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Let $T:\ P_2 \to P_1$ be a linear transformation defined by

$$T(a_0 + a_1 x + a_2 x^2) = (a_0 + a_1) - (2a_1 + 3a_2)x.$$

Find the matrix $T$ with respect to the standard bases $B = \{1, x, x^2\}$ and $B' = \{1, x\}$ for $P_2$ and $P_1$.

My Attempt

I'm not sure what to start with on this problem. I think that I should be plugging in $B$. But, then I'm not sure where $B$ belongs within the equation. Do I need to plug in the entire vector $B$ at once? For example, let $1 = a_0$, $x = a_1$, and $x^2 = a_2$? That would leave me with the answer $1 + x - 2x^2 - 3x^3$.

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ZettaSuro has given you the explanation but let me add a bit more to help you understand the problem deeply. Consider the following example of a transformation.

$ T: R^{2}\rightarrow R^{2} \mbox{ given by } T(x_{1},x_{2})=2x_{1}+3x_{2} $

All this transformation is doing is taking a vector in the plane and scaling it in each respective direction i.e $x$ and $y$. Now there is one idea that needs to be emphasized behind these types of transformations.

Any linear transformation can be represented by a matrix or a matrix is really a linear transformation from one space to another

If we change our example to the one below

$ T: R^{2}\rightarrow R^{2} \mbox{ given by } T(x_{1},x_{2})=2x_{1}^{2}+3x_{2} $

then we no longer have a linear transformation(check for yourself) and there is no representation of the transformation using a matrix.

So how do you find the matrix that corresponds to a linear transformation? All you need to do is consider what the transformation does to the basis vectors. Let's see the first example above. In this case the basis vectors are $\begin{bmatrix}0\\1 \end{bmatrix}$ and $\begin{bmatrix}1\\0 \end{bmatrix}$.

$ T \begin{bmatrix}0\\1 \end{bmatrix}=\begin{bmatrix}0\\3 \end{bmatrix} $ and $ T \begin{bmatrix}1\\0 \end{bmatrix}=\begin{bmatrix}2\\0 \end{bmatrix} $

Now, here is the key. Consider any other vector $\begin{bmatrix}a\\b \end{bmatrix}$ on the plane and apply the transformation $$ T\left(\begin{bmatrix}a\\b \end{bmatrix}\right)=T\left(a\begin{bmatrix}1\\0 \end{bmatrix} +b\begin{bmatrix}0\\1 \end{bmatrix} \right)\\ T\left(\begin{bmatrix}a\\b \end{bmatrix}\right)=aT\begin{bmatrix}1\\0 \end{bmatrix} +bT\begin{bmatrix}0\\1 \end{bmatrix} \\ T\left(\begin{bmatrix}a\\b \end{bmatrix}\right)=\begin{bmatrix} T\begin{bmatrix}1\\0 \end{bmatrix}& T\begin{bmatrix}0\\1 \end{bmatrix} \end{bmatrix} \begin{bmatrix} a\\ \\b \end{bmatrix} $$

Now we can just read the matrix from the calculation and the matrix is \begin{bmatrix} 2 & 0\\ 0 & 3\\ \end{bmatrix}

You can apply this for your case and get the matrix below \begin{bmatrix} 1 & 1 & 0\\ 0 & -2 & 3\\ \end{bmatrix}

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It may help to look at it as a set of Euclidean vectors:

$$T\begin{bmatrix}a_0\\ a_1 \\ a_2 \end{bmatrix}=\begin{bmatrix}a_0+a_1 \\ -(a_1+3a_2)\end{bmatrix}$$

This will be a $2\times3$ matrix (2 rows, 3 columns):

$$t_{11}a_0+t_{12}a_1+t_{13}a_2=a_0+a_1 + 0a_2$$

$$t_{21}a_0+t_{22}a_1+t_{23}a_2=0a_0-2a_1-3a_2$$

Now just match up the $t_{ij}$ variables to their coefficients on right hand side.

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  • $\begingroup$ Okay, so would the answer be: 1 1 0 for the first row and 0 -2 -3 for the second? (I think you may have left off the two when you made the comment above - but I think I've got it regardless). $\endgroup$
    – briteId
    Jul 17 '13 at 23:06
  • $\begingroup$ @Cee Oh, yes sorry about the typo. Yes your answer is correct. $\endgroup$ Jul 17 '13 at 23:15

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