Let's start with the very simple example and introduced the canonical basis for
the Euclidean plane.
Do you remember any vector in the Euclidean plane?
Is the point on the plane.
And this can be express as a pair of coordinates x0 and x1 on the plane.
Now, it's easy to see that if you take two vectors shape like saw,
e0 will be the pair 1, 0 and e1 will be the pair 0, 1.
Then any vector in our R2 can be expressed as a linear combination of e0 and e1.
And the coefficients,
the scalars that you will have to use in this linear combination.
Are actually the values themselves in the original vector.
The fact that the scalars in the linear combination coincide
with the values of the elements of the vector.
Is why we call this basis the canonical basis.
As a graphical example, consider the vector x of coordinates 2 and 1,
so this guy here.
And we can see very easily that we can express this as the linear combination of
two times the horizontal basis vector, canonical basis vector,
plus one time the vertical canonical basis vector.
So if we sum this 2 times 0,
plus this, we have the usual parallelogram rule, and we get back our x.
However, there are other basis that we could choose for the Euclidean plane.
For instance, the basis given by the vectors 1, 0 and 1,
1 also represents a basis for the plane.
And any vector of coordinates x0 and x1,
can be expressed as a linear combination of v0 and v1.
How do we find the coefficients in the linear combination?
Well we have a system of two equations and two unknowns, pays Alpha 0 and Alpha 1.
And we get that Alpha 0 will be equal to x0 minor x1 and
alpha 1 will be equal to x1.
So again graphically the same vector of coordinate 2 and
1 can be expressed as a linear combination of v0 and v1.
And now the coefficients are different because they will be 1 and 1.
And you can see indeed if you sum this guy plus this guy,
again with a parallelogram rule you get back x.
There are in fact an infinite number of bases for R2.
But not all pairs of vectors represent the basis for the plane.
Here's a kind of example, if you take g(0),
the vector of coordinates 1 and 0, so equal to the first canonical basis vector.
And then you take g(1) as -1 0, then it's easy to show that it's not
possible to express any point on the plane as a linear combination of g(0) and g(1).
As a matter of fact, you cannot express any vector for
which the second component is not equal to 0.
Graphically, it's easy to see because you can see that g(0) and
g(1), are actually collinear.
They're in the same line on the plane.
So they can only capture one dimensional component of the plane.
So any linear combination of g(0) and g(1) will be a point on this line.
But you will not be able to reach the other points on the plane.
We express this by saying that the vectors are not linearly independent.