Exercise on Vector Space Bases 2

In the vector space V = ℝ⁴, the vectors v₁ = (1, 0, -1, 2), v₂ = (0, 2, 1, 1), v₃ = (3, - 4, - 5, 4), and v₄ = ( - 3, 4, 1, 4) span a subspace W.

$$ W = \langle \ \vec{v}_1 , \ \vec{v}_2 , \ \vec{v}_3 , \ \vec{v}_4 \ \rangle $$

The goal is to determine the dimension and find a basis for the subspace W.

Solution

We begin by checking whether the given vectors are linearly independent.

$$ \vec{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \\ 2 \end{pmatrix}, \quad \vec{v}_2 = \begin{pmatrix} 0 \\ 2 \\ 1 \\ 1 \end{pmatrix}, \quad \vec{v}_3 = \begin{pmatrix} 3 \\ -4 \\ -5 \\ 4 \end{pmatrix}, \quad \vec{v}_4 = \begin{pmatrix} -3 \\ 4 \\ 1 \\ 4 \end{pmatrix} $$

If they are linearly independent, then they form a basis of W and the dimension is 4. To verify this, we consider the linear combination:

$$ k_1 \vec{v}_1 + k_2 \vec{v}_2 + k_3 \vec{v}_3 + k_4 \vec{v}_4 = \vec{0} $$

Substituting the vectors, we obtain:

$$ \begin{pmatrix} k_1 + 3k_3 - 3k_4 \\ 2k_2 - 4k_3 + 4k_4 \\ -k_1 + k_2 - 5k_3 + k_4 \\ 2k_1 + k_2 + 4k_3 + 4k_4 \end{pmatrix} = \vec{0} $$

This leads to the following system of equations:

$$ \begin{cases} k_1 + 3k_3 - 3k_4 = 0 \\ 2k_2 - 4k_3 + 4k_4 = 0 \\ -k_1 + k_2 - 5k_3 + k_4 = 0 \\ 2k_1 + k_2 + 4k_3 + 4k_4 = 0 \end{cases} $$

There are four unknowns. The coefficient matrix has rank 3:

$$ A = \begin{pmatrix} 1 & 0 & 3 & -3 \\ 0 & 2 & -4 & 4 \\ -1 & 1 & -5 & 1 \\ 2 & 1 & 4 & 4 \end{pmatrix} $$

Verification: The determinant of the full matrix is zero, indicating linear dependence:

$$ \det(A) = 1 \cdot \begin{vmatrix} 2 & -4 & 4 \\ 1 & -5 & 1 \\ 1 & 4 & 4 \end{vmatrix} + \dots = 0 $$

By the Rouché - Capelli Theorem, with $n = 4$ and $r = 3$, the system has infinitely many solutions:

$$ \infty^{n - r} = \infty^{1} = \infty $$

Hence, the four vectors are linearly dependent and cannot form a basis.

To extract a basis, we need to remove one of the dependent vectors from the generating set $\{ \vec{v}_1, \vec{v}_2, \vec{v}_3, \vec{v}_4 \}$.

To determine which vector to remove, we solve the homogeneous system explicitly. One particular solution is:

$$ \begin{cases} k_1 = -3k_3 \\ k_2 = 2k_3 \\ k_3 = c \\ k_4 = 0 \end{cases} $$

Letting $k_3 = 1$, we get:

$$ k_1 = -3, \quad k_2 = 2, \quad k_3 = 1, \quad k_4 = 0 $$

Which gives the relation:

$$ -3\vec{v}_1 + 2\vec{v}_2 + \vec{v}_3 = \vec{0} \quad \Rightarrow \quad \vec{v}_3 = 3\vec{v}_1 - 2\vec{v}_2 $$

So $\vec{v}_3$ can be expressed as a linear combination of $\vec{v}_1$ and $\vec{v}_2$, and can therefore be excluded from the basis.

Note: Since $\vec{v}_4$ does not appear in this dependence relation, it is linearly independent of the others. Removing it would have led to an incorrect basis.

We now define W as the span of the remaining vectors:

$$ W = \langle \vec{v}_1, \vec{v}_2, \vec{v}_4 \rangle $$

To confirm that these three vectors are linearly independent, we check:

$$ k_1\vec{v}_1 + k_2\vec{v}_2 + k_4\vec{v}_4 = \vec{0} $$

This leads to the system:

$$ \begin{cases} k_1 - 3k_4 = 0 \\ 2k_2 + 4k_4 = 0 \\ -k_1 + k_2 + k_4 = 0 \\ 2k_1 + k_2 + 4k_4 = 0 \end{cases} $$

There are three unknowns. The coefficient matrix has full rank $r = 3$:

$$ A = \begin{pmatrix} 1 & 0 & -3 \\ 0 & 2 & 4 \\ -1 & 1 & 1 \\ 2 & 1 & 4 \end{pmatrix} $$

Verification: A $3 \times 3$ minor has a non-zero determinant:

$$ \det \begin{pmatrix} 1 & 0 & -3 \\ 0 & 2 & 4 \\ -1 & 1 & 1 \end{pmatrix} = -8 \ne 0 $$

Thus, the system has a unique solution:

$$ \infty^{n - r} = \infty^0 = 1 \Rightarrow \text{only the trivial solution} $$

Therefore, $\vec{v}_1$, $\vec{v}_2$, and $\vec{v}_4$ are linearly independent and form a basis of the subspace W:

$$ B_W = \{ \vec{v}_1 , \vec{v}_2 , \vec{v}_4 \} $$

Since the basis consists of three vectors, the dimension of W is:

$$ \dim W = 3 $$

And so on.