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# Change of basis matrix

Using a change of basis matrix to get us from one coordinate system to another. Created by Sal Khan.

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• At - How are these two vectors a basis for R^3? A basis for R^3 must not only be linearly independent, but should span R^3 - which two vectors cannot! The augmented matrix of these vectors cannot have a pivot in each row. Am I missing something?
• You are right, a basis for R3 would require 3 independent vectors - but the video does not say it is a basis for R3. In fact, it is instead only a basis of a 2 dimensional subspace within R3. In this case the subspace would just be the plane given by the span of the two vectors.
• Is the standard basis Sal refers to any basis that I choose to begin my problem with?
• No, the standard basis is the orthogonal unit vectors that match up with the axes of the Cartesian coordinate system. For instance, R² has a standard basis containing two vectors. They are <1,0> and <0,1> (in that order).

Alternatively, if you had your standard basis column vectors and placed them side by side in a matrix, you'd get the identity matrix.

See: http://en.wikipedia.org/wiki/Standard_basis
• wouldn't d=[-3 -11] because 0-2=-2?
• I understand your confusion, but [d]b=[-3 11] is right. At first Sal was thinking of replacing the second row with "the second row minus 2 times the first row". However, soon after he decides to replace the second row with "2 times the first row minus the second row".

So what he actually does at is 2-0=2, and thus [d]b=[-3 11] is correct!
• Let me see if I got it right. The vectors v1 v2 defined the base, and c1 c2 defines the matrix that will multiply vector d and represent it in base B. right?
• c1 and c2 are scalars of the columns of basis B (which are like axis). So c1 and c2 represent how far you travel on each of the axis represented by basis B to get vector d in standard coordinates.
• From what I see C is change from base B to standard base( because we multiply B's coordinates by this and we get standard coordinates). But why people call it change of base from standard base to B( base C vectors as columns are change of base matrix from standard base to base C). I am confused.
• I think the idea is that, C is the CHANGE OF BASIS matrix from standard base to base B.

SO what we're doing here exactly is not changing or moving the vectors from standard base to base B, but representing them in these bases.

If we have a vector in the standard base, and we have its coordinates in this base, then to get its coordinates in a different base, we are not going to move it to that different base, but get its coordinates in that different base.

So the idea is, what we did to the initial basis to get the second basis, is change of basis,

But what we did to get the coordinates of a vector in the initial basis from its coordinates in the second basis,
(keeping in mind that a vector's coordinates in the second basis are its original coordinates in this basis) is applying the change of basis matrix to the coordinates of this vector in this same basis, we're getting the coordinates in the initial basis, not moving the vector, but getting it's coordinates knowing the change we did to the second base, so we were considering the second base the original base, but now we're applying the change on the coordinates.
• Does a coordinate matrix always have to be a column vector( in other word, is it a convention that a coordinate matrix be a matrix of a single column) ?
• The convention is most commonly as a column vector, but other conventions exist. It can also be represented as a row vector, an ordered list of numbers, etc.. For the sake of typing them out, I like to write them as row vectors such as `[v]b = [1 5 3 6]`.
• Suppose we know that a vector does not belong to a span of v1 and v2.
But still we solve equation c[a]b = a, and figure out a c1 and c2. So what does this c1 and c2 i.e. [a]b represent?
• `C[a]b = a` is the equation for a change of basis. A basis, by definition, must span the entire vector space it's a basis of. C is the change of basis matrix, and a is a member of the vector space. In other words, you can't multiply a vector that doesn't belong to the span of v1 and v2 by the change of basis matrix. Put another way, the change of basis matrix in the video will be a 2x2 matrix, but a vector that doesn't belong to the span of v1 and v2 will have 3 components. You can't multiply a 2x2 matrix with a 3x1 vector. Therefore, you can't solve for c1 and c2 at all in the scenario you gave.
• For the first 13 seconds of the video what do you mean by vk?
• It's just the last vector in his set (the kth one)
• I'm wondering about finding the coordinate vector by solving the matrix that represents the system of linear equations like you start doing at around . Say you use interchange (ie. you swap rows around while solving the matrix), do you have to keep track of where your (c1 c2 c3...) in the last column go? I mean, the matrix was originally set up in a specific order:
[ 0 1 0 c1
1 1 0 c2
0 0 1 c3 ]
If you swap a row, say the first row for the second row, your augmented matrix is now (c2 c1 c3). Do you have to undo this once you are at reduced echelon form to get the correct coordinate vector?