Looking Glass Universe | A simple condition for when the matrix inverse exists | Linear algebra makes sense @LookingGlassUniverse | Uploaded 6 years ago | Updated 3 hours ago
Check out Brilliant.org: https://brilliant.org/LookingGlassUniverse/ (it's free to sign up- you don't need to enter credit card details)
Link to my linear transformation/ matrix video: https://youtu.be/CBIO4xJ1Cok
Link to my vectors and bases video: https://youtu.be/3ZfrJ0Sk5iY
Link to my (unlisted) original version of this video: https://youtu.be/pLz_ln0ByXo
This video is about matrix inverses, and in particular, I try to give a bit more intuition for them- rather than just giving you the formula for the determinant, Cramner’s rule, the inverse of 2x2 and 3x3 matrices etc. Along the way, we cover some topics that don’t receive enough attention in linear algebra (at least, they didn’t in my math classes), like the left inverse, and why non square matrices don’t have an inverse. Finally, we will learn an intuitive condition for when the inverse exists.
Homework questions
1. What is the inverse of the following matrix:
1 2
2 5
a)
-1 -2
-2 -5
b)
1 1/2
1/2 1/5
c)
1 -2
-2 5
d)
5 -2
-2 1
2. Prove that, for a square matrix, the left inverse = the inverse
3. Prove the above for the following matrix:
1 2
2 5
by first doing these 2 questions
a) Show that M(1, 0) and M(0,1) for a basis. I.e. the old basis gets mapped to a new basis.
b) Show MLv = v for any vector (this is enough to show the M undoes L, and hence show that the left inverse is equal to the inverse).
To do this, first write v as a linear combo of M(1, 0) and M(0,1).
Hints:
3.
a) is a straightforward if you watch my first linear algebra video: https://youtu.be/3ZfrJ0Sk5iY
b) Sub the expression for v into MLv. Then use linearity!
Q2.
a) (Proving that the basis is always mapped to another basis) This is a bit tricky. I've written out the solution but stop reading at the point you think you know what to do and try it.
Assume v1,...,vn are a basis. Apply M to them to get Mv1,..., Mvn.
They're a basis if they're still linearly independent. But if they were, say:
Mv1= a Mv2+... + d Mvn
Apply L to both sides:
LMv1= L(a Mv2+... + d Mvn)
= a LMv2+... + d LMvn
using LM=identity:
v1= a v2+ ...+ d vn.
But this must be false, since v1,..., vn are linearly indep.
To see it done (SPOILER ALERT) 9:20 in https://youtu.be/pLz_ln0ByXo
b) Once you have that, proceed in a very similar way to Q3, part b)
Check out Brilliant.org: https://brilliant.org/LookingGlassUniverse/ (it's free to sign up- you don't need to enter credit card details)
Link to my linear transformation/ matrix video: https://youtu.be/CBIO4xJ1Cok
Link to my vectors and bases video: https://youtu.be/3ZfrJ0Sk5iY
Link to my (unlisted) original version of this video: https://youtu.be/pLz_ln0ByXo
This video is about matrix inverses, and in particular, I try to give a bit more intuition for them- rather than just giving you the formula for the determinant, Cramner’s rule, the inverse of 2x2 and 3x3 matrices etc. Along the way, we cover some topics that don’t receive enough attention in linear algebra (at least, they didn’t in my math classes), like the left inverse, and why non square matrices don’t have an inverse. Finally, we will learn an intuitive condition for when the inverse exists.
Homework questions
1. What is the inverse of the following matrix:
1 2
2 5
a)
-1 -2
-2 -5
b)
1 1/2
1/2 1/5
c)
1 -2
-2 5
d)
5 -2
-2 1
2. Prove that, for a square matrix, the left inverse = the inverse
3. Prove the above for the following matrix:
1 2
2 5
by first doing these 2 questions
a) Show that M(1, 0) and M(0,1) for a basis. I.e. the old basis gets mapped to a new basis.
b) Show MLv = v for any vector (this is enough to show the M undoes L, and hence show that the left inverse is equal to the inverse).
To do this, first write v as a linear combo of M(1, 0) and M(0,1).
Hints:
3.
a) is a straightforward if you watch my first linear algebra video: https://youtu.be/3ZfrJ0Sk5iY
b) Sub the expression for v into MLv. Then use linearity!
Q2.
a) (Proving that the basis is always mapped to another basis) This is a bit tricky. I've written out the solution but stop reading at the point you think you know what to do and try it.
Assume v1,...,vn are a basis. Apply M to them to get Mv1,..., Mvn.
They're a basis if they're still linearly independent. But if they were, say:
Mv1= a Mv2+... + d Mvn
Apply L to both sides:
LMv1= L(a Mv2+... + d Mvn)
= a LMv2+... + d LMvn
using LM=identity:
v1= a v2+ ...+ d vn.
But this must be false, since v1,..., vn are linearly indep.
To see it done (SPOILER ALERT) 9:20 in https://youtu.be/pLz_ln0ByXo
b) Once you have that, proceed in a very similar way to Q3, part b)
