Prove that i, j, k, i+j+k are linearly dependent, but any three of them are independent

Let $\mathbf{i}, \mathbf{j}, \mathbf{k}$ denote the unit coordinate vectors $(1,0,0)$ , $(0,1,0)$ , and $(0,0,1)$ respectively in $\mathbb{R}^3$ . Prove that $\mathbf{i}, \mathbf{j}, \mathbf{k}, \mathbf{i} + \mathbf{j} + \mathbf{k}$ are linearly dependent, but any three of them are linearly independent.

Proof. First, any four vectors in $\mathbb{R}^3$ are linearly dependent by Theorem 12.10 (page 466 of Apostol). Then we show that any set of three of these are independent.
First, $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are independent since

$x \mathbf{i} + y \mathbf{j} + z \mathbf{k} = 0 \quad \implies \quad x = y = z = 0.$

Then, $\mathbf{i}, \mathbf{j}, \mathbf{i} + \mathbf{j} + \mathbf{k}$ are independent since if

$x \mathbf{i} + y \mathbf{j} + z (\mathbf{i+j+k} ) = 0$

then

$x+z =0, \quad y + z = 0, \quad z = 0 \quad \implies \quad x = y = z = 0.$

The independence of the other two sets of three vectors follows identically $. \qquad \blacksquare$

Stumbling Robot

Prove that i, j, k, i+j+k are linearly dependent, but any three of them are independent

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