# Easy Bases Linear Algebra

1. Nov 22, 2011

### shaon0

1. The problem statement, all variables and given/known data
See Attachment.

3. 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. Nov 22, 2011

### HallsofIvy

Staff Emeritus
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, $f(x^2)= (x^2)'= 2x= 2(1+ x)- 2(1)= -2(1)+ 2(1+ x)+ 0(x^2)$. Since the basis is $\{1, 1+ x, x^2\}$, in that order, the third column of the matrix would be
$$\begin{bmatrix}-2 \\ 2 \\ 0\end{bmatrix}$$

For 6(b) there are two different ways to do this. The most direct would be to use the matrix, as given, to determine $T\left(\begin{array}{c} 1 \\ 5\end{array}\right)$ and $T\left(\begin{array}{c}1 \\ 6\end{array}\right)$ 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. Nov 22, 2011

### shaon0

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).

Last edited: Nov 22, 2011