# Show that every matrix of order >1 is the sum of two singular matrices.

Show that every matrix of order >1 is the sum of two singular matrices.

Let
\begin{align*} A= \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \vdots & \vdots \\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix} \end{align*} and assume that A is of order $>1$.

I think that singular matrices $B,C$ in this problem

\begin{align*} B= \begin{bmatrix} 0 & 0 & \cdots & 0\\ a_{2,1} & a_{2,2} & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots \\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix}, \: \: \: C=\begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ 0 & 0 & \cdots & a_{2,n}\\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix} \end{align*} Then, it satisfies $B+C=A$ and $B,C$ are singular matrices.

However, I don't know why A's order $>1$..

Any help is appreciated!!

Thank you!

• A "matrix" of order $1$ is just a real number and the only singular "matrix" with order $1$ is $0$, but $0+0\ne 1$ , so $1$ (for example) does not have the desired representation. – Peter Dec 9 '17 at 13:55
• You got the idea, but your answer is slightly incorrect: $a_{2,2}$ appears in both $B$ and $C$. – Jean-Claude Arbaut Dec 9 '17 at 13:56
• For order greater than $1$, it is enough to replace one row or column by zeros and to subtract this matrix from the given matrix. – Peter Dec 9 '17 at 14:00
• Thank you!! I modify my answer. And, my curiosity about problem's condition is solved. – bluejyellow Dec 9 '17 at 14:02

Note that for order 1 $$det(a)=0 \iff a=0$$
Thus I can only write $0$ as sum of two singular matrices $0+0=0$.