Question

let v(vector space) be a collection of all 2*3 matrices with real entries such that v={[a11 a12 a13:a21 a22 a23]| a11 +a23 =1}
determine whether the following vector spaces axioms hold on v .........still to post questions

Answer 1

\[V=\left\{\left[\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{matrix}\right]|a_{11}+a_{23}=1\right\}\]correct?

Answer 2

questions .....
(a)\[ni 0 vector in V\] such that \[alpha + 0 vector = 0 vector + alpha = alpha,forall alpha in V\]

Answer 3

\[exists 0 vector in V\]

Answer 4

Given\[V=\left\{\left[\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{matrix}\right]|a_{11}+a_{23}=1\right\}\]Find if\[\exists\vec0\in V:\vec\alpha+\vec0=\vec0+\vec\alpha=\vec\alpha,\forall\vec\alpha\in V\]right?

Answer 5

I don't see how this could be true given that \(a_{11}+a_{23}=1\)

Answer 6

oops...y is it not showing the equations??

Answer 7

try reloading

Answer 8

itz now ok....the plzvshow me da steps for saying itz not true

Answer 9

in trying to get a zero vector I would have something like\[\vec0=\left[\begin{matrix}a_{11}&0&0\\0&0&1-a_{11}\end{matrix}\right]\]but clearly this will not give\[\vec0+\vec\alpha=\vec\alpha\]I don't know if that is enough proof for your prof but it convinces me lol

Answer 10

i understand a lil bit ....but just dont get it how did u get that 0 (vector) matrix

Answer 11

well I tried to make all elements zero, but given that\[a_{11}+a_{23}=1\]we have that\[a_{23}=1-a_{11}\]so all entries will be of the form\[\vec\alpha=\left[\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&1-a_{11}\end{matrix}\right]\]

Answer 12

0⃗ +α⃗ =α⃗ from this equation i dont get it ..how will we represent vector 0...do we use other vector or just take 0 (vector ) as a 2*3 matrix containing all zeros

Answer 13

normally yes, \[\vec0=\left[\begin{matrix}0&0&0\\0&0&0\end{matrix}\right]\]but since in this case \[a_{11}+a_{23}=1\implies a_{23}=1-a_{11}\]we know that all vectors will have the form\[\left[\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&1-a_{11}\end{matrix}\right]\]so the best we could do is\[\left[\begin{matrix}a_{11}&0&0\\0&0&1-a_{11}\end{matrix}\right]\]which doesn't help, so there exists no zero vector\[\vec0:\vec\alpha+\vec0=\vec\alpha\]

Answer 14

i got prof ...thanks

Answer 15

i had encountered a problem again when trying to implement 0+a=a ....it does not give me the solution dnt understand why

Answer 16

that's because there is no zero vector in this space

Answer 17

firstly should i take a +b=a then solve for b

Answer 18

right, that should work

Answer 19

\[\vec\alpha+\vec\beta=\vec\alpha\]\[\left[\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&1-a_{11}\end{matrix}\right]+\left[\begin{matrix}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&1-b_{11}\end{matrix}\right]=\left[\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&1-a_{11}\end{matrix}\right]\]\[\left[\begin{matrix}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&1-b_{11}\end{matrix}\right]=\left[\begin{matrix}0&0&0\\0&0&0\end{matrix}\right]\]this implies that\[b_{11}=1-b_{11}=0\]which is impossible, a contradiction. Hence there is no zero vector in this vector space.

Answer 20

with the first 1 im ok....and the on same statement for verifying -a+a=0 ...
does it holds for b23 =-a23

Answer 21

if you want to verify -a+a=0 the you only need the fact that each element in a is the negative of each element in -a

Answer 22

if you want you could write\[a_{23}=1-a_{11}\]\[-a_{23}=a_{11}-1\]\[a_{23}-a_{23}=1-a_{11}-(a_{11}-1)=0\huge~~~~~~~~~~~~~~~~~~~\checkmark\]but that is unnecessary.

Answer 23

This also proves that the space is not closed under addition since we know that the zero vector is not in it. Hence this is not even a vector space.