determine if the given vectors span $\mathbb{R}^4$

Question

Determine if the given vectors span $\mathbb{R}^4$

${(1, 1, 1, 1), (0, 1, 1, 1), (0, 0, 1, 1), (0, 0, 0, 1)}$.

I'm completely confused on this question. My textbook gives a different problem but in $\mathbb{R}^3$. How would i go about this?

Answer 1

You want to find constants $a,b,c,d \in \mathbb{R}$ such that every vector $(x,y,z,w) \in \mathbb{R}^4$ can be written as a linear combination of the given vectors. That is, \begin{align}(x,y,z,w)&=a(1,1,1,1)+b(0,1,1,1)+c(0,0,1,1)+d(0,0,0,1)\\ &=(a,a+b,a+b+c,a+b+c+d). \end{align} Equating components, we have: $$x=a, \quad y=a+b, \quad z=a+b+c, \quad w=a+b+c+d.$$

Hence, $$a=x, \quad b=y-x, \quad c=z-y, \quad d=w-z.$$

Therefore, the given list of vectors spans $\mathbb{R}^4$ (since the vector $(x,y,z,w)$ was arbitrary).

Alternatively, you can check to see that the given vectors are linearly independent, and then, since there are $\dim(\mathbb{R}^4)=4$ of them, they must be a basis for $\mathbb{R}^4$. They therefore span $\mathbb{R}^4$.

Answer 2

1. Arrange the vectors as rows of a matrix.
2. Compute the determinant. It is particularly easy to compute as the matrix is upper triangular, so the determinant is just the product of the diagonal entries. It is equal to $1$.
3. Conclude that the rows of the matrix are linearly independent.
4. Any Four linearly independent vectors span $\mathbb{R}^4$.

Answer 3

Not needed any sophisticated knowledge, just a simple understanding: the most right vector gives 1 dimension, 2nd rightmost 2nd dimension, 2nd left 3rd dimension and first one 4th dimension. If you do not believe, subtract two consetive vectors - and get the standard basis exactly.