Find all 2 x 2 matrices A for which E(sub 7) =R^2.

Does E(sub 7) mean there are 7 eigenspaces? Can someone clarify this problem further?

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- Dec 1st 2009, 12:18 PMnoles2188Eigenspaces and eigenvectors
Find all 2 x 2 matrices A for which E(sub 7) =

**R**^2.

Does E(sub 7) mean there are 7 eigenspaces? Can someone clarify this problem further? - Dec 1st 2009, 01:14 PMtonio
- