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Find the $2 \times 2$ matrix ${A}$ such that ${A}^2 = {A}$ and ${A} \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$

I have tried to express A as a matrix with variables a,b,c, and d, but it gets too messy with the equations. Any help is appreciated.

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Notice that $A \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$ and also $A \begin{pmatrix} 6 \\ 2 \end{pmatrix} = A^2 \begin{pmatrix} 7 \\ -1 \end{pmatrix} = A \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$.

Thus $A \begin{pmatrix} 7 &6 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 6& 6 \\ 2& 2 \end{pmatrix}$.

And therefore $A = \begin{pmatrix} 6& 6 \\ 2& 2 \end{pmatrix}\begin{pmatrix} 7 &6 \\ -1 & 2 \end{pmatrix}^{-1} = \frac{1}{10}\begin{pmatrix} 9& 3 \\ 3& 1 \end{pmatrix}$.

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Hint: we have

$$A\begin{pmatrix} 7\\-1\end{pmatrix}=\begin{pmatrix} 6\\2\end{pmatrix}$$

And therefore

$$\begin{align}A\begin{pmatrix} 6\\2\end{pmatrix} &=A^2\begin{pmatrix} 7\\-1\end{pmatrix}\\ &=A\begin{pmatrix} 7\\-1\end{pmatrix}\\ &=\begin{pmatrix} 6\\2\end{pmatrix} \end{align}$$

So if $A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$ we have

$$\begin{cases} 7a-b=6\\7c-d=2\\6a+2b=6\\6c+2d=2\end{cases}$$

And this equivalent to two systems of two equations with two unknowns $(a,b)$ on one hand and $(c,d)$ on the other and I am sure you can solve.

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First, $A^2-A=0$ means $A$ must have eigenvalues 1 or 0. $A$ is not a zero matrix, so the two eigenvalues of $A$ cannot be all 0. The two eigenvalues cannot be all 1 either for otherwise $A$ would either be (1) the identity matrix which is certainly the case (2) a nondiagonalizable matrix but judging from the polynomial $A$ satisfies this cannot be the case. This leaves the only possibility that $A$ has eigenvalues 0 and 1, and so it's of rank 1.

Since $A\begin{pmatrix}7\\-1\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$ its range is $\text{Span\{(3, 1)\}}$ so $A=\begin{pmatrix}3a&3b\\a&b\end{pmatrix}$ for some $a$ and $b$. Its null space is spanned by $\begin{pmatrix}7\\-1\end{pmatrix}-\begin{pmatrix}6\\2\end{pmatrix}=\begin{pmatrix}1\\-3\end{pmatrix}$. So \begin{eqnarray}\begin{pmatrix}3a&3b\\a&b\end{pmatrix}\begin{pmatrix}1\\-3\end{pmatrix}=0\end{eqnarray} implies $a=3b$. Finally \begin{eqnarray}\begin{pmatrix}9b&3b\\3b&b\end{pmatrix}\begin{pmatrix}7\\-1\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}\end{eqnarray} implies that $b=\frac{1}{10}$. So $A=\frac{1}{10}\begin{pmatrix}9&3\\3&1\end{pmatrix}$

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