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I was given the following problem:

Let $S = \{v_1, \ldots, v_4\} \subset \mathbb{R}^4$ where \begin{align*} v_1 = (-1, 0, 1, 2), ~ v_2 = (3, 4, -2, 5), ~ v_3 = (0, 4, 1, 11), v_4 = (1, 4, 0, 9) .\end{align*} Describe implicitly the subspace $W = \langle S\rangle$. That is, find a homogeneous system of equations for which the solution space is $W$.

Should it not be a heterogeneous system of equations and not a homogeneous one? For the span of $S$ is nothing but the linear combination of its vectors. It follows that any $\textbf{x} = (x_1, ..., x_4) \in W$ is defined by the heterogeneous system whose associated matrix is

\begin{align*} \begin{pmatrix} -1 & 3 & 0 & 1 &|& x_1 \\ 0 & 4 & 4 & 4 &|& x_2 \\ 1 & -2 & 1 & 0 &|& x_3 \\ 2 & 5 & 11 & 9 &|& x_4 \end{pmatrix} .\end{align*}

This is equivalent to saying any$\textbf{x} \in W$ is of the form

$$ \textbf{x} = \begin{pmatrix} -\lambda_1 + 3 \lambda_2 + \lambda_4 \\ 4\lambda_2 + 4 \lambda_3 + 4\lambda_4 \\ \lambda_1 -2 \lambda_2 + \lambda_3 \\ 2\lambda_1 + 5 \lambda_2 + 11 \lambda_3 + 9\lambda_4 \end{pmatrix} $$

where $\lambda_1, \ldots, \lambda_4$ are scalars.

Digitallis
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lafinur
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  • The solution set of a heterogeneous linear system of equations is, in general, not a subspace. (An easy way to see this is that 0 is not contained in that set ) – K. Jiang May 22 '23 at 12:40
  • First, $v_4$ is not linearly independent from the other three basis vectors chosen. Second, google is your friend check out this link – Gerald May 22 '23 at 14:02

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