How come this video doesn't have hundreds of thousands of views !!!. All students learning Linear Algebra must see this. Maybe more catchy title would make it appear in google searches
It just clicked for me that a 2x4 matrix is a mapping from R4 to R2. And assuming that the four columns of the 2x4 matrix span R2, then for the given linear system, there is a plane of solutions from the domain (R4), and our job is to find that plane. It is a plane because if the matrix is 2x4 and its columns span R2, then we know there are two free variables, and those two free variables allow us to trace out a plane. And because this plane doesn't intersect (0,0,0,0), we know that it is not a vector space. And now that I think about this some more, if we had this same problem in R3 instead of R4, the null space would be a line instead of a plane (again not passing through the origin); and we are allowed to move the b vector (a single point in R2) in two directions on the range (R2), and a line in R3 has 2 degrees of freedom, so its movements correspond to the movements of the b vector.
I feel like I am really close to understanding the relationship between the general solution, nullspace, and linear dependence but I can't put the concepts together. At 18:00, Pavel refers to an idea from an older video where scaling one vector and adding it to another provides solutions along a line. In that video (Linear Algebra 6b: Alternative Definition of Linear Dependence), the linearly dependent vectors in geometric space were collected together under alpha(a + b - c) which he called a "fancy zero" since the combination gives the zero vector. How does this idea factor into the diagram shown at 18:00? How does the nullspace solution and linear dependence factor into the solutions given by p + alpha(a)? In other words, what would our knowledge about the nullspace add to the diagram such that it would give us all possible solutions and what is the significance of the fact that the null space vectors are linearly dependent? I know this is all connected somehow but I can't quite piece it together!
Hi, I an trying to figure out the same thing: So...I DO THINK that 1) P and Q ARE NOT particular solutions. SOLUTION in this case IS A GEOMETRIC VECTOR WHOSE TIP LIES ON THE LINE DRAWN. 2) aA here is the NOT THE NULL SPACE of ALL THE VECTORS WHOSE TIP LIES ON THE LINE DRAWN - because adding/subtracting it (to/from P or Q and other alike) will LEAVE US STILL WITH A GEOMETRIC VECTOR WHOSE TIP LIES ON THE LINE DRAWN 3) Draw two vectors (B1 & B2) with an angle between them from the origin to the line. THESE WILL SPAN OUR SPACE. ALL THE VECTORS WHOSE TIP LIES ON THE LINE DRAWN ARE A SUBSET of our SPANNED 2D space, but NOT A SUBSPACE. 4) MAKE SURE to draw B1 & B2 so that : B1+A=B2. Now move A to the origin SO IT BECOMES A PART OF OUR "GAME". Now, The vectors A, B1 and B2 are Linearly Dependent and thus have a NULLSPACE: (1 of B1) + (1 of A) + (-1 of B2) = ZERO VECTOR 5) Lets use B1 as A PARTICULAR SOLUTION: B1+a(B1+A-B2). ONLY NOW MAYBE WE CAN SAY/REPLACE B1 with P/Q and CLAIM THAT a(B1+A+B2) is the NULLSPACE and B1/P/Q are particular solutions But I would like clarification about that as well
Enjoying the videos Professor Grinfeld I have a silly question You highlighted that there were multiple choices for the particular solution, is this true for the null space vectors as well? For example I saw [3 6 1 -1] as a vector that could have been chosen. Perhaps I am also putting too much focus on the expression. :) Thank you for posting these videos they are great, I feel like all my prior encounters with this subject have involved memorizing how to push matrices around. Your approach I feel manages to feel both more abstract and concrete. Thanks Evan
until now i thought the particular solution to the linear system would be unique would be unique. Oh boy got a lot to learn. Is there any way to get around it with ease. Is it true to say that if there is a nxn, invertible transformation matrix for linear system Ax=b. x and b both being of dimension Rn, x has a unique solution...
