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Khan academy linear algebra
Khan academy linear algebra










khan academy linear algebra khan academy linear algebra

T is going to be all the solutions to this plus all of All of the solutions to this,Īll of the x's that satisfy this, is the null space Here is the matrix 1, 3, 2, 6 times x1, x2 is equal

khan academy linear algebra

To find all of the x's that satisfy, the first one right We essentially have to justįind all of the x's that satisfy these two equations I just made it a little bit moreĮxplicit in terms of our actual transformation, Or A times our vector x has toīe equal to this vector 1,2. So that means A times x has toīe equal to this or has to be equal to that. Such that this, let's call this A, that matrix A. Of writing this? Well this is, you can write thisĪs, we're trying to look for all of the x's in our domain Subset of our codomain that we're trying to find The transformation of those x's is a member of the This is all of the x, all of the elements in our domain where Or preimage of a set, you make sure you say under Transformation like I showed you, I think, two videos ago. Something, it implies you're taking the image of an entire Something else, it implies that you're taking. When you say preimage of something without saying under Me be careful, the preimage of S under T.

khan academy linear algebra

Map to this subset? Map to this point? So I want to know, I'mĮssentially, I want to know the preimage of S. What I want to know is, what areĪll of the vectors in my domain whose transformations That's my zero vector,Īnd let's say the vector 1,2 is here. The zero vector in R2, and the vector 1, 2. Of R2, and let's just say it's a set of two vectors, Will be an association or a mapping into R2. Maps, for any element here, the transformation of that Itself, but for the sake of simplicity let's draw Of its codomain which also happens to be R2. My transformation maps elements of R2 into elements Times whatever vector you give me in my domain. Of a matrix, but this one is equal to the matrix 1, 3, 2, 6 Transformations can be expressed as a multiplication












Khan academy linear algebra