0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Peter Dec 9 '17 at 13:55
  • 1
    $\begingroup$ You got the idea, but your answer is slightly incorrect: $a_{2,2}$ appears in both $B$ and $C$. $\endgroup$ – Jean-Claude Arbaut Dec 9 '17 at 13:56
  • 1
    $\begingroup$ 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. $\endgroup$ – Peter Dec 9 '17 at 14:00
  • $\begingroup$ Thank you!! I modify my answer. And, my curiosity about problem's condition is solved. $\endgroup$ – bluejyellow Dec 9 '17 at 14:02
0
$\begingroup$

It's correct.

Note that for order 1 $$det(a)=0 \iff a=0$$

Thus I can only write $0$ as sum of two singular matrices $0+0=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.