[Math] Do all singular $n\times n$ matrices form a vector subspace when $n\ge2$

linear algebramatrices

I want to prove or disprove that the set of all $n\times n$ singular matrices form a vector subspace of $M_{nn}$ when $n\geq 2$. So, let:

$$
A_{n,n} =
\begin{pmatrix}
a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\
a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n,1} & a_{n,2} & \cdots & a_{n,n}
\end{pmatrix}
,\text{ where }a_{i,j}\in \mathbb R$$

I can't just take the case where $A_{n,n} = $ the zero vector and show the space is non-empty, closed under addition and scalar multiplication, can I?
this is only one case of $\det(A) =0$
so how can I prove/disprove this for all cases where $\det(A) =0$

Thanks in advance.

Best Answer

Hint. What do you get when you add $$\pmatrix{1&0\cr0&0\cr}\quad\hbox{and}\quad \pmatrix{0&0\cr0&1\cr}\ ?$$

Note that

  • to prove a set is closed under addition you must show that if $M,N$ are in the set then $M+N$ is always in the set;
  • to show a set is not closed under addition, you must find one example of $M,N$ in the set, such that $M+N$ is not in the set.
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