $\mathbf{x}$ such that
\[\mathbf{x}\neq0, \quad \mathbf{x}\neq \mathbf{v}_1, \quad \mathbf{x}\neq \mathbf{v}_2,\] and
\[A\mathbf{x}=\mathbf{0}.\]

(Stanford University, Linear Algebra exam problem)
 
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Solution.

We are asked to find a vector $\mathbf{x}$ in the null space of $A$, which are not $\mathbf{0}, \mathbf{v}_1, \mathbf{v}_2$.

Recall that the null space is a vector space. Thus, any linear combination of vectors in the null space is still in the null space.

Since $\mathbf{v}_1, \mathbf{v}_2$ are basis of the null space of $A$, they are in particular vectors in the null space of $A$.
Thus, for example,
\[\mathbf{x}=2\mathbf{v}_2=\begin{bmatrix}
-4 \\
1 \\
0 \\
0 \\
0
\end{bmatrix}\] is an element of the null space and it is not equal to $\mathbf{0}, \mathbf{v}_1, \mathbf{v}_2$.

Another example is
\[\mathbf{x}=\mathbf{v}_1+\mathbf{v_2}=\begin{bmatrix}
-6 \\
1 \\
-3 \\
-2 \\
1
\end{bmatrix}.\]

In general, you can prove that any vector of the form
\[\mathbf{x}=c_1\mathbf{v}_1+c_2\mathbf{v}_2,\] where $c_1, c_2$ are scalars such that $(c_1, c_2)\neq (0,0), (1,0), (0, 1)$, satisfied the required conditions.


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