Is a combination of the vectors in a basis of a space also a basis of that space? Suppose $B = \{x, y, z\}$ is a basis of a subvector space $V$ in $\mathbb{R^n}$. 


*

*Is the set of vectors $C = \{x+y, y+z, z+x\}$ a basis of $V$? 

*Is the set of vectors $C = \{2\cdot y + z + 3\cdot x, 3\cdot y + z - x, y - 4\cdot x\}$ a basis of $V$?

*Is the set of vectors $C = \{x + z, y - z\}$ a basis of $V$?

*Is the set of vectors $C = \{y, y + z, z + y + x, x\}$ a basis of $V$?
I always provide some sort of attempt at answering the questions I post here, but I really don't even have a clue where to begin here. Is there some sort of connection between the combination of vectors in a basis? Any help would be greatly appreciated. 
 A: Guide:


*

*First thing to check, since $B$ has $3$ vectors, to be a basis, it must have exactly $3$ elements. If it does not, it can't be a basis.

*Next, to be a basis, check that they are linearly independent.
For example for the first set.
$$a(x+y)+b(y+z)+c(z+x)=0$$
$$(a+c)x+(a+b)y+(b+c)z=0$$
Since $\{x,y,z\}$ is a basis, we have
$$a+c=0$$
$$a+b=0$$
$$b+c=0$$
Try to find whether they have only the trivial solution, if it does, then it is linearly independent and hence it forms a basis. If it is not linearly independent, then it is not a basis.
First answer:

 Yes 

A: Hint (for the first one): let $C=\{u,v,w\}$ where $\,u=x+y\,$, $\,v=y+z\,$, $\,w=z+x\,$, then by simple algebra $\,x=(u-v+w)/2\,$, $\,y=(u+v-w)/2\,$, $\,z=(-u+v+w)/2\,$. The latter gives the explicit change of basis transformation, and also proves that any linear combination of $\,x,y,z\,$ can be written as a linear combination of $\,u,v,w\,$.
A: Note that the subspace  V is a three dimensional space.
Here are the questions and answers. 
1)Is the set of vectors C = {x+y, y+z, z+x} a basis of V?
Yes,because the matrix of this transformation is non-singular.
2) Is the set of vectors C = {2y + z + 3x, 3y + z - x, y - 4x} a basis of V?
Yes,because the matrix of this transformation is non-singular.
3)Is the set of vectors C = {x + z, y - z} a basis of V?
No, because you need three vectors in your basis.
4)Is the set of vectors C = {y, y + z, z + y + x, x} a basis of V?
No, these four vectors can not be linearly independent in a three dimensional space.
A: Consider the matrix:
$$B=\begin{bmatrix}
x_1&y_1&z_1\\
x_2&y_2&z_2\\
\vdots&\vdots&\vdots\\
x_n&y_n&z_n
\end{bmatrix}$$
$\mathcal B $ is a basis of V if and only if the matrix $B$ has rank $3$. A set of three linear combinations of these vectors corresponds to multiplying $B$ on the right by a $3\times 3$ matrix, and this set is a basis of V if and and only if it is multiplied by an invertible matrix.


*

*The first set results from a multiplication by the matrx $M=\begin{bmatrix}1&0&1\\1&1&0\\0&1&1\end{bmatrix}$, which has determinant $2$, so yes, we obtain  a basis of $V$.

*For the second set, we  multiply by $M=\begin{bmatrix}3&-1&0\\2&3&1\\1&1&-4\end{bmatrix}$, which has determinant $-24$.

*For the third set, we can't obtain a basis since it hasn't the required number of vectors. However it results from multiplying on the right by the $3\times 2$ matrix $\;\begin{bmatrix}1&0\\0&1\\1&-1\end{bmatrix}$, of rank $2$, so we can say the obtained vectors are linearly independent.

*The fourth set of vectors has too many vectors to be a basis.

A: A  matrix takes a basis to a basis if and only if nonsingular (and in this case actually changes basis from the basis formed by the columns of the matrix to the standard basis).   
In 1) the matrix is $$\begin{bmatrix} 1&0&1\\ 1&1&0\\ 0&1&1\end{bmatrix}$$.  The rank is $3$, so the answer is yes. 
In 2) the matrix is $$\begin{bmatrix}3&-1&-4\\ 2 &3&1\\ 1&1&0 \end{bmatrix}$$.  Again, by row reduction, or computing the determinant (=0),  one can see that it's a singular matrix.   So no. 
As noted by @saulspatz, you can dismiss 3) and 4).
