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Three vector spaces are given, and each one has a set that is potentially the subspace of the vector space.

$$ (1) \quad V = \mathbb{R}^4 \quad W=\begin{Bmatrix} \begin{bmatrix} a \\ -b \\ 2a+b \\ a \end{bmatrix}\space \bigg| \space a,b \in \mathbb{R} \end{Bmatrix} $$
$$ (2) \quad V = M_{3, 2}(\mathbb{R}) \quad W=\begin{Bmatrix} \begin{bmatrix} 0 & y \\ 6b & |y| \\ 0 & x \end{bmatrix}\space \bigg| \space x, y \in \mathbb{R} \end{Bmatrix} $$
$$ (3) \quad V = \mathcal{P}(\mathbb{R}) \quad W=\Big\{\space a - b(3-x)+c(x+5x^3) \space \bigg| \space a, b, c \in \mathbb{R} \space \Big\} $$

I am pretty sure all three W's are subspaces of corresponding V's. However, my professor told me to check it once again, as it is unlikely that all are subspaces.

I checked the three points:

  • If the Zero vector belongs to W.
  • If w1 and w2 are vectors in W, their sum must also be in W.
  • If w is a vector in W and c is some scalar in V, their product must also be in W.

However, this gave me a result I am not satisfied with.

Did I do something wrong?

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    $\begingroup$ The third point is false; it should be : if $w$ is a vector of $W$ and $\lambda$ a scalar, then $\lambda w$ is a vector of $W$. $\endgroup$
    – Abezhiko
    Mar 19 at 10:48
  • $\begingroup$ The second case is not a subspace : indeed, taking two matrix $A_\pm$ with $y = \pm1$, then one has $A_+ + A_- \not\in W$, because the 2nd component is $1 + (-1) = 0$, but the 4th one is $|1| + |-1| = 2 \neq |0|$. $\endgroup$
    – Abezhiko
    Mar 19 at 10:59
  • $\begingroup$ $\bullet$ "If w1 and w2 are vectors in W, their product must also be in W." - no, the product of vectors is not defined in general. Only the product of a vector with a scalar is always defined in a vector space. $\endgroup$
    – Dietrich Burde
    Mar 19 at 11:21
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    $\begingroup$ @AlexAramyan $$ \begin{pmatrix} 0 & \pm1 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} \in W \;\mathrm{but}\; \begin{pmatrix} 0 & 1 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & -1 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 2 \\ 0 & 0 \end{pmatrix} \not\in W $$ because the absolute value breaks down the linear structure. $\endgroup$
    – Abezhiko
    Mar 19 at 12:02
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    $\begingroup$ @AlexAramyan That's indeed the idea. However, be careful with the "column notation" without referring ot a basis; $(0,y)+(0,-y)$ has no meaning in the six-dimensional vector space $V$. $\endgroup$
    – Abezhiko
    Mar 19 at 13:31

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