Determine Sub-Vector Spaces of W & V

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The discussion centers on determining whether the sets W and V are sub-vector spaces and calculating their dimensions. Set W, defined as W={\begin{pmatrix} a &2b \\ c-b &b+c-3a \end{pmatrix}|a,b,c\epsilon \mathbb{R}}, is confirmed to be a subspace with a dimension of 3, spanned by the vectors (1,0,0,-3), (0,2,-1,1), and (0,0,1,1). For set V, defined as V=ax^{2}+bx+c|(a-2b)^{2}=0, the discussion emphasizes verifying closure conditions and finding a linear relationship among a, b, and c to establish if it is a subspace.

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Yankel
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Hello all,

I have two sets:

\[W={\begin{pmatrix} a &2b \\ c-b &b+c-3a \end{pmatrix}|a,b,c\epsilon \mathbb{R}}\]

\[V=ax^{2}+bx+c|(a-2b)^{2}=0\]I need to determine if these sets are sub vector spaces and to determine the dim.

I think that W is a sub space and dim(W)=3 (am I right?)

I don't know how to approach V...any help will be appreciated
 
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The set of all 2x2 real matrices can be viewed as a vector space of dimension 4.

In this view, W becomes the space spanned by:

(1,0,0,-3), (0,2,-1,1) and (0,0,1,1).

Is this a linearly independent set?

For the second set, it might be good to verify the closure conditions, first (the 0-polynomial is obviously a member of $V$).

Suppose:

$p(x) = ax^2 + bx + c$
$q(x) = dx^2 + ex + f$

are both in $V$.

Does $(p+q)(x) = (a + d)x^2 + (b+e)x + (c+f)$ satisfy:

$[(a+ d) - 2(b+e)]^2 = 0$?

If $k$ is any real number, does:

$(ka - 2kb)^2 = 0$ (that is, is $(kp)(x) \in V$)?

(Hint: if $r^2 = 0$, then $r = 0$).

Can you find a LINEAR relationship between $a,b,c$?

Finally, PROVE your answer by exhibiting a basis, if $V$ is indeed a subspace.
 

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