# Let say K and L are two different subspace real vector space V. If given dim(K)=dim(L)=4, how to determine minimal dimensions are possible for V ?

May 4, 2018

5

#### 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 $\left\{{k}_{1} , {k}_{2} , {k}_{3} , {k}_{4} , {l}_{1}\right\}$ is a linear independent set 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
$\left(\begin{matrix}\alpha \\ \beta \\ \gamma \\ \delta \\ 0\end{matrix}\right)$ and $\left(\begin{matrix}\mu \\ \nu \\ \lambda \\ 0 \\ \phi\end{matrix}\right)$

It is easy to see that the vectors

$\left(\begin{matrix}1 \\ 0 \\ 0 \\ 0 \\ 0\end{matrix}\right)$,$\left(\begin{matrix}0 \\ 1 \\ 0 \\ 0 \\ 0\end{matrix}\right)$,$\left(\begin{matrix}0 \\ 0 \\ 1 \\ 0 \\ 0\end{matrix}\right)$and $\left(\begin{matrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{matrix}\right)$

form a basis of $K$. Append the vector $\left(\begin{matrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{matrix}\right)$, and you will get a basis for the entire vector space,