I think the answer to your question is the number of columns in null space is equal to number of dimensions of set of vectors.Disclaimer : This may be not the right answer , this is just what i thought
Thank you very much for all the work you put in the videos. I watch them all and think they are very good. The series of lectures by Strang and the lesser known WildLinAlg by Wildberger are a good compliment to yours. Your videos pull everything together for me. I get a much deeper understanding and grasp of the material. One thing. wouldn't it be a good thing to name the alpha and the beta something like alpha_1 and beta_1 when used in combination with a different particular solution? I will not be able to comment and thank you on every video but rest assured that I am watching and looking forward to the next one.
Hi Herman, Thank you for your comment. I think that the Strang (my teacher) and Wildberger videos are masterpieces. You can roughly think of Strang's approach as algebraic, Wildberger's as geometric and mine as the meeting of the two. I understand your alpha_1 comment - and I'm glad you made it! On the one hand, you're correct: why mix things together and risk confusion? On the other hand, I find that my students are too focused on the expressions themselves - as if the were some real thing - rather than on the more real objects that they represent. This is the case not just in linear algebra but in all of mathematics - too much focus on manipulating expressions rather than on the ideas they represent. Case in point, the general solution is *not* an expression, but rather a collection of vectors. We have no better choice than to capture them with an expression, but that's all it is - a symbolic representation of a set of vectors. With that in mind, when one looks at an expression, one should immediately imagine the set of vectors it represents, and alpha in both cases means "any number whatsoever". So is there really a need to have any_number_whatsoever and any_number_whatsoever_1? Pavel
To me this looks like a four dimensional space spanned by x, y, z and t. There are only two constraints represented by two equations. So this system is under specified or under constrained.
Go to LEM.MA/LA for videos, exercises, and to ask us questions directly.
How come this video doesn't have hundreds of thousands of views !!!. All students learning Linear Algebra must see this. Maybe more catchy title would make it appear in google searches
Thanks. Any title with the word handsome in it is fine by me.
just king is this Professor
It just clicked for me that a 2x4 matrix is a mapping from R4 to R2. And assuming that the four columns of the 2x4 matrix span R2, then for the given linear system, there is a plane of solutions from the domain (R4), and our job is to find that plane. It is a plane because if the matrix is 2x4 and its columns span R2, then we know there are two free variables, and those two free variables allow us to trace out a plane. And because this plane doesn't intersect (0,0,0,0), we know that it is not a vector space. And now that I think about this some more, if we had this same problem in R3 instead of R4, the null space would be a line instead of a plane (again not passing through the origin); and we are allowed to move the b vector (a single point in R2) in two directions on the range (R2), and a line in R3 has 2 degrees of freedom, so its movements correspond to the movements of the b vector.
I feel like I am really close to understanding the relationship between the general solution, nullspace, and linear dependence but I can't put the concepts together. At 18:00, Pavel refers to an idea from an older video where scaling one vector and adding it to another provides solutions along a line. In that video (Linear Algebra 6b: Alternative Definition of Linear Dependence), the linearly dependent vectors in geometric space were collected together under alpha(a + b - c) which he called a "fancy zero" since the combination gives the zero vector. How does this idea factor into the diagram shown at 18:00? How does the nullspace solution and linear dependence factor into the solutions given by p + alpha(a)? In other words, what would our knowledge about the nullspace add to the diagram such that it would give us all possible solutions and what is the significance of the fact that the null space vectors are linearly dependent? I know this is all connected somehow but I can't quite piece it together!
Hi, I an trying to figure out the same thing: So...I DO THINK that
1) P and Q ARE NOT particular solutions. SOLUTION in this case IS A GEOMETRIC VECTOR WHOSE TIP LIES ON THE LINE DRAWN.
