Linear Algebra, cube & dimensions > 3

To calculate the number of edges: as you say there are $2^n$ corners. Each one is connected to n other corners. Adding all of these up counts each edge twice, so there are $2^{n-1}n$ edges, which equals $32$ for $n=4$. Another way to count edges is to define $E(n)$ as the number of edges in $n$ dimensions. If you think of the $n+1$ dimensional cube as connecting the corresponding corners of two $n$ dimensional cubes, the recurrence is $E(n+1)=2E(n)+2^n$

Square faces are made by starting with a square in the one of the two $n$ cubes plus the translation of the edges of the $n$ cube in the new dimension. So $S(n+1)=2S(n)+E(n)$.

$3$-cube faces are made by starting with a $3$-cube in one of the two n-cubes plus translation of the squares in the new dimension. So $C(n+1)=2C(n)+S(n)$.

As you mentioned, a vertex (or a 0-face) is just a choice of string $(x_1,\ldots, x_n)$, where $x_i\in \{0,1\}$. If you think back to dimensions 2 or 3 (or even 1!), you'll see that an edge (i.e., a 1-face) is determined by two vertices which differ in a single slot. So e.g. on the unit square, the left edge goes between $(0,0)$ and $(0,1)$ -- we may name this edge more succinctly as $(0,*)$. This method generalizes, too: the bottom square (2-face) of the 3-cube is $(*,*,0)$, etc. Note that the $n$-cube will have $k$-faces for $0\leq k\leq n$. See if you can find a formula for how many there are!

This isn't exactly linear algebra, it's more like combinatorics. Also, this problem is actually a decent introduction to the idea of higher dimensions, if you try to visualize things in 4 dimensions. Search the internet for some representations of 4-cubes, and try to understand why they look the way they do.

This is the 27th problem in Exercises 1.1 of the book, Introduction to Linear Algebra, Gilbert Strang. I hope my answer will be helpful for future readers.

Question 1: How many corners does a cube have in 4 dimensions?

Solution: A corner of a cube in $N$ dimenison(s) can be written as $(x_1, x_2, \cdots, x_N)$ and for each $x$, the choice is either $1$ or $0$. Then the number of corners equal to $2^N$. Therefore, the number of corners in 4 dimesions is $16$.

Question 2: How many edges does a cube have in 4 dimensions?

Solution:

Notice that there are two useful invariants.

• Each edge is shared by exactly $2$ corners.
• Each corner is connected to $N$ corners.

Then the number of edges of a cube in $N$ dimensions is $2^N \cdot N \cdot \frac{1}{2} = 2^{N-1}\cdot N$. Therefore, the number of edges is $2^{4-1} \cdot 4 = 32$.

Question 3: How many 3D faces does a cube have in 4 dimensions?

Solution: Let's first consider the question, "How man 2D faces does a cube have in 3 dimensions?" This answer is $3 * 2 = 6$. $3$ means 3 dimensions and $2$ means positive and negative directions.

Now, by a similar reasoning, the number of 3D faces is $4*2=8$.

Question 4: How many 2D faces does a cube have in 4 dimensions?

Solutions:

Notice that there are two useful invariants.

• Starting from a corner, the number of faces between edges is $C_{N}^{2}$.
• Each face is shared by $4$ corners.

Then the numebr of 2D faces of a cube in $N$ dimensions is $C_{N}^{2} \cdot 2^{N} \cdot \frac{1}{4} = C_{N}^{2} \cdot 2^{N-2}$. Therefore, the number of 2d faces of a cube in 4 dimensions is $6*16/4=24$.