Question

Is the set of all 2x2 matrices of the form (row1:[a b] row2:[a+b 0]) a subspace of the set of all 2x2 matrices?

Answer 1

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Answer 2

like this?

\[M= \left[ \begin{array}{c}
a & b
\\ a+b & 0
\end{array} \right] \]



(( reference /use
\[\text{\[M= \left[ \begin{array}{c}
a & b
\\ a+b & 0
\end{array} \right] \] } \] ))

Answer 3

yes

Answer 4

hard to write like that :P

Answer 5

no

Answer 6

Rephrased code:
\[\mbox{\[M=}\color{blue}{\mbox{\left[}}\color{green}{\mbox{\begin{array}{c}}}\mbox{ a & b \\ a+b & 0 }\color{green}{\mbox{\end{array}}} \color{blue}{\mbox{\right]}}\mbox{\] }\]

Answer 7

i mean coloured*

Answer 8

:'(

Well darn.

Answer 9

addition is coordinate wise

Answer 10

\[\left[\begin{array}{cc} \alpha & \beta \\ \gamma & \delta \\ \end{array}\right]+\left[\begin{array}{cc} \epsilon & \zeta \\ \eta & \theta \\ \end{array}\right]=\left[\begin{array}{cc} \alpha+\epsilon & \beta+\zeta \\ \gamma+\eta & \delta+\theta \\ \end{array}\right]\]

Answer 11

OH! thanks. Brain completely jumbled that up.

Answer 12

\[M_1 + M_2 = \left[ \begin{array}{c} a_1 & b_1 \\ a_1+b_1 & 0 \end{array} \right] + \left[ \begin{array}{c} a_2 & b_2 \\ a_2+b_2 & 0 \end{array} \right] \]
\[ = \left[ \begin{array}{c} (a_1+a_2) & (b_1+b_2) \\ (a_1+b_1)+(a_2+b_2) & 0\end{array} \right] \]

So that does work. And so will multiplication. also
\[M= a\left[ \begin{array}{c} 1 & 0 \\ 1 & 0 \end{array} \right] +b\left[ \begin{array}{c} 0 & 1 \\ 1 & 0 \end{array} \right]\]

so it's a subspace?

Answer 13

~vector multiplication

\[rM= r\left[ \begin{array}{c} a & b \\ a+b & 0 \end{array} \right]= \left[ \begin{array}{c} ra & rb \\ r(a+b) & 0 \end{array} \right] \]

Answer 14

those all seem like trivial things to do, however. I was just basing this off another proof I found, but it doesn't seem particularly proof-y....

Answer 15

\[=\left[\begin{array}{cc} ra & rb \\ ra+rb & 0\\ \end{array}\right]\]

now it is of the form as the original matrix


should probably include this little step

Answer 16

also, one should show that you can get the zero vector (zero matrix in this case)

Answer 17

Does the ability to separate it into

\[M= a\left[ \begin{array}{c} 1 & 0 \\ 1 & 0 \end{array} \right]+b\left[ \begin{array}{c} 0 & 1 \\ 1 & 0 \end{array} \right]\]
automatically show that it's a subspace? Or do you have to show those other operations as well? It almost seems more like you're defining those properties onto the matrices more than you're proving that it works - or is it just that this is a simple example?

Lastly,
\[ 0M= 0\left[ \begin{array}{c} a & b \\ a+b & 0 \end{array} \right]=\left[ \begin{array}{c} 0(a) & 0(b) \\ 0(a+b) & 0 \end{array} \right]=\left[ \begin{array}{c} 0 & 0 \\ 0(a)+0(b) & 0 \end{array} \right]=\left[ \begin{array}{c} 0 & 0 \\ 0 & 0 \end{array} \right]\]
?

Answer 18

That last bit is wrong methinks. That's just the multiplication. Would you have to define
a=a+b=b =0 ?

Answer 19

for your first question...No

you are just showing that your matrix can be written as a linear combination of those two vectors. That does not show it is a subspace

Answer 20

right...just let a=b=0 and you get the zero vector

Answer 21

Ok, thanks!

So, for instance, if it were
\[M= \left[ \begin{array}{c} a+1 & b \\ a+b & 0 \end{array} \right]\]
since the zero matrix isn't in M, then that would not a subspace of the set of all 2x2 matrices?

Answer 22

Thanks for all the help, @Zarkon ^_^

Answer 23

**
Is the set W of all 2x2 matrices of the form (row1:[a b] row2:[a+b 0]) a subspace of the set of all 2x2 matrices?
**

I would say yes. W is a subspace iff
The zero vector, 0, is in W.
If u and v are elements of W, then the sum u + v is an element of W;
If u is an element of W and c is a scalar from K, then the product cu is an element of W;


you have the zero matrix, (set a = b= 0)

and, as posted above, any linear combination of u, and v (both in W) result in a matrix contained in W.

Answer 24

@AllTehMaffs

right...that set \(M\) you just made does not contain the zero vector so it can't be a subspace