Proving T is a Subspace of M23: Linear Algebra Problem

[Warpenguin]

Hi there. I started learning about subspaces in linear algebra and I came across a question which I'm unsure how to solve. I understand that there are 'rules' which need to be passed in order for something to be a subspace, but I have no idea how to start with this problem:

Consider the set M23 of all 2 × 3 matrices with real entries under the usual operations of matrix addition and scalar multiplication.
Let
T=([a b c] : a + c= 0 and b + d + f =0)
[d e f]
Prove that T is a subspace of M23
(T is a 2x3 matrix if i made it unclear)
I know that T must contain a zero vector and I know that there must be closer of scalar multiplication and addition.

Can anyone help?

 

[micromass]

So there are three things that you need to do:

• T contains the zero matrix. Is this true.
• T is closed under composition. Thus if
  
  \left(\begin{array}{ccc} a & b & c\\ d & e & f\end{array}\right)~\text{and}~\left(\begin{array}{ccc} a^\prime & b^\prime & c^\prime \\ d^\prime & e^\prime & f^\prime\end{array}\right)
  
  are in T, then
  
  \left(\begin{array}{ccc} a & b & c\\ d & e & f\end{array}\right) + \left(\begin{array}{ccc} a^\prime & b^\prime & c^\prime \\ d^\prime & e^\prime & f^\prime\end{array}\right)
  
  is in T
• T is closed under scalar multiplication, thus if
  
  \left(\begin{array}{ccc} a & b & c\\ d & e & f\end{array}\right)
  
  is in T, then for each \alpha
  
  \alpha\left(\begin{array}{ccc} a & b & c\\ d & e & f\end{array}\right)
  

Let's start with the first. Why is the zero matrix in T?

 

[Warpenguin]

Well i'd say that it has a zero matrix because a + c = 0 (which can be 0 + 0 = 0) and b + d + f = 0 (i.e 0 + 0 + 0 =0) i.e a and d = 0. Is this correct?

 

[micromass]

Well i'd say that it has a zero matrix because a + c = 0 (which can be 0 + 0 = 0) and b + d + f = 0 (i.e 0 + 0 + 0 =0) i.e a and d = 0. Is this correct?

Yes, that is correct.
Now, why is it closed under addition?

 

[Warpenguin]

I'm not sure if this is the way to do it but I would say
u=[a1]
[d1]
v=[a2]
[d2]

i.e if u and v are vectors in T, then u+v will be a vector in T.

u+v=[a1+a2]
[d1+d2]
a1+a2+d1+d2=0
(a1+a2)+(d1+d2)=0
0 + 0 = 0

I'm not sure if it can be written like that

 

[micromass]

Why do you write u and v like that?? u and v are 3x2 matrices! But you seem to write them as 2x1 matrices...

 

[Warpenguin]

Thats what i meant when i said i wasn't sure lol. So would it be correct if i wrote u and v as 3x2 matrices containing all the elements of T and writing them as a1+a b1 +b2 etc.. what would be the next step? can i still use this after:
a1+a2+d1+d2=0
(a1+a2)+(d1+d2)=0
0 + 0 = 0
or must i use all elements from u+v?

 

[micromass]

Thats what i meant when i said i wasn't sure lol. So would it be correct if i wrote u and v as 3x2 matrices containing all the elements of T and writing them as a1+a b1 +b2 etc.. what would be the next step? can i still use this after:
a1+a2+d1+d2=0
(a1+a2)+(d1+d2)=0
0 + 0 = 0
or must i use all elements from u+v?

You must write u and v as

u= \left(\begin{array}{ccc} a_1 & b_1 & c_1\\ d_1 & e_1 & f_1\end{array}\right)~ \text{and}~ v=\left(\begin{array}{ccc} a_2 & b_2 & c_2\\ d_2 & e_2 & f_2 \end{array}\right)

Now, what is u+v??

 

[Warpenguin]

Yes that's what I meant to say xD

u+v=[a1+a2 b1+b2 c1+c2]
[d1+d2 e1+e2 f1+f2]

 

[micromass]

Yes that's what I meant to say xD

u+v=[a1+a2 b1+b2 c1+c2]
[d1+d2 e1+e2 f1+f2]

Yes. Now, does

\left(\begin{array}{ccc} a_1+a_2 &amp; b_1+b_2 &amp; c_1+c_2\\ d_1+d_2 &amp; e_1+e_2 &amp; f_1+f_2\\<br /> \end{array}\right)

belong to T? That is, is (a_1+a_2)+(c_1+c_2)=0 and (b_1+b_2)+(d_1+d_2)+(f_1+f_2)=0??

 

[Warpenguin]

Yes because (a1+a2)=0, (c1+c2)=0, (b1+b2)=0, (d1+d2)=0, (f1+f2)=0
0 + 0 = 0 and 0+0+0=0

 

[micromass]

OK, now do the same thing with scalar multiplication.

 

[Warpenguin]

Hmm like this?

ka + kc = 0
k(a + c) = 0
k(0) = 0

kb + kd + kf = 0
k(b + d + f) = 0
k(0) = 0

 

[micromass]

Hmm like this?

ka + kc = 0
k(a + c) = 0
k(0) = 0

kb + kd + kf = 0
k(b + d + f) = 0
k(0) = 0

That's it!

 

[Warpenguin]

Awesome! Thanks so much! I understand this much better than i did several hours ago :D