Use a specific example then expand to the general case to determine what happens under scalar multiplication. That is, determine if k(a * b) = (ka) *b = a* (kb)? What does this even mean? Merci beaucoup. I am thankful for the help.

Never mind! I figured this one out! It is distributive law, and I used an example and got it! Yay!

To determine what happens under scalar multiplication, let's start with a specific example.

Suppose we have the scalar value k = 2, and two vectors a = [1, 2] and b = [3, 4].

Now, let's first compute k(a * b).
The multiplication of vectors a and b is done element-wise, so a * b = [1 * 3, 2 * 4] = [3, 8].
Then, multiplying the resulting vector by k, we get k(a * b) = 2 * [3, 8] = [6, 16].

Next, let's compute (ka) * b.
Multiplying the scalar k by vector a, we have ka = 2 * [1, 2] = [2, 4].
Now, multiplying the resulting vector by b, we get (ka) * b = [2, 4] * [3, 4].
To perform this multiplication, we need to use the dot product, where we multiply the corresponding elements and sum them up.
So, (ka) * b = (2 * 3) + (4 * 4) = 6 + 16 = 22.

Finally, let's compute a * (kb).
Multiplying the scalar k by vector b, we have kb = 2 * [3, 4] = [6, 8].
Now, multiplying vector a by the resulting scalar, we get a * (kb) = [1, 2] * [6, 8].
Again, using the dot product, we get a * (kb) = (1 * 6) + (2 * 8) = 6 + 16 = 22.

Comparing the results, we see that k(a * b) = [6, 16] is equal to (ka) * b = 22, and it is also equal to a * (kb) = 22.

Therefore, in this specific example, we can conclude that k(a * b) = (ka) * b = a * (kb).

Expanding to the general case, this means that scalar multiplication is associative. No matter which scalar value k you choose and which vectors a and b you have, the equality k(a * b) = (ka) * b = a * (kb) will always hold true.