Change of basis matrix(linear algebra)

[leeewl]

Hi I'm stuck on this problem and I could not find similar examples anywhere.. any help would be greatly appreciated, thank you.

Homework Statement

Compute the change of basis matrix that takes the basis
[itex]V1 = \begin{bmatrix} -1 \\ 3 \end{bmatrix}[/itex] [itex]V2 = \begin{bmatrix} 2 \\ 5 \end{bmatrix}[/itex]
of R² to the basis
[itex]W1 = \begin{bmatrix} 2 \\ 5 \end{bmatrix}[/itex] [itex]W2 = \begin{bmatrix} 3 \\ 7 \end{bmatrix}[/itex]
I have done this first part, the change of basis matrix is [itex]A = \begin{bmatrix} 2 & 1 \\ -1 & 0 \end{bmatrix}[/itex]

next part I don't quite know how to start:
Consider v = V1 + 2(V2) [itex]\in[/itex] R²: Determine the column vector [itex]\begin{bmatrix} a \\ b \end{bmatrix}[/itex] which represents v with respect to the basis {W1, W2}

The Attempt at a Solution

Do I turn v1 into [itex]V1 = \begin{bmatrix} -1 \\ 3 \end{bmatrix}[/itex] and V2 into [itex]V2 = \begin{bmatrix} 4 \\ 10 \end{bmatrix}[/itex] and then try and find a linear combination that gives me {W1, W2}?

 

[radou]

Set up the linear combination and make it equal to your vector in order to find the coefficients a, b.

 

[vela]

Hi I'm stuck on this problem and I could not find similar examples anywhere.. any help would be greatly appreciated, thank you.

Homework Statement

Compute the change of basis matrix that takes the basis
[itex]V1 = \begin{bmatrix} -1 \\ 3 \end{bmatrix}[/itex] [itex]V2 = \begin{bmatrix} 2 \\ 5 \end{bmatrix}[/itex]
of R² to the basis
[itex]W1 = \begin{bmatrix} 2 \\ 5 \end{bmatrix}[/itex] [itex]W2 = \begin{bmatrix} 3 \\ 7 \end{bmatrix}[/itex]
I have done this first part, the change of basis matrix is [itex]A = \begin{bmatrix} 2 & 1 \\ -1 & 0 \end{bmatrix}[/itex]

next part I don't quite know how to start:
Consider v = V1 + 2(V2) [itex]\in[/itex] R²: Determine the column vector [itex]\begin{bmatrix} a \\ b \end{bmatrix}[/itex] which represents v with respect to the basis {W1, W2}

The Attempt at a Solution

Do I turn v1 into [itex]V1 = \begin{bmatrix} -1 \\ 3 \end{bmatrix}[/itex] and V2 into [itex]V2 = \begin{bmatrix} 4 \\ 10 \end{bmatrix}[/itex] and then try and find a linear combination that gives me {W1, W2}?

The fact that [itex]\vec{v} = 1\vec{v}_1 + 2\vec{v}_2[/itex] means that with respect to the [itex]\{\vec{v}_1,\vec{v}_2\}[/itex] basis,
[tex]\vec{v} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}_{\{\vec{v}_1,\vec{v}_2\}}[/tex]Use the matrix A to convert the coordinates from one basis to the other.

By the way, I don't think your matrix A is correct.

 

[leeewl]

Thank you for your answers. I made a mistake in the op. [itex]V1 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}[/itex] and not (-1, 3) so my matrix A should be correct.
Is multiplying A by (coefficients of v1, v2) [itex]v = \begin{bmatrix} 1 \\ 2 \end{bmatrix} [/itex] all I really need to do to?
Then my answer is \begin{bmatrix} 4 \\ -1 \end{bmatrix}

 

[vela]

It's easy enough to check. Is [itex]4\vec{w}_1-\vec{w}_2[/itex] equal to [itex]\vec{v}_1+2\vec{v}_2[/itex]?

If your matrix is correct, then yes, that's all you have to do. That's why it's called a change of basis matrix.