0
$\begingroup$

Let

$V=\mathrm{span}\{(1,0,0,1),(0,1,0,1),(0,0,1,-1)\}$

and

$W=\mathrm{span}\{(1,2,0,1),(2,0,-1,2),(1,1,1,1)\}$

How I can find the basis of the subspaces sum and the subspaces intersection? There is a general way to do that? I know only the Grassmann formula to find the dimension.

I'm new and I don't know how to proceed.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

Since $V$ and $W$ are each generated by three linearly independent vectors, it follows that $\dim V = \dim W = 3$.

In general, when you have two finite-dimensional subspaces $V$ and $W$, their sum $V + W$ will be finite-dimensional as well and will be spanned by the union of any two bases of $V$ and $W$. So in this case, we know that that the list

$ (1,0,0,1), (0,1,0,1), (0,0,1,-1), (1,2,0,1), (2,0,-1,2), (1,1,1,1)$

spans $V + W$. Notice that the first three vectors in this list are linearly independent, so $\dim (V + W) \geq 3$. Since $V + W \subseteq \mathbb{R}^{4}$, we also have $\dim (V + W) \leq 4$. You can then show that

$(1,0,0,1), (0,1,0,1), (0,0,1, -1), (1,2,0,1)$

are linearly independent in $V + W$, so they must be a basis for $V + W$. This in turn implies that $\dim V + W = 4$, so $V + W = \mathbb{R}^{4}$. Now using Grassmann formula, we see that

$\dim (V \cap W) = \dim V + \dim W - \dim (V + W) = 2$

so we only need to find two linearly independent vectors in $V \cap W$ and we'll be done.

First, notice that $(1,1,1,1) \in V$ because

$(1,1,1,1) = (1,0,0,1) + (0,1,0,1)+ (0,0,1,-1).$

Thus, $(1,1,1,1) \in V \cap W$.

Notice also that $(1,0,0,1) \in W$, because

$(1, 0,0,1) = (-\frac{1}{5})(1,2,0,1) + \frac{2}{5}(2,0,-1,2) + \frac{2}{5}(1,1,1,1).$

Thus, $(1,0,0,1) \in V \cap W$.

Since $(1,1,1,1), (1,0,0,1)$ are linearly independent, they form a basis for $V \cap W$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .