Let V = 2-by-2 matrix over Q and let W = {A E V|det(A)=0}

[druzizzle619]

I need some help with this proof, as I am bad at doing these things:

Let V = 2-by-2 matrix over the rationals and let W = { A E V | det(A) = 0}. Show that W is not a subspace of V.
For the proof, I understand that I have to show that either the sum of two of the vectors or a scalar multiplied by a vector is not in the vector space, but I don't know how to go about it. Help please?

 

[pka]

Re: Linear Algebra Proof

Let V = 2-by-2 matrix over the rationals and let W = { A E V | det(A) = 0}. Show that W is not a subspace of V.

This may be counterintuitive. But to prove some proposition is not true we must produce a counter-example. Consider these two matrices:
\(\displaystyle \left[ {\begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} } \right]\,\& \,\left[ {\begin{array}{lr} 1 & { - 1} \\ 0 & 0 \\ \end{array} } \right]\).

Are both of these in the set W?
What about their sum?

 

[druzizzle619]

Re: Linear Algebra Proof

Oh, alright then. Would a proof by contradiction work in that sense?

 

[pka]

Re: Linear Algebra Proof

Oh, alright then. Would a proof by contradiction work in that sense?[/quote]
NO! I have told you: Find a counter-example.
A very basic principle of scientific logic is: A negative cannot be proven.
What we can do is to produce a counter-example.
It happens to be the difference in a universal in counter-distinction to the existential.
This goes to the heart of proof theory.