# How to check whether a set is a subspace of a vector space?

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?

• 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 $W$. Mar 19 at 10:48
• 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 $|1| + |-1| = 2 \neq |0|$. Mar 19 at 10:59
• $\bullet$ "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. Mar 19 at 11:21
• @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. Mar 19 at 12:02
• @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 $V$. Mar 19 at 13:31