Finding the basis of a specific subspace

Question

Can someone help me to find the basis for $(1-y-z,y,z)$ please?

I have $(-1,1,0)$ and $(-1,0,1)$, but I don't know how to add $1$ to $x$.

Answer 1

I am assuming you mean the subspace spanned by the vector $$\begin{pmatrix} 1-y-z \\ y \\ z\end{pmatrix},$$ where $y,z\in\mathbb{R}$.

Note that vector is really just

$$\begin{pmatrix} 1-y-z \\ y \\ z\end{pmatrix}=\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix}+\begin{pmatrix} -1 \\ 1 \\ 0\end{pmatrix}y+\begin{pmatrix} -1 \\ 0 \\ 1\end{pmatrix}z.$$

If you take $y=z=0$, you have the vector $\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix}$, which is in this subspace, and linearly independent of the other two vectors you found. If it is a subspace than all scalar multiples of this vector must also be in the space.

Also, if your question is whether the set of vectors of the given form is a subspace, it is not because it is not closed under addition nor scalar multiplication.

On the other hand, if you are trying to find the subspace generated by all vectors of this form, it would be all of $\mathbb{R}^{3}$. Note, taking $y=1$ and $z=0$, we have the vector $$\begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}.$$ If we take $y=0$ and $z=1$, we have the vector $$\begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}.$$ So each standard unit vector would be in this subspace in that case.

Answer 2

I'm not sure what you meant in "add $1$ to $x$" for what you found is already near basis.

The subspace you said should be $\{(1-y-z,y,z)|y,z\in\mathbb{R}\}$ over $\mathbb{R}$ (or other fields). Then the dimension of this subspace is $2$ and you may find $\{(-1,2,0), (-1,0,2)\}$ as a basis, for example.

You should mind your basis element is also in this subspace, but $(-1,1,0),(-1,0,1)$ are not.