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I'm currently looking at problem #2. I wrote $W$ in the form

$$W=a\begin{bmatrix} 2\\ 1\\ 4\end{bmatrix}+b\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}+c\begin{bmatrix}-1\\ 1\\ 1\end{bmatrix}$$

and then I took the determinant of the matrix formed by these vectors, that is, $$\det\begin{bmatrix}2 & 1 &-1 \\ 1& 0& 1\\ 4& 1&1 \end{bmatrix}=0$$

Yet the study guide solutions say

$$ W=Col \begin{bmatrix}2 & 1 &-1 \\ 1& 0& 1\\ 4& 1&1 \end{bmatrix}$$ is a subspace of $\mathbb{R}^3$. I don't understand how it can be if these vectors are not linearly independent as the determinant is zero. What am I missing?

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    $\begingroup$ The determinant being zero just means that the three vectors are not linearly independent. But they still span a subspace of $\mathbb R^3$, the representation of the vectors in this subspace as a linear combination of the three vectors is just not unique. The dimension of this subspace is smaller than $3$ - that's the only thing the determinant tells you. $\endgroup$
    – Lukas
    Mar 16, 2021 at 21:58
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    $\begingroup$ How should one go about determining whether a set is a subspace? $\endgroup$ Mar 16, 2021 at 22:02
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    $\begingroup$ There are three things to check: is the zero-vector an element of the span of the three vectors (yes, it is, e.g. by choosing $a=b=c=0$ - but there are more possibilities in this case). And you have to check if: $v, w \in W \Rightarrow v+w \in W$ and $w \in W, a \in \mathbb R \Rightarrow av \in W$. $\endgroup$
    – Lukas
    Mar 16, 2021 at 22:04
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    $\begingroup$ The zero vector all by itself is a subspace of every vector space. It will almost never span the entire space. $\endgroup$ Mar 16, 2021 at 22:11

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The fact that they are not linearly independent does not mean that W is not a subspace. It is only mean that they are not the basis of the subspace - they aren't the minimal set that span w. . In order to show that W is subspace, you only need to take two vector, u and v in w, and show that u+v contains in w.

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Okay so you got that $W$ is exactly equal to the solution set of this equation.

$$W=a\begin{bmatrix} 2\\ 1\\ 4\end{bmatrix}+b\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}+c\begin{bmatrix}-1\\ 1\\ 1\end{bmatrix}$$

But this is saying that $W$ is equal to the span of those three vectors. Anytime you have the a subset defined by the span of a set of vectors, that subset is a subspace.

Theorem: Let $W = span\{v_1,..,v_m\}$ be in $\mathbb{R}^n$. Then $W$ is a subspace of $\mathbb{R}^n$

Now, what YOU did was check to see if those vectors are linearly independent. Since $det=0$, they are not linearly independent. That just means that $W$ is not going to be a $3$ dimensional subspace (and so it isn't equal to the whole space, but rather it is a PROPER subspace).

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    $\begingroup$ Thanks! This makes sense. $\endgroup$ Mar 16, 2021 at 22:15
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    $\begingroup$ It better! I'm teaching linear algebra this semester... my students have a test tomorrow ugh!! $\endgroup$
    – user637978
    Mar 16, 2021 at 22:15
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    $\begingroup$ Haha, I have an exam today! $\endgroup$ Mar 16, 2021 at 22:16
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You have already written $W= a.x+b.y+c.z$, where $x=[2~ 1 ~4]^t$ $y= [1~ 0~ 1]^t$ and $z= c[-1~ 1~ 1]^t$, $a,b,c \in \mathbb{R}$.

That simply means $W$ is given by the column space of the mentioned matrix and that is what you guide's solution addresses. Note that column space of the matrix is a subspace of $\mathbb{R}^3$, since the matrix can be thought of a linear transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$ and colum space of the matrix is the range of the linear transformation.

Note that "determinant of matrix is zero" means that the vectors $x$, $y$, $z$ are not linearly independent. However, it does not deny the fact that $W$ is spanned by them. Of course, the set of vectors $\{x,y,z\}$ is not the minimal spanning set of $W$, i.e., not a basis of $W$. If it was, you would have obtained the determinant to be non-zero.

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    $\begingroup$ This makes perfect sense. The column space is the set of all linear combinations of the columns, which is equal to the span of the constituent column vectors, which is a subspace. So this is obvious now. $\endgroup$ Mar 16, 2021 at 22:14
  • $\begingroup$ I'm a bit confused on part (b) of the question above. How would I determine if that vector is a subspace of $ \mathbb{R}^4$? $\endgroup$ Mar 16, 2021 at 22:36

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