Exercise on Vector Space Bases 5

Let $V = \mathbb{R}^3$ be a vector space of dimension $\dim(V) = 3$. Consider the subspace $W = \langle \vec{v}_1, \vec{v}_2, \vec{v}_3 \rangle$ generated by the vectors $\vec{v}_1 = (2,\,0,\,1)$, $\vec{v}_2 = (1,\,1,\,2)$, and $\vec{v}_3 = (3,\,-1,\,0)$. The task is to determine the dimension and identify a basis for the subspace $W$.

The given vectors are:

$$ \vec{v}_1 = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}, \quad \vec{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}, \quad \vec{v}_3 = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix} $$

These vectors form a generating set for $W$:

$$ W = \langle \vec{v}_1, \vec{v}_2, \vec{v}_3 \rangle $$

To determine whether they form a basis, we must check their linear independence.

The vectors are linearly independent if the equation

$$ k_1 \vec{v}_1 + k_2 \vec{v}_2 + k_3 \vec{v}_3 = \vec{0} $$

has only the trivial solution $k_1 = k_2 = k_3 = 0$.

Substituting the vectors yields:

$$ \begin{pmatrix} 2k_1 + k_2 + 3k_3 \\ k_2 - k_3 \\ k_1 + 2k_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$

This leads to the system:

$$ \begin{cases} 2k_1 + k_2 + 3k_3 = 0 \\ k_2 - k_3 = 0 \\ k_1 + 2k_2 = 0 \end{cases} $$

Solving the system: from the second equation, $k_2 = k_3$; from the third, $k_1 = -2k_2$. Substituting into the first:

$$ 2(-2k_2) + k_2 + 3k_2 = -4k_2 + k_2 + 3k_2 = 0 $$

This simplifies to $0 = 0$, confirming that the system has infinitely many solutions. Hence, the vectors are linearly dependent and do not form a basis for $W$.

To reduce the set to a basis, we can eliminate one vector. Since $\vec{v}_3 = 2\vec{v}_1 - \vec{v}_2$, we can discard $\vec{v}_3$ and consider the remaining set:

$$ \{ \vec{v}_1, \vec{v}_2 \} $$

Since $\vec{v}_3$ lies in the span of $\vec{v}_1$ and $\vec{v}_2$, this pair still generates the subspace:

$$ W = \langle \vec{v}_1, \vec{v}_2 \rangle $$

Now, let’s verify whether $\vec{v}_1$ and $\vec{v}_2$ are linearly independent:

$$ k_1 \vec{v}_1 + k_2 \vec{v}_2 = \vec{0} $$

Substituting the vectors:

$$ \begin{pmatrix} 2k_1 + k_2 \\ k_2 \\ k_1 + 2k_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$

Which gives the system:

$$ \begin{cases} 2k_1 + k_2 = 0 \\ k_2 = 0 \\ k_1 + 2k_2 = 0 \end{cases} $$

From the second equation, $k_2 = 0$, and thus $k_1 = 0$. The only solution is the trivial one, so the vectors are linearly independent.

Therefore, $\vec{v}_1$ and $\vec{v}_2$ form a basis for the subspace $W$:

$$ B_W = \{ \vec{v}_1, \vec{v}_2 \} $$

Since the basis consists of two vectors, the dimension of the subspace $W$ is:

$$ \dim(W) = 2 $$

And so on.