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Why can't a set of four vectors in $\mathbb{R}^3$ be linearly independent?

I know that if the determinant of the vectors together is not $0$ then the vectors are linearly independent. But this is not relevant to this case of a non square matrix.

Is the answer "because there can only ever be 3 pivots when reduced" a good answer?

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  • $\begingroup$ You can use the determinant method by adding a zero column to the matrix to compensate $\endgroup$ – clark Jun 20 '16 at 12:08
  • $\begingroup$ Consider the case where you have 3 linearly independent vectors in $\mathbb{R}^{3}$. In this case you can already define every point in the space in terms of those three vectors. How could you add another vector that is linearly independent to the other 3? There is no way since they "complete" the space already. $\endgroup$ – Carser Jun 20 '16 at 12:14
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    $\begingroup$ So basically the 4th vector will always be a linear combination of the other 3 if the other 3 span $\mathbb{R}^3$ $\endgroup$ – Patrick Lee Jun 20 '16 at 12:22
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This is a consequence of the Steinitz exchange lemma, which says that for a vector space $ V $, if $ L $ is a linearly independent set and $ S $ is a spanning set, then we have $ |L| \leq |S| $. Since we may take $ S $ to be a basis of $ V $, this implies that we must have $ |L| \leq \dim V $.

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The dimension of $R^3$ is 3 i.e its basis has 3 vectors. So additional vectors would be just linear combination of 3 vectors. Any set of 3 L. I vectors will be a basis. If you say 4 vectors are linearly independent in R^3 then it would mean they will be part of basis. Hence dimension of R^3 will become 4 which is not so

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    $\begingroup$ Thank you this makes more sense now. Since I can visualize what it means. $\endgroup$ – Patrick Lee Jun 20 '16 at 12:20
  • $\begingroup$ You can describe every vector is R^3 with help of 3 vectors like (1, 0, 0 ), (0, 1, 0), (0, 0, 1)... Any 4th vector can be written in terms of these $\endgroup$ – Gathdi Jun 20 '16 at 14:44
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Let's create new names for the unit vectors of $\mathbb{R}^3$: $v_1=(1,0,0)$, $v_2=(0,1,0)$, $v_3=(0,0,1)$

Consider the basis $\{v_1,v_2,v_3\}$. I will proceed with a proof by contradiction. Assume that there exists a set of $4$ linearly independent vectors in $\mathbb{R}^3$, $\{w_1, w_2, w_3, w_4\}$.

We can then represent $w_1$ as a linear combination the elements in the set $\{v_1,v_2,v_3\}$.

$$w_1=a_1\cdot v_1+a_2\cdot v_2+a_3\cdot v_3$$

Since $w_1\neq(0,0,0)$, we know that $a_i\neq 0$ for some $i$. After reordering $v_1$, $v_2$, and $v_3$, we can assume that $a_1\neq0$ Then,

$$v_1=\frac{w_1}{a_1}-\frac{a_2}{a_1}\cdot v_2-\frac{a_3}{a_1}\cdot v_3$$

Since we can represent $v_1$ as a linear combination of the elements in $\{w_1, v_2, v_3\}$, the subset $\{w_1, v_2, v_3\}$ generates $\mathbb{R}^3$. Now we can represent $w_2$ as a linear combination of the elements in $\{w_1, v_2, v_3\}$.

$$w_2=b_1\cdot w_1+c_2\cdot v_2+c_3\cdot v_3$$

We know that some $c_i\neq0$ since if both $c_1$ and $c_2$ was equal to $0$, $w_1$ and $w_2$ would be linearly dependent (which contradicts our initial assumption). After reordering, we can assume that $c_2\neq 0$ So,

$$v_2=\frac{w_2}{c_2}-\frac{b_1}{c_2}w_1-\frac{c_3}{c_2}v_3$$

We now see that the subset $\{w_1, w_2, v_3\}$ generates $\mathbb{R}^3$. Repeating this process one more time, we can see that the subset $\{w_1, w_2, w_3\}$ generates $\mathbb{R}^3$. So, we can represent $w_4$ as a linear combination of the elements in the set $\{w_1, w_2, w_3\}$, proving that they must be linearly dependent.

Source: Linear Algebra, Third Edition, by Serge Lang

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