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Let $A$ be a matrix with linearly dependent rows.

What can we infer about the the solutions of:
$$Ax = 0$$

$$Ax = b$$

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1 Answer 1

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In the first case, you have infinite solution. In particular, said that $n$ is the size of $A$ and $r < n$ is the rank of $A$, then there exists a linear subspace $\mathcal{X_0}$, with dimension $n - r$, such that every $x \in \mathcal{X_0}$ is a solution of $Ax = 0$

In the second case, you can have different scenarios.

  1. If $b$ cannot be expressed as a linear combination of all the columns of $A$, then the system is impossible (there is no solution).
  2. If $b$ can be expressed as a linear combination of all the columns of $A$, then let $x'$ the vector which contains all the coefficients of this linear combination. Then you have an affine subspace $\mathcal{X}$ such that:

    a. for each $x \in \mathcal{X}$ there exists an $x_0 \in \mathcal{X_0}$ such that $x = x' + x_0$

    b. $Ax = Ax' + Ax_0 = b + 0 = b$, then $x$ is a solution of the system.

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