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Let linear transformation is defined as

$\mathcal{A}(1,1,1)=(1,0,0)$

$\mathcal{A}(1,-1,0)=(1,1,1)$

$\mathcal{A}(1,0,1)=(1,1,1)$

Find matrix of $\mathcal{A}$ and inverse (not in matrix representation, if exists).

Attempt:

Transformation $\mathcal{A}$ can be represented as $\mathcal{A}:\mathbb{R^3}\rightarrow \mathbb{R^3}$ which can be represented by $\mathcal{A}:\mathcal{P_2}\rightarrow \mathcal{P_2}$ where $\mathcal{P_2}$ is the space of polynomials $p(x)=a+bx+cx^2$.

$\mathcal{A}(1,1,1)=(1,0,0)\Rightarrow A \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix}= \begin{bmatrix} \alpha \\ 0 \\ 0 \\ \end{bmatrix}$

How to find matrix of $\mathcal{A}$?

Is this approach (with polynomial vector space) correct?

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  • $\begingroup$ What does "not in matrix representation" means? $\endgroup$ – Shoutre Dec 21 '15 at 17:03
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    $\begingroup$ With respect to which basis do you want to find the matrix representation? $\endgroup$ – Alekos Robotis Dec 21 '15 at 17:04
  • $\begingroup$ @Antonios-Alexandros Robotis , In standard basis. $\endgroup$ – user300048 Dec 21 '15 at 17:05
  • $\begingroup$ @Shoutre, General representation, with domain and co-domain. $\endgroup$ – user300048 Dec 21 '15 at 17:06
  • $\begingroup$ Can you find the matrix with respect to the basis $(1,1,1), (1,-1,0), (1,0,1)$? $\endgroup$ – Alekos Robotis Dec 21 '15 at 17:06
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Let $$ A=\begin{bmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3\\ \end{bmatrix} $$

we have: $$ A \begin{bmatrix}1\\1\\1 \end{bmatrix}=\begin{bmatrix}1\\0\\0 \end{bmatrix} \Rightarrow \begin {cases} a_1+b_1+c_1=1\\ a_2+b_2+c_2=0\\ a_3+b_3+c_3=0\\ \end {cases} $$

$$ A \begin{bmatrix}1\\-1\\0 \end{bmatrix}=\begin{bmatrix}1\\1\\1 \end{bmatrix} \Rightarrow \begin {cases} a_1-b_1=1\\ a_2-b_2=1\\ a_3-b_3=1\\ \end {cases} $$

$$ A \begin{bmatrix}1\\0\\1 \end{bmatrix}=\begin{bmatrix}1\\1\\1 \end{bmatrix} \Rightarrow \begin {cases} a_1+c_1=1\\ a_2+c_2=1\\ a_3+c_3=1\\ \end {cases} $$

Find $b_i=1+a_i$ and $c_i=1-a_i$ from the last two systems, so that $b_i+c_i=2$, and substitute in the first system. You find:

$$ A= \begin{bmatrix} -1&0&2\\ -2&-1&3\\ -2&-1&3\\ \end{bmatrix} $$ that clearly is not invertible.

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Hint: Notice that if you call $$ e_1 = \begin{pmatrix} 1 \\ 1 \\1 \end{pmatrix} , \quad e_2 = \begin{pmatrix} 1 \\ -1 \\0 \end{pmatrix} , \quad e_3 = \begin{pmatrix} 1 \\ 0 \\1 \end{pmatrix} $$ We have that $$ e_1 + e_2 - e_3 = \begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix} $$ $$ e_1 - e_3 = \begin{pmatrix} 0 \\ 1 \\0 \end{pmatrix} $$ $$ - e_1 -e_2 + 2 e_3 = \begin{pmatrix} 0 \\ 0 \\1 \end{pmatrix} $$

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Let S be the matrix for $\mathcal{A}$.

Then $S\begin{bmatrix}1&1&1\\1&-1&0\\1&0&1\end{bmatrix}=\begin{bmatrix}1&1&1\\0&1&1\\0&1&1\end{bmatrix}$,

so $S=\begin{bmatrix}1&1&1\\0&1&1\\0&1&1\end{bmatrix}\begin{bmatrix}1&1&1\\1&-1&0\\1&0&1\end{bmatrix}^{-1}=\begin{bmatrix}1&1&1\\0&1&1\\0&1&1\end{bmatrix}\begin{bmatrix}1&1&-1\\1&0&-1\\-1&-1&2\end{bmatrix}=\begin{bmatrix}1&0&0\\0&-1&1\\0&-1&1\end{bmatrix}$.

Notice that $\mathcal{A}(1,-1,0)=\mathcal{A}(1,0,1)$, so $\mathcal{A}$ is not invertible

(and S is not invertible since it has two identical rows).

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