Khan Academy: Orthogonal Complements: Unique Rowspace Solution to Ax=b

Sal dissects basic algebraic equations of the form Ax=B. He encourages viewers to fully understand the equations before learning how to manipulate them. Furthermore, he teaches students how to solve the equations systematically. The...

An instructional video that gives an example of how to find a solution to the equation Ax = b. In the example, the matrix A is 2 x 2. Shows that when the solution is a member of the row space of A for a given vector b, a member of the...

An instructional video that gives an example of how to find a solution to the equation Ax = b. In the example, the matrix A is 2 x 2. Shows that when the solution is a member of the row space of A for a given vector b, a member of the...

An instructional video that gives an example of how to find a solution to the equation Ax = b. In the example, the matrix A is 2 x 2. Shows that when the solution is a member of the row space of A for a given vector b, a member of the...

A video lesson proving that there is a unique solution in the row space of A to the equation Ax = b, where A is some m x n matrix, b is a vector that is a particular member of the column space of A, and x is a member of Rn. The solution...

A video proving that the orthogonal complement of the orthogonal complement of some subspace V must be a member of the original subspace V.

Video introducing the orthogonal complement of a subspace and showing that it is also a subspace. Proves that the null space of a matrix is equivalent to the orthogonal complement of its row space by showing that every member of the null...

A video lesson illustrating that the orthogonal complement of the nullspace is equal to the row space and vice versa.

Showing that if V is a subspace of Rn, then dim(V) + dim(V's orthogonal complement) = n

The second video on Benford's Law from the Khan Academy. This video connects Benford's Law to the logarithmic scale in an effort to explain why it holds true, particularly in data that grows exponentially. [6:27]