Find the change-of-basis matrix PB ←c from the given basis C to the given basis B of the vector

Find the change-of-basis matrix PB ←c from the given basis C...

Key Concepts

Basis

A basis for a vector space is a set of linearly independent vectors such that every vector in the space can be uniquely expressed as a linear combination of these basis vectors. Understanding bases is fundamental because it provides a framework for representing the elements of the vector space in terms of coordinates relative to the basis.

Change-of-Basis Matrix

The change-of-basis matrix is the matrix that converts coordinate vectors from one basis to another. It is constructed by expressing each vector of the target basis as a linear combination of the vectors from the initial basis, where the coefficients form the columns of the matrix. This matrix is invertible, which reflects the fact that both sets of basis vectors span the space and are linearly independent.

Coordinate Transformation

Coordinate transformation involves recalculating the coordinates of a vector when the basis is changed. This process uses the change-of-basis matrix to convert the original coordinate vector into its representation in the new basis. Understanding this transformation is essential for solving problems involving different representations of vectors.

Matrix Invertibility

The invertibility of a change-of-basis matrix is crucial because it confirms that the transformation between two bases is reversible. Since each basis consists of linearly independent vectors, the associated matrix is nonsingular, ensuring that every coordinate vector can be uniquely mapped between the bases.

Transcript

00:01 To solve this problem, we can simply compute the inverse of the matrix obtained in problem 19, but we can also solve separately again.

00:14 So first i write the first vector of basis c, which is 1, negative 1, 1.

00:23 Then we need to find the vector, which has 3 components c1, 3, 3, 3, 3.

00:31 That when it's multiplied by the vectors of the basis p gives us the this vector so just called to c1 fly by the first vector 5 -0 plus c2 3 0 plus c3 8 2 and then we need to write a system of equation so here is x y z she won't buy the x of the first vector plus c2 multiplied by the x of the second and c3 multiply by the x of the third vector is going to give us the x on the left hand side so 2 c1 plus 3 c2 plus 8 c 3 is equal to 1 minus 5 c1 plus we don't have the c2 for this equation so i only see 3 minus 2 3 is equal to negative 1 and then 5 c2 minus 9 c3 is equal to negative 1 and then 5 c2 minus 9 c3 is equals 1.

02:15 So we can just simplify this one, say c1 is equal to 1 minus 2c3, divided by 5.

02:38 And then replace c1 here.

02:44 Then we will have a system of equations with two equations.

02:50 And then again, we do the same.

02:52 Substitute c2 or c3 the one equation one solution and then we can solve c1 c2 c3 the other solution would be to use gaussian elimination so here we will have an augmented matrix here to three eight here is one then if 5 -0 -2 -0 -5 -9 and here we need to 1 and 1 and what we need to do is to eliminate one side of the matrix so if this is the diagonal for example we need to multiply the first row by number and then add it to the third row as to so when it's added to the third row this is zero and added to the second row so this is zero when we do this we get c1 c2 c3 and the answer for the first equation that i wrote here is a very vector w1 relative to basis b, which is one fifth and fifth zero...

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