Linear Algebra
"All The Mathematical Methods I Learned In My University Math Degree Became Obsolete In My Lifetime" - Keith Devlin
Visualizing the geometry of a 2x2 Linear Transformation.: Enter the entries for a 2x2 matrix, and see it's effect on the standard basis, arbitrary vectors, areas, and it's eigenspaces (https://www.desmos.com/calculator/n0xzbzzprx)
Visualizing the span of two vectors in \(\mathbb{R}^2\).: Shows the span of two vectors in \(\mathbb{R}^2\) (https://www.desmos.com/calculator/fpbqag6qku)
Solving Higher Dimensional System of Equations: When we have more than 2 variables, it takes more work to find the solution to the system than graphing the 2 lines. The following uses the reduced row echelon form technique to solve the system. Right now it is set to solve the system:
2x+3y-2z=10
-2x+y+z=-1
4y=8
You should see when you evaluate that you get as a solution x=1, y=2, z=-1. You can set A to represent whatever system you want, for example is the system is
2x-y=-3
2x+2y=6
Then we should let A=matrix(QQ, [[2,-1,-3],[-2,2,6]])
Which has solution x=2, y=1.
Linear Combinations and Span: Is (3, -1, 13)^T in the span of (1,1,-3)^T, (2, 0, 5)^T? If it were, we should be able to find, a and b such that a(1,1,-3)^T+b(2,0,5)^T=(3,-1,13)^T. Such an a would satisfy:
1a+2b=3
1a+0b=-1
-3a+5b=13
We can solve this with sage:
Giving us a=-1,b=2.
On the other hand, if we wanted to know if (1,1,1)^T were a linear combination of (1,1,-3)^T and (2, 0, 5)^T, we would have:
1a+2b=1
1a+0b=1
-3a+5b=1
We can solve this with sage:
The last row suggests 0=1, which is impossible. Thus (1,1,1)^T is not a linear combination of (1,1,-3)^T and (2, 0, 5)^T and consequently not in their span.
Linear Independence: Is {(1,2,3)^T, (4,5,0)^T} linear independent? To be linearly independent means that there is a unique linear combination of these vectors for any element in their span, in particular (0,0,0)^T.
In other words, they are linearly independent if and only if a(1,2,3)^T+b(4,5,0)^T=(0,0,0)^T has a unique solution. That is:
1a+4b=0
2a+5b=0
3a+0b=0
Has a unique solution:
Which it does, with a=b=0.
On the other hand, to see if {(1,2)^T, (2,3)^T, (-1,1)^T} is linearly independent, we solve:
We see that a=-5c, b=3c are valid solutions. Thus 0(1,2)^T+0(2,3)^T+0(-1,1)^T=(0,0)^T is a valid solution, but -10(1,2)^T+6(2,3)^T+2(-1,1)^T=(0,0)^T is as well. Thus there is not a unique solution and these vectors are linearly dependent.
Visualizing the span of two vectors in \(\mathbb{R}^2\).: Shows the span of two vectors in \(\mathbb{R}^2\) (https://www.desmos.com/calculator/fpbqag6qku)
Solving Higher Dimensional System of Equations: When we have more than 2 variables, it takes more work to find the solution to the system than graphing the 2 lines. The following uses the reduced row echelon form technique to solve the system. Right now it is set to solve the system:
2x+3y-2z=10
-2x+y+z=-1
4y=8
You should see when you evaluate that you get as a solution x=1, y=2, z=-1. You can set A to represent whatever system you want, for example is the system is
2x-y=-3
2x+2y=6
Then we should let A=matrix(QQ, [[2,-1,-3],[-2,2,6]])
Which has solution x=2, y=1.
Linear Combinations and Span: Is (3, -1, 13)^T in the span of (1,1,-3)^T, (2, 0, 5)^T? If it were, we should be able to find, a and b such that a(1,1,-3)^T+b(2,0,5)^T=(3,-1,13)^T. Such an a would satisfy:
1a+2b=3
1a+0b=-1
-3a+5b=13
We can solve this with sage:
Giving us a=-1,b=2.
On the other hand, if we wanted to know if (1,1,1)^T were a linear combination of (1,1,-3)^T and (2, 0, 5)^T, we would have:
1a+2b=1
1a+0b=1
-3a+5b=1
We can solve this with sage:
The last row suggests 0=1, which is impossible. Thus (1,1,1)^T is not a linear combination of (1,1,-3)^T and (2, 0, 5)^T and consequently not in their span.
Linear Independence: Is {(1,2,3)^T, (4,5,0)^T} linear independent? To be linearly independent means that there is a unique linear combination of these vectors for any element in their span, in particular (0,0,0)^T.
In other words, they are linearly independent if and only if a(1,2,3)^T+b(4,5,0)^T=(0,0,0)^T has a unique solution. That is:
1a+4b=0
2a+5b=0
3a+0b=0
Has a unique solution:
Which it does, with a=b=0.
On the other hand, to see if {(1,2)^T, (2,3)^T, (-1,1)^T} is linearly independent, we solve:
We see that a=-5c, b=3c are valid solutions. Thus 0(1,2)^T+0(2,3)^T+0(-1,1)^T=(0,0)^T is a valid solution, but -10(1,2)^T+6(2,3)^T+2(-1,1)^T=(0,0)^T is as well. Thus there is not a unique solution and these vectors are linearly dependent.