Is the sum of singular and nonsingular matrix always a nonsingular matrix?

Not true even for positive matrices: $$ \begin{pmatrix}1 & 1\\2 & 2\end{pmatrix}+ \begin{pmatrix}3 & 2\\2 & 1\end{pmatrix}= \begin{pmatrix}4 & 3\\4 & 3\end{pmatrix}. $$

No. Consider the matrices $$ A = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}, B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$ Then:

• $A$ is regular.
• $B$ is singular.
• $A + B = \binom{2\,0}{0\,1}$ is regular.
• $A - B = A+(-B) = \binom{0\,0}{0\,1}$ is singular.

Let me tell one particular way of generating lots of examples. We will find $A$ such that $A +I$ will be singular. You can easily adapt this method to use with any non-singular matrix instead of identity. We will work backwards to get solutions.

Take a matrix with two identical rows as $A+I$. This gives the condition that $$A+\pmatrix{1 &0&0\cr0&1&0\cr 0&0&1\cr}=\pmatrix{a & b &c \cr a &b & c\cr * & * & *\cr}$$ First two rows of $A$ are forced. To ensure singularity of $A$ make the last row identical to 2nd row:

$$\pmatrix{a-1 & b& c\cr a &b-1 & c\cr a & b-1 & c} +\pmatrix{1 &0&0\cr0&1&0\cr 0&0&1\cr}=\pmatrix{a & b &c \cr a &b & c\cr * & * & *\cr} $$ Now we can work backwards to get the values to be used in place of stars: $$\pmatrix{a-1 & b& c\cr a &b-1 & c\cr a & b-1 & c} +\pmatrix{1 &0&0\cr0&1&0\cr 0&0&1\cr}=\pmatrix{a & b &c \cr a &b & c\cr a & b-1 & c+1\cr} $$ Now replace $a,b,c$ with your ATM pin number, your friend's age, and your annual salary in Euros respectively, you will get a solution.