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Let $$U_1 = \left\{ \left( \begin{array}{cc} x_1\\ x_2\\ x_3\\ x_4\\ \end{array} \right) \in \mathbb{R} :-x_1-x_2+x_3=0 \right\}$$

$$U_2 = span \left( \left( \begin{array}{cc} -1 \\ 1 \\ 1 \\ 2 \\ \end{array} \right), \left( \begin{array}{cc} 2 \\ -1 \\ -1 \\ 2 \\ \end{array} \right) \right) $$

, find the bases for $U_1, U_2, U_1 \cap U_2, U_1 + U_2$ and give the dimension of each subspace.

Now $U_1$ can be written as $$\left\{ \left( \begin{array}{cc} x_1\\ x_2\\ x_1 + x_2\\ x_4\\ \end{array} \right) \right\}$$ and so a basis would be $$B= \left( \left( \begin{array}{cc} 1 \\ 0 \\ 1 \\ 0 \\ \end{array} \right), \left( \begin{array}{cc} 0 \\ 1 \\ 1 \\ 0 \\ \end{array} \right), \left( \begin{array}{cc} 0 \\ 0 \\ 0 \\ 1 \\ \end{array} \right) \right)$$

the vectors in the span of $U_2$ form a basis since they are linearly independent.

Now to find the basis of $U_1 + U_2$ I could take the span of all the basis vectors in $U_1$ and $U_2$, but since there are only four linearly independent vectors in that span, only four of those vectors form the basis of $U_1 + U_2$.

I cannot find a basis for the intersection of $U_1$ and $U_2$. I can only determine the dimension of that subspace by using the dimension formula for subspaces. So $U_1 \cap U_2$ must have dimension 1, but I cannot find a spanning vector. Can anybody help please?

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I tend to use a bare-hands approach like this:

Suppose $x \in U_1 \cap U_2$. Since $x \in U_2$, this implies that

$$x= s \left( \begin{array}{cc} -1\\ 1\\ 1\\ 2\\ \end{array} \right) +t \left( \begin{array}{cc} 2\\ -1\\ -1\\ 2\\ \end{array} \right) = \left( \begin{array}{cc} -s + 2t\\ s-t\\ s-t\\ 2s+2t\\ \end{array} \right) $$

Since we require this to belong to $U_1$, we need $x_3 = x_1 + x_2$, so that $s=2t$.

So we can just take (for example) $t=1$ and $s=2$. This gives a vector in the intersection:

$$x = \left( \begin{array}{cc} 0\\ 1\\ 1\\ 6\\ \end{array} \right)$$

Since this is non-zero, it is a spanning set (in fact a basis) for the intersection.

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