Tell if vectors form a plane through the origin

linear algebramatrices

So I'm wondering, how can I tell if vectors form a plane through the origin in $\mathbb{R^3}$, a line through the origin in $\mathbb{R^3}$, a line that may not go through the origin, etc.?

Say we have the following matrix A:
$$\begin{pmatrix}
1&-2&-4\\
3&-6&-16\\
0&0&-4\\
\end{pmatrix}$$

Since we are looking at the form $Ax = b$, solving for echelon form results in the following:

$$\begin{pmatrix}
1&-2&-4&b_1\\
0&0&-4&-3b_1+b_2\\
0&0&0&3b_i-b_2+b_3\\
\end{pmatrix}$$

Now, we have 3 variables so it's in $\mathbb{R^3}$, yet, the third row indicates that that $3b_1-b_2+b_3$ needs to be $0$. How would I be able to tell if it goes through the origin as a plane, a line, etc?

EDIT: Forgot to mention $b$ is of the form $Ax = b$

EDIT 2:
To be clear, I am asking whether the sets of vectors b that makes Ax=b consistent is a plane through the origin, a line through the origin, or a line that does not go through the origin, etc.

Would appreciate some help, thanks in advance!

Best Answer

EDIT: So you want to know about the set of vectors $b$ that make $Ax=b$ consistent.

A system of linear equations $Ax=b$ is consistent if and only if $b$ is in the range of $A$ (the column space of $A$). That means that $b$ can be written as the linear combination of the columns of $A$, which is one of the two fundamental ways to view the matrix multiplication $Ax$.

The column space of a matrix always includes the origin, since $A\vec{0}=\vec{0}$. (for appropriate dimensions on the two zero vectors)

Determining the dimension of the column space means determining the maximum number of columns of $A$ that can form a linearly independent set.

For this specific $A$, it is easy to see that column 2 is a multiple of column 1, but that columns 1 & 3 are linearly independent, so the dimension is 2 (a plane)

In general, there are multiple ways to find the dimension of the column space for a matrix, but in general the way that I would do it is to use elementary row operations to put $A^T$ in reduced row echelon form. The nonzero rows will be a basis for the row space of $A^T$, which is the column space of $A$.

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