Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space

Problem 131

Let $V$ be the following subspace of the $4$-dimensional vector space $\R^4$.
$V:=\left\{ \quad\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \in \R^4 \quad \middle| \quad x_1-x_2+x_3-x_4=0 \quad\right\}.$ Find a basis of the subspace $V$ and its dimension.

Solution.

Any vector $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$ in $V$ satisfies $x_1-x_2+x_3-x_4=0$, or equivalently $x_1=x_2-x_3+x_4$.

Thus we have
\begin{align*}
\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
&=\begin{bmatrix}
x_2-x_3+x_4 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}\\
&=x_2\begin{bmatrix}
1 \\
1 \\
0 \\
0
\end{bmatrix}
+x_3\begin{bmatrix}
-1 \\
0 \\
1 \\
0
\end{bmatrix}
+x_4\begin{bmatrix}
1 \\
0 \\
0 \\
1
\end{bmatrix}.
\end{align*}

Let
$\mathbf{u}_1=\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \mathbf{u}_2=\begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \mathbf{u}_3=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}.$ The above computation shows that any vector $\mathbf{x}$ in $V$ can be written as a linear combination of the vectors $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$.
Hence the set $\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3 \}$ is a spanning set for the subspace $V$.

We claim that $\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3 \}$ is a linearly independent set.
Consider
$a_1 \mathbf{u}_1+ a_2 \mathbf{u}_2+a_3\mathbf{u}_3 =\mathbf{0} \tag{*}.$ We show that this equation has only the zero solution $a_1=a_2=a_3=0$.
The equation (*) can be written as
$\begin{bmatrix} a_1-a_2+a_3 \\ a_1 \\ a_2 \\ a_3 \end{bmatrix}=\mathbf{0}.$ Comparing entries, we obtain $a_1=a_2=a_3=0$.
Thus the equation (*) has only the zero solution and hence the vectors $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ are linearly independent.

Therefore, the set $\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3 \}$ is linearly independent spanning set for $V$, thus it is a basis for the subspace $V$.
Since the basis consists of $3$ vectors, the dimension of the subspace $V$ is $3$.

In summary, we found a basis
$\left\{\quad\mathbf{u}_1=\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix},\quad \mathbf{u}_2=\begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix},\quad \mathbf{u}_3=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}\quad \right\}.$ for the subspace $V$ and the dimension of $V$ is $3$.

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