Determine if a set is closed under scalar multiplication | Linear algebra I

In this video I went through an example from an intro linear algebra course dealing with closure under scalar multiplication. The problem was to determine if the set U of all 2x2 matrices with third entry equal to 0 is closed under scalar multiplication.

If something is closed, you can't get out. A set being 'closed under scalar multiplication' means that you can't get outside of that set just with multiplication by a constant. In this case, since multiplying a matrix by a constant won't change the fact that it has a zero in the third entry, this set U is closed under scalar multiplication.

Formally how we see if a set is closed under scalar multiplication is by taking some arbitrary element in the set, let's call it v, and see if the vector k times v is also in the set, for any real number (constant) k. If scaling by any constant isn't enough to get you out of the set, then the set is closed under scalar multiplication.

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[theboombody]

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[raitup]

Great explanation! You should do more videos of Linear algebra

[itdepends]

Hey Jessie. Would you be able to share what tools/apps you used to make this video? Really nice and easy to follow

[brandoncazares]

Thanks for this video Jessie. It's very helpful and helped me.

[ivang]

yay thank you, I needed this for modern algebra ahaha.... kind of didn't pay attention in linear

[tommywalker]

How would we prove u with addition?i love how simple you made it to understand.

[worldwideoffline]

What's a common case where something IS NOT closed under scalar multiplication?

[souravnatta]

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[HeavyMachinery]

if k is negative, then is it still closed? i think it will change the order of the result matrix and thus not in closed. for example: lets say we have matrix
[ 1 2 3
-1 -2 -3 ],
Order change
Multiple with -1. please guide me here.

[Quiyum]

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