# 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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