Consider 3 linearly independent, n-dimensional vectors. Adding a fourth under this circumstance is not linearly independent. Why? Suppose that $\bf x$, $\bf y$, and $\bf z$ are three linearly independent, $n$-dimensional vectors. Define $\bf k$ by
$$
{\bf k} = \alpha_1 {\bf x} + \alpha_2 {\bf y} + \alpha_3 {\bf z},
$$
where $\alpha_1, \alpha_2, \alpha_3 \in \mathbb{R} \backslash\{0\}$.
Are the vectors, $\bf x$, $\bf y$, $\bf z$, and $\bf k$ linearly independent?
I believe the answer is no, however I don't quite understand why. How can we have a linear independence when all alphas cannot be zero? Does it not require alphas as zero due to it being the trivial solution to $\sum^n_{i=1} \alpha_i {\bf x}_i = 0_n$?
 A: Hint: $k + (-\alpha_x) x + (-\alpha_y)y + (-\alpha_z)z = 0_n$ and the coefficients $1, -\alpha_x, -\alpha_y,$ and $-\alpha_z$ are non-zero. 
Edit:
$\{x, y, k\}$ are linearly independent. 
To see this, imagine that $\{k,x,y\}$ were linearly dependent. Then there exist some $c_1, c_2, c_3$ not all $0$ such that 
$c_1 k + c_2 x + c_3 y = 0$. Then, 
$$
c_1(\alpha_x x + \alpha_y y + \alpha_z z) + c_2 x + c_3 y = 0.
$$
But then you will have a linear combination of $\{x,y,z\}$ equal to $0_n$.. which will contradict that $\{x,y,z\}$ are linearly independent.
A: Your belief is correct since 
$$\alpha_xx+\alpha_yy+\alpha_zz-k=0$$
and not all the coefficients of $\;x,y,z,k\;$ above are zero.
A: Tom and Timbuc already gave accurate and formal answer/hint. But linear independence can also be viewed informally as follows: if ${\bf v}_1,\ldots,{\bf v}_n$ are linearly independent vectors, then each ${\bf v}_i$ cannot be "dependent" on the other vectors. The fact that ${\bf k} = \alpha_1 {\bf x} + \alpha_2 {\bf y} + \alpha_3 {\bf z}$ means that $\bf k$ "depends" on $\bf x$, $\bf y$, and $\bf z$. So ${\bf x}, {\bf y}, {\bf z}, {\bf k}$ cannot be linearly independent.
But the above is only an informal way of understanding linear independence. To answer such a question on a test, you have to give formal answers that are in line with Tom and Timbuc's arguments.
