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Let $f:P_2[x]\rightarrow M_{2\times 2}(\mathbb{R})$ is a linear transformation defined as $f(a+bx+cx^2)= \begin{bmatrix} b+c & a \\ b & c \\ \end{bmatrix}$ Find the matrix of $f$ with respect to basis $ ({\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 0 & 0 \\ \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 0 \\ \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}})$ and $(1,x,x^2)$. Find a basis and dimension of $Im (f)$ and $Ker(f)$.

In a standard basis for $P_2[x]$, the mapping is: $$f \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix}= \begin{bmatrix} 0 + b + c \\ a + 0 + 0 \\ 0 + b + 0 \\ 0 + 0 + c \end{bmatrix}\Rightarrow f=\begin{bmatrix} 0 & b & c \\ a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$$ Question: Is the matrix of $f$ the same for both basis $ ({\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 0 & 0 \\ \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 0 \\ \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}})$ and $(1,x,x^2)$? If not, what is the method for finding the matrix of $f$ in both cases?

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For the standard basis of $P_2[x]$, your answer is wrong: there can't be $a,b,c $ in the matrix, which should be: $$\begin{pmatrix} 0&1&1\\1&0&0\\0&1&0\\0&0&1 \end{pmatrix}.$$ Now with the given basis of $P_2[x]$, first set $$E_1=\begin{pmatrix} 1&0\\0&0 \end{pmatrix},\enspace E_2=\begin{pmatrix} 1&1\\0&0 \end{pmatrix},\enspace E_3=\begin{pmatrix} 1&1\\1&0 \end{pmatrix},\enspace E_4=\begin{pmatrix} 1&1\\1&1 \end{pmatrix}. $$ Then note that \begin{align*} f(1)=\begin{pmatrix} 0&1\\0&0 \end{pmatrix}&=E_2-E_1, &f(x)=\begin{pmatrix} 1&0\\1&0 \end{pmatrix}&=E_3-E_2+E_1,\\ f(x^2)=\begin{pmatrix} 1&0\\0&1 \end{pmatrix}&=E_4-E_3+E_1.\\ \end{align*} whence the matrix of $f$: $$\begin{pmatrix} -1&1&1\\1&-1&0\\0&1&-1\\0&0&1 \end{pmatrix}.$$

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First of all, something about the notations. I would rather write

$$[f]_E^F \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix}= \begin{bmatrix} 0 + b + c \\ a + 0 + 0 \\ 0 + b + 0 \\ 0 + 0 + c \end{bmatrix}\Rightarrow [f]_E^F=\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

instead of $f$ itself, since $f$ is not a matrix but an operator. $[f]_E^F$ is the matrix representation of $f$ with respect to the bases $E = \{1,x,x^2\}$ for $V = P_2$ and $F = \left\{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right\}$ for $W = M_{2 \times 2}$.

The above matrix representation $[f]_E^F$ of $f$ with respect to $E$ and $F$ is constructed column-by-column sense where the $j$-th column of the matrix is determined by:

  1. $v_j = x^{j-1}$ is the $j$-th vector in the basis $E$.
  2. $f(v_j)$ is in the vector space $W$.
  3. $f(v_j)$ can be written into a linear combination of the basis $F$.
  4. $[f(v_j)]_F$ contains the coefficients of $f(v_j)$ with respect to the basis $F$.
  5. $[f(v_j)]_F$ is the $j$-th column of $[f]_E^F$.

Now can you do the same with the basis $F$ replaced by $G = \left\{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\right\}$?

This is the direct way of doing this question. Another way is to use the matrix representation of composite of linear operators and the change of coordinate matrix. You can look for more if interested.

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Your question is fuzzy: $f$ has one matrix with respect to the two given bases. So there is only one case.

Let's call the matrix we want to find $M_f$ and write the coefficients of the two given vectors down:

The vector $a + bx + cx^2$ with respect to the basis $(1,x,x^2)$ is $(a,b,c)$.

The vector $\begin{bmatrix} b+c & a \\ b & c \\ \end{bmatrix}$ with respect to the basis $({\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 0 & 0 \\ \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 0 \\ \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}})$ is $(b+c-a,a-b,b-c,c)$.

So, $M_f (a,b,c)^T = (b+c-a,a-b,b-c,c)^T$.

If you denote the $3$ columns of $M_f$ by $c_1, c_2, c_3$ then you want $c_i$ to be such that

$$ a c_1 + b c_2 + c c_3 = (b+c-a,a-b,b-c,c)$$

How to find the $c_i$?

Note that a linear map is uniquely determined by what it maps basis vectors to. In this case, we have the basis vectors

$$ 1 = (1,0,0)^T = b_1, x = (0,1,0)^T=b_2, x^2 = (0,0,1)^T = b_3$$

and

$$ M_f(b_1) = (-1,1,0,0), M_f(b_2) = (1,-1,1,0), M_f(b_3)=(1,0,-1,1)$$

Can you read the columns of $M_f$ off of this?

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