2) aA here is the NOT THE NULL SPACE of ALL THE VECTORS WHOSE TIP LIES ON THE LINE DRAWN - because adding/subtracting it (to/from P or Q and other alike) will LEAVE US STILL WITH A GEOMETRIC VECTOR WHOSE TIP LIES ON THE LINE DRAWN
3) Draw two vectors (B1 & B2) with an angle between them from the origin to the line. THESE WILL SPAN OUR SPACE. ALL THE VECTORS WHOSE TIP LIES ON THE LINE DRAWN ARE A SUBSET of our SPANNED 2D space, but NOT A SUBSPACE.
4) MAKE SURE to draw B1 & B2 so that : B1+A=B2. Now move A to the origin SO IT BECOMES A PART OF OUR "GAME". Now, The vectors A, B1 and B2 are Linearly Dependent and thus have a NULLSPACE: (1 of B1) + (1 of A) + (-1 of B2) = ZERO VECTOR
5) Lets use B1 as A PARTICULAR SOLUTION: B1+a(B1+A-B2). ONLY NOW MAYBE WE CAN SAY/REPLACE B1 with P/Q and CLAIM THAT a(B1+A+B2) is the NULLSPACE and B1/P/Q are particular solutions
But I would like clarification about that as well
Enjoying the videos Professor Grinfeld
I have a silly question
You highlighted that there were multiple choices for the particular solution, is this true for the null space vectors as well?
For example I saw [3 6 1 -1] as a vector that could have been chosen.
Perhaps I am also putting too much focus on the expression. :)
Thank you for posting these videos they are great, I feel like all my prior encounters with this subject have involved memorizing how to push matrices around. Your approach I feel manages to feel both more abstract and concrete.
Thanks
Evan
Had to double back, just watched the comment video, my question is answered :)
Thanks again
Evan
until now i thought the particular solution to the linear system would be unique would be unique. Oh boy got a lot to learn. Is there any way to get around it with ease. Is it true to say that if there is a nxn, invertible transformation matrix for linear system Ax=b. x and b both being of dimension Rn, x has a unique solution...
How do you know how many terms go in the Null Space? Here you have 2. Why don't you need to add a third and when would you need to add a third?
+The Flagged Dragon This is a great question! It is discussed in great detail in subsequent videos.
I think the answer to your question is the number of columns in null space is equal to number of dimensions of set of vectors.Disclaimer : This may be not the right answer , this is just what i thought
Thank you very much for all the work you put in the videos. I watch them all and think they are very good. The series of lectures by Strang and the lesser known WildLinAlg by Wildberger are a good compliment to yours. Your videos pull everything together for me. I get a much deeper understanding and grasp of the material. One thing. wouldn't it be a good thing to name the alpha and the beta something like alpha_1 and beta_1 when used in combination with a different particular solution? I will not be able to comment and thank you on every video but rest assured that I am watching and looking forward to the next one.
Hi Herman,
Thank you for your comment.
I think that the Strang (my teacher) and Wildberger videos are masterpieces. You can roughly think of Strang's approach as algebraic, Wildberger's as geometric and mine as the meeting of the two.
I understand your alpha_1 comment - and I'm glad you made it! On the one hand, you're correct: why mix things together and risk confusion?
On the other hand, I find that my students are too focused on the expressions themselves - as if the were some real thing - rather than on the more real objects that they represent. This is the case not just in linear algebra but in all of mathematics - too much focus on manipulating expressions rather than on the ideas they represent. Case in point, the general solution is *not* an expression, but rather a collection of vectors. We have no better choice than to capture them with an expression, but that's all it is - a symbolic representation of a set of vectors.
With that in mind, when one looks at an expression, one should immediately imagine the set of vectors it represents, and alpha in both cases means "any number whatsoever". So is there really a need to have any_number_whatsoever and any_number_whatsoever_1?
Pavel
@4:27 why did you use the coefficient as vector in R2? Shouldn't each vector be represented by x,y,z,t in R4 space?
To me this looks like a four dimensional space spanned by x, y, z and t. There are only two constraints represented by two equations. So this system is under specified or under constrained.
Wouldn't it be nice if my teacher had explained this (if he even understood it)?