dimension of intersection of hyperplanes

What are the possible dimensions of intersection of k-number of hyperplanes in $\mathbb{R}^n$ ?

I look at some examples in lower dimension but I cannot come with a nice cases according to which the dimension varies.

Thanks for your valuable time.

By hyperplane I assume you mean an affine space of dimension $n-1$, which can be defined by a single linear equation in $n$ variables.

In general, when a hyperplane intersects an affine space of dimension $m$ there are three possibilities:

1. The hyperplane includes the affine space, so the intersection still has dimension $m$.
2. The hyperplane is parallel to the affine space, so the intersection is empty.
3. (The usual case) There is a proper intersection, and the intersection has dimension $m-1$, one less than the affine space.

In summary, each hyperplane leaves the number of dimensions unchanged or reduced by one, or makes the result empty (with an undefined number of dimensions).

So if $k>=n$ the intersection can have any number of dimensions from $0$ to $n-1$ (as Linus S. said in a comment) or empty.

If $k<n$ then the intersection can have any number of dimensions from $n-k$ to $n-1$, or empty. It is not possible to get $0$ through $n-k-1$ dimensions or $n$ dimensions. For example, in three dimensions you cannot get two planes to get the intersection of a point ($0$ dimensions) or all of space ($3$ dimensions): you get a line ($1$ dimension, if the planes intersect), a plane ($2$ dimensions, if the planes are identical), or nothing (empty, if the planes are parallel).

You could summarize this by saying that the intersection has $\max(n-k,0)$ through $n-1$ dimensions or is empty, and empty can occur only if $k>1$.

You can see this algebraiclly as follows. Let $X$ be a $K$-Vector Space ,($K$ is an arbitrary field and $X$ has not necessarily finite dimension), and $Hi$ be hyperplanes of $X$ for $i=1,...,k$, (equivalently $dim(X$$\diagup$$Hi)=1$ for $i=1,...,k$ ). If $A$ is the intersection of the $k$ hyperplanes, let the map $f:$$X$$\diagup$$A$$\to$$(X$$\diagup$$H1$$,...,$$X$$\diagup$$Hk)$ with $f(x+A)=(x+H1,...,x+Hk)$. It is an easy matter to check that this is a well defined map, linear and 1-1. Consequently $X$$\diagup$$A$ can be embedded in the vector space $(X$$\diagup$$H1$$,...,$$X$$\diagup$$Hk)$ and because $dim$$(X$$\diagup$$H1$$,...,$$X$$\diagup$$Hk)$$=k$ we conclude that $dim$($X$$\diagup$$A)$$\le$$k$. If in particular $X$ has dimension $n$ , then $dim$($X$$\diagup$$A)$$=dimX-dimA$ and therefore $dimA$$\ge$$dimX-k=n-k$