I am given the following

$V = \mathbb R^4$

$W = \{(w,x,y,z)\in \mathbb R^4|w+2x-4y+2 = 0\}$

I have to prove or disprove that $W$ is a subspace of $V$.

Now, my linear algebra is fairly weak as I haven't taken it in almost 4 years but for a subspace to exist I believe that:

1) The $0$ vector must exist under $W$

2) Scalar addition must be closed under $W$

3) Scalar multiplication must be closed under $W$

I don't think the first condition is true because if I were to take the vector, there is no way I can get the zero vector back. Is that correct or am I doing something very wrong?

  • $\begingroup$ Why the downvote? $\endgroup$
    – mrnovice
    Mar 5 '17 at 9:13
  • 1
    $\begingroup$ Possible duplicate of How to prove that a subspace is in a vector space? $\endgroup$ Mar 5 '17 at 9:15
  • $\begingroup$ No. I did not know that question existed. Sorry. Also, that question uses transposes. Mine doses not. $\endgroup$ Mar 5 '17 at 9:17
  • $\begingroup$ Yeah. $0$ vector is not in $W$, so its not even a vector space. $\endgroup$
    – rookie
    Mar 5 '17 at 9:18
  • $\begingroup$ It is a bit different question, you can figure out infinite number of simple examples of a set which is not a vector space :-) For this one, you are right, the first condition fails. $\endgroup$ Mar 5 '17 at 9:21

You're right in what you think, but as how you argue that I am not sure what you mean by "getting the zero vector back"? Back...from where?

Anyway, if we take $\;(w,x,y,z)=(0,0,0,0)\;$ , this vector belongs to $\;W\;$ iff

$$0+2\cdot0-4\cdot0+2=0\iff 2=0$$

and since the last equality is false we get that $\;(0,0,0,0)\notin W\implies W\;$ is not a subspace and you were right.

  • $\begingroup$ I mean like if I plug in 0 into the vector, I don't get 0. I get 2 if w x and y are all 0. $\endgroup$ Mar 5 '17 at 9:20
  • $\begingroup$ @SubhashisChakraborty you don't plug in 0 into a vector, you plug in the zero vector into the defining equation of $W$. (An important notational difference.) $\endgroup$ Mar 5 '17 at 17:21
  • $\begingroup$ Ahh thanks for clearing that up. $\endgroup$ Mar 5 '17 at 23:17

No, it is not a subspace.

It is because you have to verify your first point: "$0\in W$".

But $(0,0,0,0)\notin W$ because

$$0+2\times 0-4\times 0+2\ne 0.$$

  • $\begingroup$ Thanks. My linear algebra is fairly rusty so I just needed to be sure. $\endgroup$ Mar 5 '17 at 9:27

Let me give you a more general result: Consider a linear system of equations $$\begin{cases} a_{11}x_1 + a_{12}x_2 + \ldots + a_{1n}x_n &= b_1\\ a_{21}x_1 + a_{22}x_2 + \ldots + a_{2n}x_n &= b_2\\ &\vdots&\\ a_{m1}x_1 + a_{m2}x_2 + \ldots + a_{mn}x_n &= b_m \end{cases},$$ with $m \leq n$. Suppose one of the $b_i \neq 0$, then clearly this system of equations does not form a subspace, since in the $i$th equation, we would find that (filling in the zero-vector $(0, 0, \ldots, 0)$ that $0 = b_i \neq 0$, which is not possible.

However, if all of the $b_i$ are zero, then this forms a subspace: clearly the zerovector is a solution to all equations and hence to the system. If both $(y_1, \ldots, y_n), (z_1, \ldots, z_n)$ are solutions, then so is $(y_1 + z_1, \ldots y_n + z_n)$ (in order to see this: we fill in this vector and use distributivity of the product over the summation). Analogously, $(\lambda y_1, \ldots, \lambda y_n)$ for some $\lambda \in \mathbb{R}$ will form a solution (note that $(\lambda y_1, \ldots, \lambda y_n) = \lambda (y_1, \ldots, y_n)$).

  • $\begingroup$ This makes a lot of sense. Thanks Why is it required to assume m is less than or equal to n though? $\endgroup$ Mar 5 '17 at 9:56
  • $\begingroup$ Note that there are only $n$ variables which we want to find. So if we take $m > n$, we have an overdetermined system of linear equations. If there would be some $b_i \neq 0$, it is possible that there is no solution (however, then we have the empty set, which is not a vectorspace either, so in this case it does not really matter). In case all the $b_i = 0$, then there will be equations which are linearly dependent, so we have some redundant equations. Therefore, we can assume $m \leq n$. $\endgroup$
    – Student
    Mar 5 '17 at 10:29
  • $\begingroup$ However, you might need some result from linear algebra to show this. The dimension of $\mathbb{R}^n$ is $n$, so if we would write this system of linear equations in a matrix, then we could rowreduce this matrix and because of the dimension, we will always have that the rank of the matrix would be $n$ or less. This means that if we would have $m$ equations (which translates to $m$ rows in the matrix), we would find zerorows (and this corresponds to redundant equations in the system). $\endgroup$
    – Student
    Mar 5 '17 at 10:31
  • $\begingroup$ The keypoint is that we can assume, without loss of generality, that $m \leq n$ (however, without this assumption, the things I wrote are still valid). $\endgroup$
    – Student
    Mar 5 '17 at 10:32
  • $\begingroup$ That makes sense. Thanks a lot :) . $\endgroup$ Mar 5 '17 at 10:40

I feel I should indicate that condition 2 and condition 3 fails as well. The vector $v = (2, 0 , 0 , 0)$ belongs in the subset, but neither $2v$ nor $v+v$ belongs in the subset.

It is thus not a subspace.

  • $\begingroup$ Yes I realize that as well. Thank you. $\endgroup$ Mar 5 '17 at 23:18

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