Proof by mathematical induction for matrices Let $N=((n_{ij}))$ be a $n\times n$ matrix with entries $n_{ij}= 1$ for all $1\le i, j\le n$.
$(i)$ Show that $N^2 = nN$.
I would like to use mathematical induction for the proof.
For my base case, I let the square matrices to be $2 \times 2$, and then I assume it is true for all square $n \times n$ matrices.
However, when I am going to prove it is also true for all square matrices $(n+1)\times (n+1)$, I was stuck in presenting my proof, like how can I present my proof without drawing completely the square matrices $(n+1) \times (n+1)$ out?
 A: For some column, row, and $n\times n$ matrices consisting of only ones $x$, $y$, and $A$ respectively, your $(n+1)\times (n+1)$ matrix will have the form below. Now, you can square the matrix and use your induction hypothesis in the upper left slot. Be sure to note that this $1$ is not a scalar, but a $1\times 1$ matrix.
$
\begin{bmatrix}
A_{n\times n} & x\\
y & 1
\end{bmatrix}
$
A: This is not the induction you may be looking for, but it is an implicitly inductive argument:
Let $e_{j}$ be the vector with 1 in the jth position while zeros everywhere else. Note that $v_{1} = \begin{pmatrix} 1 \\ 1 \\\vdots \\1 \\1 \end{pmatrix} = e_{1} + e_{2} + \cdots + e_{n}$ is an eigenvector of $N$ associated with the eigenvalue $\lambda_{1} = n$.  And $v_{2} = e_{1} - e_{2}, v_{3} = e_{2} - e_{3}, \cdots, v_{n} = e_{n-1} - e_{n}$ are the $n -1$ eigenvectors associated with $\lambda_{2} = 0$. Because $v_{1}, v_{2}, \cdots, v_{n}$ form a linearly independent set, the minimal polynomial must be $m_{N}(x) = (x - \lambda_{1})(x - \lambda_{2}) = (x -n)x$.
Then by Cayley-Hamilton Theorem, we have $O_{n\times n} = (N -nI)N \implies N^2 = nN$.
