Basis and dimension for subspaces of $\mathbb{R^3}$. I am supposed to find a basis and dimension for $J$ and $W$ of $\mathbb{R^3}$ where $J=\{(a,b,c)^T : a+b+c=0\}$ and $W=\{(a,b,c)^T : a=b=c\}$. I know $(a,b,c)^T = \begin{bmatrix}a\\b\\c\end{bmatrix}$, but I am used to row-reducing for finding a basis/dimension. Here I just have three arbitrary coefficients. I can't really row-reduce it seems like because I have a $3\times 1$ vector. Is there an easy way of going about this problem or am I missing something?
 A: You should be able to see very easily that $W$ will be 1-dimensional, a basis would be $(1,1,1)^T$ and clearly any vector with the property that $a=b=c$ is a scalar multiple of this basis vector.
For $J$, if you really want to think of this in terms of row reducing then you need to think of this as a system of simultaneous equations where the only equation you have is $a+b+c=0$ and then find the infinitely many solutions to this system you will have, it will become clear that the space has dimension 2 since we have one equation in a space of three dimensions so need 2 parameters to express the solutions.
A: As a first step, the subspace $W$ must be one-dimensional. What is an obvious basis vector then?
Next, note that $J$ is an orthogonal subspace to $W$ (they only intersect at the zero vector). So you need to find two vectors which are orthogonal to vectors in $W$. You could do this by taking the dot product with your basis vector from $W$. That will give you a set of equations to solve. Then you can row-reduce those if you like.
A: Observe $J$ is a plane in $\mathbb{R}^3$:
Hint:
Can you see how $\begin{bmatrix}1\\-1\\0\end{bmatrix}$ and $\begin{bmatrix}1\\0\\-1\end{bmatrix}$ are related to $J$?
