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Easy Bases Linear Algebra

  • Thread starter shaon0
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Homework Statement


See Attachment.

The Attempt at a Solution


For the first two questions, I know I have to sub in values for a and b. But, i'm not sure what the output matrix is.
The 3rd q, to prove B is Basis for R^3. I just have to row reduce [(1,5);(1,6)] to get [(1,0);(0,1)] right? But like before, I don't know how to do b).

Any help is appreciated.
 

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  • #2
HallsofIvy
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To find the matrix representing a given linear transformation, in a given ordered basis, apply the linear transformation to each basis vector in turn. Write the result as a linear combination of the basis vectors. The coefficients give the columns of the matrix.

For example, [itex]f(x^2)= (x^2)'= 2x= 2(1+ x)- 2(1)= -2(1)+ 2(1+ x)+ 0(x^2)[/itex]. Since the basis is [itex]\{1, 1+ x, x^2\}[/itex], in that order, the third column of the matrix would be
[tex]\begin{bmatrix}-2 \\ 2 \\ 0\end{bmatrix}[/tex]


For 6(b) there are two different ways to do this. The most direct would be to use the matrix, as given, to determine [itex]T\left(\begin{array}{c} 1 \\ 5\end{array}\right)[/itex] and [itex]T\left(\begin{array}{c}1 \\ 6\end{array}\right)[/itex] and then write the results as linear combinations of those new basis vectors. Another would be to construct the "change of basis" matrix and multiply by that.
 
  • #3
48
0
To find the matrix representing a given linear transformation, in a given ordered basis, apply the linear transformation to each basis vector in turn. Write the result as a linear combination of the basis vectors. The coefficients give the columns of the matrix.

For example, [itex]f(x^2)= (x^2)'= 2x= 2(1+ x)- 2(1)= -2(1)+ 2(1+ x)+ 0(x^2)[/itex]. Since the basis is [itex]\{1, 1+ x, x^2\}[/itex], in that order, the third column of the matrix would be
[tex]\begin{bmatrix}-2 \\ 2 \\ 0\end{bmatrix}[/tex]


For 6(b) there are two different ways to do this. The most direct would be to use the matrix, as given, to determine [itex]T\left(\begin{array}{c} 1 \\ 5\end{array}\right)[/itex] and [itex]T\left(\begin{array}{c}1 \\ 6\end{array}\right)[/itex] and then write the results as linear combinations of those new basis vectors. Another would be to construct the "change of basis" matrix and multiply by that.
Thanks HallsOfIvy; I still don't understand how to do the other two though. My approach is sub in values for a and b which correspond to the co-effs. of the given basis. Then assemble a matrix from them. Eg: f[(5,0);(1,3)]=[(1,1);(2;-1)] which gives me [(-17,4);(15,-3)] for c).
 
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