TITLE: Can we use mathematical induction when induction basis is 'too' broad?
QUESTION [6 upvotes]: Imagine I wanted to use the induction principle to prove that a certain proposition was true for diagonal matrices. The induction is done in the dimension of the matrix.
So, the induction basis is for $n=1$. However, at this basis, there is no difference between a diagonal matrix and a non-diagonal matrix. So, how can we be sure that the reason it's right is due to being a diagonal matrix, and not because it's non-diagonal matrix. Shouldn't the basis be $n=2$?
Any help would be appreciated.

REPLY [10 votes]: The induction principle you are trying to use is:

Suppose that $P(1)$ is true and that if $P(n)$ is true then $P(n+1)$ is true.  Then $P(n)$ is true for all $n$.

In this case, $P(n)$ is the statement

For all diagonal matrices of dimension $n$, [...something...]

You are confused about showing that $P(1)$ is true, because every matrix of dimension $1$ is diagonal.  The good news is that you needn't be worried about this - if you show that the statement is true for all $1\times 1$ matrices, then you've proved $P(1)$.
You worry that:

So, how can we be sure that the reason it's right is due to being a diagonal matrix, and not because it's non-diagonal matrix

But you needn't worry about this.  The principle of induction says nothing about the reason that something is true.
In your case, it turns out that all $1\times 1$ matrices have your property, but once you switch to $2\times 2$ matrices, only the diagonal matrices have that property.  But you've proved that if all $n\times n$ diagonal matrices have the property, then all $(n+1)\times(n+1)$ diagonal matrices have that property.  So you can get from case $1$ to case $2$ as follows:

All $1\times1$ matrices have the property $\Rightarrow$
All diagonal $1\times 1$ matrices have the property (as you know, this is actually equivalent) $\Rightarrow$
All diagonal $2\times2$ matrices have the property (using the induction rule) $\Rightarrow$
All diagonal $3\times3$ matrices have the property (using the induction rule again) $\Rightarrow$
and so on