# Linear Independence and Subspaces

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?

• 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. Mar 16, 2021 at 21:58
• How should one go about determining whether a set is a subspace? Mar 16, 2021 at 22:02
• 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 $w \in W, a \in \mathbb R \Rightarrow av \in W$. Mar 16, 2021 at 22:04
• The zero vector all by itself is a subspace of every vector space. It will almost never span the entire space. Mar 16, 2021 at 22:11

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.

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).

• Thanks! This makes sense. Mar 16, 2021 at 22:15
• It better! I'm teaching linear algebra this semester... my students have a test tomorrow ugh!!
– user637978
Mar 16, 2021 at 22:15
• Haha, I have an exam today! Mar 16, 2021 at 22:16

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.

• 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. Mar 16, 2021 at 22:14
• I'm a bit confused on part (b) of the question above. How would I determine if that vector is a subspace of $\mathbb{R}^4$? Mar 16, 2021 at 22:36