Vectors and subspace For which real numbers $w, x, y, z$ do the vectors: $(w, 1, 1, 1)$, $(1, x, 1, 1)$, $(1, 1, y, 1)$, $(1, 1, 1, z)$ NOT form a basis of $\mathbb{R}^4$? For each of the values that you found, what is the dimension of the subspace of $\mathbb{R}^4$ that they span?
 A: The fact that the vectors does not form a basis of $\mathbb{R}^4$ means that they are not all linearly independent, which in turn is equivalent to: 
$$det(A) = \begin{vmatrix} w & 1 & 1 & 1 \\ 1 & x & 1 & 1 \\ 1 & 1 & y & 1 \\ 1 & 1 & 1 & z \end{vmatrix} = 0$$ 
The determinant can be caluclated by expanding it using cofactors:
$$det(A) = w\begin{vmatrix} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{vmatrix} - \begin{vmatrix} 1 & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{vmatrix} + \begin{vmatrix} 1 & x & 1 \\ 1 & 1 & 1 \\ 1 & 1 & z \end{vmatrix} - \begin{vmatrix} 1 & x & 1 \\ 1 & 1 & y \\ 1 & 1 & 1 \end{vmatrix}$$
You can easily compute these determinants and see that:
$$det(A)=w(xyz+2-x-y-z)-(yz+2-y-z-1)+(x+z+1-xz-2)-(xy+2-x-y-1)=wxyz-wx-wy-wz-xy-xz-yz+2(w+x+y+z)-3$$
So all numbers $(w,x,y,z)$ satisfying the condition are solution to the problem:
$$wxyz-wx-wy-wz-xy-xz-yz+2(w+x+y+z)-3 = 0$$ 
The dimensionality of the subspace spanned by this vector is equal to the number of linearly indepndant vectors of the 4, or in other words the rank of the matrix $A$.
