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Im new to linear algebra, so please just dont blast me.

I have intuitively understood why two independent vectors in $\mathbb R^3$ can't generate all the vector space, by using geometrical intuition. But for dimensions $> 3$ i cannot understand why.

Where i can find a formal definition ( a principle or a law) that explains "why two independent vectors in $\mathbb R^3$ can't generate all the vector space" ?

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  • $\begingroup$ Any textbook of linear algebra. $\endgroup$ – user228113 Jan 7 '18 at 17:30
  • $\begingroup$ The result you're looking for is the dimension theorem $\endgroup$ – Omnomnomnom Jan 7 '18 at 18:31
  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$ – user Feb 3 '18 at 23:57
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It's because for a given finite-dimensional vector space, all of the bases have the same cardinality (namely, the dimension of the vector space). This is a standard result proven in virtually all linear algebra texts. In the case of $\mathbb{R}^3$, the space is obviously three-dimensional.

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    $\begingroup$ Might be worth mentioning that for example, a basis of $\mathbf{R}^3$ is $e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)$. $\endgroup$ – Antonios-Alexandros Robotis Jan 7 '18 at 17:32
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    $\begingroup$ This theorem has a name ? I would read a proof :( $\endgroup$ – Poiera Jan 7 '18 at 18:30
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Suppose you have the following two vectors in $\mathbb{R}^3$:

$$e_1=(1,0,0)$$

$$e_2=(0,1,0)$$

thus you can't obtain $e_3=(0,0,1)$ by linear combinations of $e_1$ and $e_2$ thus you can't span all the space with the two vectros.

Since given two any linearly independent vectors you can assume them a basis, the same argument is valid in general.

And it is also valid in $\mathbb{R}^n$ assuming $e_1,...e_{n-1}$.

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Just to add to the above answer. Note that if 2 vectors are orthogonal to another vector $v_3=(x,y,z)$ they don't "generate" it. Now if you have $v_1$ and $v_2$ does the following set of linear equations have any nontrivial solutions?

$$v_1\cdot v_3=v_{11}x+v_{12}y+v_{13}z=0\\ v_2\cdot v_3=v_{21}x+v_{22}y+v_{32}z=0$$

Then there is a vector $v_3$ in $\mathbb{R}^3$ not generated by $v_1$ and $v_2$

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If you take the span, which is the set of all possible linear combinations, of $2$ linearly independent vectors $u_1$, $u_2$ in $\mathbb{R}^3$ they will form a plane $P_1$ through the origin $(0,0,0)$. We say a basis for $P_1$ is $\{u_1, u_2\}$ which is a $2$-dimensional subspace of $\mathbb{R}^3$, being a copy of $\mathbb{R}^2$, (note this basis is not unique, we can find an infinite number of bases that span $P_1$).

Now if you take the span of $2$ more linearly independent vectors $v_1$, $v_2$ in $\mathbb{R}^3$ which form a plane $P_2$ through the origin, with $P_2$ not equal to $P_1$, then these two planes shall intersect in a line $L$ going through the the origin. This line will be a $1$-dimensional subspace, being a copy of $\mathbb{R}^1$, of $\mathbb{R}^3$. Now we have $4$ vectors in a $3$-dimensional space, and so they cannot all be linearly independent since $4$ is greater than the dimension of the space. This means the $1$-dimensional subspace has to be the span of some vector $u'$ which is some linear combination of $u_1$ and $u_2$, and also the span of some vector $v'$ which is some linear combination of $v_1$ and $v_2$. Then we have $\{u'\}$ and $\{v'\}$ both bases for $L$.

So how do we form a basis. Say we pick $\{u_1, u_2\}$, then as long as one of $v_1$ or $v_2$ don't lie on $L$ (for then it lies in the span of $u'$) we can append this vector, say $v_1$, to the set to give $\{u_1,u_2,v_1\}$ as a basis for $\mathbb{R}^3$.

Similar reasoning applies for $\mathbb{R}^4$: $3$ linearly independent vectors $w_1$, $w_2$, $w_3$ in $\mathbb{R}^4$ will form a $3$-dimensional subspace (which is a copy of $\mathbb{R}^3$) of $\mathbb{R}^4$. If we add one more linearly independent vector $w_4$ we get a basis for the whole of $\mathbb{R}^4$. Now generalise to higher dimensions.

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