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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? • 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? • 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!
• 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.
(1 vote)
• 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? • Once you have written the matrix in reduced row echelon form, you should be able to determine what the solution is by inspection, because interchanging the rows does not change the ordering of the columns and therefore does not change the ordering of the variables you are solving for(since each variable corresponds to a column, due to how matrix multiplication is defined).

Hence, it is not necessary to rearrange the rows back into their original order. In fact, doing so would only make things worse, because interchanging the rows after writing A in its reduced row echelon form would undo some of the work that was already done to solve the system.
• At wouldn't you say Sal is pretty lucky that the reduced row echelon form only had 2 pivots? Or said in another way that the reduced row echelon form had all zeroes in the bottom row? Otherwise d couldn't have been represented in terms of B! • I wouldn't call it lucky but rather good planning on Sal's part for choosing d to be in the span of v1 and v2. Had he not planned ahead, as I suspect he did, then yes, Sal did get quite lucky that the row reduced form of the augmented matrix didn't end up with a pivot in the final row/column, as that would've led to d being found outside the span of v1 and v2.
(1 vote)
• 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) ?
(1 vote) • 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?
(1 vote) • `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. 