I'm reading the Linear algebra book by Gilbert Strang. In the first chapter, there's a line (see last two lines in the picture) saying that "vectors $c\mathbf{v}$ lie along a line". I am not able to understand the meaning of this line. Have we constrained (This is the paragraph containing that line.) ourselves to make v a vector of 2 dimensions here? Because, in that case, we can plot all the points and that would create a line. Is my understanding correct?
What’s the meaning of “vectors cv lie along a line”
linear algebra
Related Solutions
The core of studying matrices is to study linear transformations between vector spaces. These can be realized as matrix multiplication on the left (or right) of column (or row) vectors.
If we are in this setup: $x\mapsto Ax$ for a column vector $x$ and appropriate matrix $A$, then the image of the linear transformation will be spanned by the columns of $A$.
The kernel of the transformation (nullspace) is the set of all $x$ such that $Ax=0$ is important for understanding the solutions to some matrix equations. You probably have already learned that if $x_0$ is a solution to $Ax=b$, then every other solution is given by $x_0+k$ where $k$ is in the nullspace.
This all has analogous explanation on the other side. If we are in this setup: $x\mapsto xA$ for a row vector $x$, then the image of the linear transformation is now spanned by the rows of $A$.
Talking about the nullspace of $A^T$ is just a fancy way of dressing up the "left nullspace" of $A$, since $xA=0$ iff $A^T x^T=0$. The nullspace is now the set of all $x$ such that $xA=0$, and you can draw the same conclusions about solutions to $xA=b$.
In short, these four spaces (really just two spaces, with a left and a right version of the pair) carry all the information about the image and kernel of the linear transformation that $A$ is affecting, whether you are using it on the right or on the left.
The linear combinations of the vectors $v,w$ can be denoted $span(v,w)=kv+cw$, where $v,w \in V$ and $k,c \in \mathbb{F}$ (scalars).
If the two vectors are linearly independent, informally consider the follow scenario:
Let $V = \mathbb{R^2}$. Let two vectors $v,w \in V$. Set $v = (x,0)$ and $w = (0,y)$. There is no scalar $k \in \mathbb{F}$, that when multiplied by the vector $v = (x,0)$, yields a vector of the form $w = (0,y)$. The two independent vectors, when combined linearly as $kv+cv$, make up a whole plane or $\mathbb{R^2}$. You can think of these vectors as the $x-axis$ and $y-axis$ coming together to form the $xy-plane$. When taking the set of all linear combinations of the vectors $v$ and $w$ we can obtain any other vector of the form $(x,y)$, and thus $span(v,w) = \mathbb{R^2}$.
Formally, the only combination $kv+cw$ that yields the $0$ vector is by setting each $k,v \in \mathbb{F}$ equal to $0$. Is is clear that the only scalars that satisfy the linear combination $k(x,0) + c(0,y) = (0,0)$ for the $0$ vector are $k = c = 0$, which implies that the two vectors in $\mathbb{R^2}$ are independent and span the plane.
Now, consider a separate case, where the vectors $v,w \in V$ are $(x,0)$ and $(2y,0)$, respectively. Let $span(v,w) = a(x,0) + b(2y,0)$ which denotes all possible linear combinations of the vectors. However, it is apparent that the $0$ can be obtained by more than just the "obvious" way which means setting $a = b = 0$ to obtain $(0,0) + (0,0) = (0,0)$. Setting $y = \frac{-1}{2}x$, and we get $a(x,0)+b(-x,0)$ which equals $0$ for $a = b = 0$ or $a = b = 1$, which shows that the two vectors are dependent and as a result do not span $\mathbb{R^2}$, so their linear combinations do not fill a plane.
It's not that $v$ and $w$ can't lie on the same line. It's just that if we have two vectors are on the same line, then they are two linearly dependent vectors in $\mathbb{R^2}$, and as a result, do not fill a plane.
Best Answer
In this example $v = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$, and we can multiply it by some constant $c$ to get $cv = \begin{bmatrix} c \\ c \end{bmatrix}$. As you've said, if you plot all of these points then you'll end up with a line, the line $y = x$.
We haven't constrained ourselves to 2 dimensions here, as an example if $v$ was instead $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$ then we've added an extra dimension by it's still the same line.