Two Linear Transformation Problems

**DarrenM** · September 14th 2011, 08:07 AM

I think I've got a rather glaring hole in my understanding of this material. I've read and reread the text, watched the MIT Open Courseware lectures, and beat my head against this material for quite a while now. Any help would be much appreciated. Without further ado...

1) Define a linear transformation $L: P_{3} \rightarrow P_{2}$ by $L(f(x) = f'(x) + \int_0^6 \! f(x) \, \dif x$ . Consider ordered bases $B = {1, x, x^2}$ and $C = {1, x}$ for $P_{3}$ and $P_{2}$ respectively.

a) Find the representation matrix $A$ of $L$ with respect to the given bases $B$ and $C$ .
b) Express $[L(a_{0} + a_{1}x + a_{2}x^2)]_C$ in terms of $a_{0}, a_{1}, a_{2}$

2) Let $\mathbb{R}^{2 x 2}$ be the vector space of all 2 x 2 matrices (over $\mathbb{R}$ . Consider
$E_{11} = \begin{array}{|cc|}1&0\\0&0\end{array}$ , $E_{12} = \begin{array}{|cc|}0&1\\0&0\end{array}$ , $E_{21} = \begin{array}{|cc|}0&0\\1&0\end{array}$ , $E_{22} = \begin{array}{|cc|}0&0\\0&1\end{array}$ ; and $M = \begin{array}{|cc|}3&4\\5&6\end{array}$ .

It is known that $E = E_{11}, E_{12}, E_{21}, E_{22}$ is an ordered basis for $\mathbb{R}^{2 x 2}$ . Define $L: \mathbb{R}^{2 x 2} \rightarrow \mathbb{R}^{2 x 2}$ by $L(x) = Mx$ for all $x$ = $\begin{array}{|cc|}x_11&x_12\\x_21&x_22\end{array} \in \mathbb{R}^{2 x 2}$ .

Edit: Couldn't get the format right for that x matrix. Is should be x_{11}, x_{12}, x_{21}, x_{22}.

a) Find $[M]_E$ , the coordinate vector of $M$ with respect to the ordered basis $E$ .
b) Is L a linear transformation?
c) Find the representation matrix of L with respect to $E$ , if L is a linear transformation.

For 1), I found $L(1) = 6, L(x) = 19, and L(x^2)= 2x+72$ . Then I tried to set up a matrix with the bases of C and those values. I interpreted the bases of C as the standard bases (1, 0), (0, 1). I really don't think that's right, so I'm kind of stuck there.

For 2), I can't even make headway on the first part. I set up the problem like I would to find a coordinate vector, but it comes out as a 2 x 10 matrix. Trying to account for the zero columns still leaves me with a 2 x 6. I believe that this is a linear transformation, but as I said, I'm quite stumped.

Any help would be very much appreciated.

**zoek** · September 14th 2011, 11:19 AM

for 2a):

Since $M = 3E_{11}+4E_{12}+5E_{21}+6E_{22}$ ,

the coordinate vector of $M$ with respect to the ordered basis $E$ is $[M]_E = (3,4,5,6).$

for 2b):

$L(x+y) = M(x+y) = Mx+My = L(x)+L(y)$ ,
$L(\lambda x) = M(\lambda x) = \lambda Mx = \lambda L(x)$ ,

$\forall \lambda \in \mathbb{R}$ and $\forall x,y \in \mathbb R^{2x2}$ so $L$ is a linear transformation.

for 2c):

$L(E_{11}) = ME_{11}=$ $\left(\begin{array}{cc}3&4\\5&6\end{array}\right) \left(\begin{array}{cc}1&0\\0&0\end{array}\right) = \left(\begin{array}{cc}3&0\\5&0\end{array}\right)= 3E_{11}+0E_{12}+5E_{21}+0E_{22}$

so the first column of representation matrix of $L$ with respect to $E$ is $\left(\begin{array}{cc}3\\ 0\\ 5\\ 0\end{array}\right)$ .

In a similar way you can find the rest three columns.

**DarrenM** · September 15th 2011, 05:17 AM

Thank you very much. I feel incredibly dense for not having seen $M = 3E_{11}+4E_{12}+5E_{21}+6E_{22}$ , especially since I had been staring at $M = c_{1}E_{11} + c_{2}E_{12} + c_{3}E_{21} + c_{4}E_{22}$ for so long trying to make sense of a coordinate vector in terms of a 2 x 2 matrix.

Once I read that comment the third part fell into place.

As for problem 1, it turns out I was correct. Evaluating the linear transformation with the given ordered bases of $P_{3}$ , and then putting those in an augmented matrix with the ordered bases of $P_{2}$ (interpreted as (1, 0) and (0,1)) gives the matrix representation.

Thanks very much for the help.

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