elbluffz1

2022-01-24

Let say K and L are two different subspace real vector space V. If given $\mathrm{dim}\left(K\right)=\mathrm{dim}\left(L\right)=4$, how to determine minimal dimensions are possible for V?

Expert

Explanation:
Let the four vectors ${k}_{1},{k}_{2},{k}_{3}$ and ${k}_{4}$ form a basis of the vector space K. Since K is a subspace of V, these four vectors form a linearly independent set in V. Since L is a subspace of V different from K, there must be at least one element, say ${l}_{1}$ in L, which is not in K, i.e, which is not a linear combination of ${k}_{1},{k}_{2},{k}_{3}$ and ${k}_{4}$.
So, the set $\left\{{k}_{1},{k}_{2},{k}_{3},{k}_{4},{l}_{1}\right\}$ s a linear independent et of vectors in V. Thus the dimensionality of V is at least 5!
In fact, it is possible for the span of $\left\{{k}_{1},{k}_{2},{k}_{3},{k}_{4},{l}_{1}\right\}$ to be the entire vector space V - so that the minimum number of basis vectors must be 5
Just as an example, let V be ${\mathbb{R}}^{5}$ and let K and V consists of vectors of the forms
$\left(\begin{array}{c}\alpha \\ \beta \\ \gamma \\ \delta \\ 0\end{array}\right)$ and $\left(\begin{array}{c}\mu \\ \nu \\ \lambda \\ 0\\ \varphi \end{array}\right)$
It is easy to see that the vectors
$\left(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}\right)$
form a basis of K Append the vector $\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}\right)$ , and you will get a basis for the entire vector space,

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