Linear algebra starts with cv + dw.
Take two vectors. Stretch each one by a number, then add. That is a linear combination, and the whole subject is built on it.
Think: directions you can give
You are guiding someone across a field. You may only use two kinds of step: step v and step w.
"Take 3 steps of v, then 2 steps of w." Where do they end up? At 3v + 2w. Every instruction of that kind is a linear combination.
The two moves
- Scale: 3v = (6, 3). Each component is multiplied by 3. A negative number turns the arrow around.
- Add: put the tail of w on the tip of v. The sum goes from the start to the final tip. In numbers, add component by component.
Why the word "linear"? Scale one vector by every possible number and its tip traces a straight line through the origin.
Worked example
You already use this
The point (4, 5) is itself a combination: 4 steps along (1, 0) plus 5 steps along (0, 1). Coordinates are just the numbers c and d for the two standard directions.
The question behind everything
Which points can you reach? Think of each result as the dot at the tip of its arrow. Try all c and d and the dots fill the whole plane, because v and w point in different directions.
If w pointed along v, every dot would stay on one line. Slide 3 is about exactly that.
Finding c and d for a target means solving 2c − d = 4 and c + d = 5. On slide 4 this turns into Ax = b.
Worked examples
Example 1 · Compute a combination
v = (1, 1), w = (1, −1), c = 2, d = 3. Find cv + dw.
Example 2 · Find the numbers
Write (3, 0) as c·(1, 1) + d·(1, −1).
Problems
Try each one before you open the solution.
Compute 2v − w for v = (3, 1) and w = (1, 4).
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2v = (6, 2). 2v − w = (6 − 1, 2 − 4) = (5, −2).
Find c and d with c·(1, 2) + d·(0, 1) = (3, 8).
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The first component gives c = 3. The second: 2·3 + d = 8, so d = 2.
Can combinations of (2, 4) and (−1, −2) ever reach (1, 0)?
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No. (−1, −2) = −½·(2, 4), so both vectors lie on the line y = 2x. Every combination stays on that line, and (1, 0) is not on it.