Complete the proof of the following theorem. Show that the described bijection is indeed a bijection; show that this bijection is both an injection and a surjection.
Theorem: For any integers $k$ and $n$ with $n$ $\geq$ $k$ $>$ $1$, the number of partitions of $n$ into $k$ parts, $p_{k}(n)$, is the number of partitions of $(k-1)n$ into $k$ parts that are all less than $n$.
Proof:
Let $\lambda$ be a partition of $n$ into $k$ parts. Since $k$ exceeds $1$, each part is less than $n$. Put the Ferrers diagram of $\lambda$ in a k$\times$n grid of squares so as to cover the northwest corner. Note that there are $kn$ squares in all, $n$ of which are occupied by squares of $\lambda$. Therefore, the complementary squares, when rotated $180^{\circ}$, form the Ferrers diagram of a partition of $(k-1)n$ into $k$ parts, each of which is less than $n$. It is easy to see that the function defined in this way is indeed a bijection from the set of partitions of $n$ into $k$ parts onto the set of partitions of $(k-1)n$ into $k$ parts, all of which are less than $n$.
I am not entirely sure what this is saying, but to I just show that this is just simply a transpose of a matrix?
What I have:
Let the elements of the matrix $A$=$a_{ij}$, where $i$ represents the row number and $j$ represents the column number. So $A^{t}$= $a^{ji}$, and so the transformation is one-to-one. A function is onto if and only if for every $y$ in $A^{t}$ there exists at least one $x$ in $A$ such that $f(x) = y$. Taking, the empty parts of the matrix to be $0$, if we transpose the matrix, then it will be onto.
Sorry if this is bad.