Prove $A-I $ is not invertible.

Question

If A=($a_{ij}$) is a nxn matrix of real numbers such that $\sum_j^n$ $a_{ij}$=1 for each i, show that the matrix A-I is not invertible. I am not very good at proofs and have previously heard you cannot prove a negative so I am wondering if this is solvable.

Answer 1

I'm assuming you mean the sum from $j=1$ to $n$ is equal 1, i.e. the matrix is stochastic. If this is the case, consider what happens when you multiply $A-I$ by a vector where all entries are one. Certainly it is not true that you cannot prove a negative statement.

Answer 2

I have no idea what you mean about "can't prove a negative"; there are numerous negative statements that can be proven.

Here's a hint for this problem: a square matrix $B$ is not invertible if and only if there is a nonzero $x$ such that $Bx=0$. Consider $B=A-I$. By multiplying it out you find that $Bx$ is a vector whose $i$th component is $\sum_{j=1}^n b_{ij} x_j$. Since you subtract a $1$ in each row, $\sum_{j=1}^n b_{ij} = 0$ for every $i$. So can you choose $x_j$ such that $\sum_{j=1}^n b_{ij} x_j = \sum_{j=1}^n b_{ij} = 0$?