Consider the vectorspace V_3(R),R is field ofreal numbers.Let S and T be the subspace of V_3(R) spanned by (1,1,1) and (1,2,1) respectively dim(S+T)=

Question

I know that basis is the smallest spanning set in sense of cardinality

It is also given that S and T are subspace of V_3 So dimension of S and T is almost 3

What should I do next?

Answer 1

Let us fix some notation first..

$v = (1, 1, 1)$ and $w = (1, 2, 1)$.

$$ S = \mbox{span}(v) = \{ (x, x, x) \in \mathbf{R}^3 : x \in \mathbf{R} \} $$

$$ T = \mbox{span}(w) = \{ (y, 2 y, y) \in \mathbf{R}^3 : y \in \mathbf{R} \} $$

Clearly, $S$ and $T$ are one-dimensional subspaces of $R^3$. Geometrically, they represent straight lines passing through the vectors $v$ and $w$ respectively.

So, $\mbox{dim}(S) = 1$ and $\mbox{dim}(T) = 1$.

Also, $$ S \cap T = \{ (0, 0, 0) \}.$$

The two straight lines intersect only at the origin.

Hence, $$\mbox{dim}(S \cap T) = 0.$$

Finally, we compute $\mbox{dim}(S + T)$ as follows:

Since $S \cap T = \{ \mathbf{0} \}$, $$ \mbox{dim}(S + T) = \mbox{dim}(S) + \mbox{dim}(T) = 1 + 1 = 2 $$