I've recently got into Linear Algebra again after a long break. I've been watching Gilbert Strang's lectures and reading his book. I understand the concepts of linear Independence, basis and span of vectors. I'm also aware that $n$ linearly independent vectors give every other vector in $R^n$. However, it's not clear to me, intuitively, that (for example) two linearly independent vectors give every possible vector (i.e span a plane) in $R^2$. If I have a $2$ by $2$ matrix that consists of following vectors:
$$\begin{pmatrix}1\\0\\\end{pmatrix}\begin{pmatrix}0\\1\\\end{pmatrix}$$
Then it's clear that I can get any other vector. However it's not as clear if, say, two vectors are: $$\begin{pmatrix}1\\9\\\end{pmatrix}\begin{pmatrix}11\\25\\\end{pmatrix}$$
My idea was that I can reduce this to the identity matrix, therefore, from upper example, the linear combination will give me every possible vector. But i'm not sure if i'm allowed to do that. Is my idea correct? Is there a more intuitive way to understand this?
Edit : to clarify what I mean by "allowed". Does reducing the matrix ($A$) to it's row reduced echelon ($R$) form and then taking it's combinations (of $R$) yields the same result as if I take linear combinations (of $A$) without reducing it first? if so, why?
Best Answer
It's not at all obvious that your two vectors span all of the plane, so this is a great question.
A usual proof goes like this:
"The span of those vectors is a 2-dimensional subspace of a 2-dimensional space. Hence, by a decomposition theorem usually proved pretty early, it must equal the whole space."
That's not very helpful, even if you know the decomposition theorem I'm talking about. So here's an alternative approach.
If your two vectors both have $0$ in the first entry, then they are in fact not independent. So at least ONE of them has a nonzero first entry. Call that $a$, and the other one $b$. In your example, $$ a = \pmatrix{1\\9}, b = \pmatrix{11\\25} $$ Now form a linear combination of $a$ and $b$ that looks like this:
$$ a' = \frac{1}{a_1} a, b' = b. \tag{1} $$ Now $a'$ has a $1$ as its first entry. Observe that any combination of $a'$ and $b'$ is ALSO a combination of $a$ and $b$ (just substitute in formula 1).
In your example, $a' = a, b' = b$, because $a_1$ just happens to be $1$.
Now let $$ b'' = b' - b'_1 a' = \pmatrix{11\\25} - 11 \pmatrix{1\\9} = \pmatrix{0\\-74} $$
Now let's get that bottom entry of $b''$ nicer: replace $$ a''' = a''; b''' = \frac{1}{b''_2}b'' = \frac{1}{-74} \pmatrix{0\\-74}= \pmatrix{0\\1} $$
And finally, let $$ a'''' = a''' - a'''_2 b''' = \pmatrix{1\\9} - 9 \pmatrix{0\\1} = \pmatrix{1\\0}. $$
And now we have shown that the two standard basis vectors can be written as linear combinations of $a$ and $b$, and therefore (by back-substituting), any combination of the standard basis vectors can also be written as a combination of $a$ and $b$. Hence all of $\Bbb R^2$ is in the span of $a$ and $b$.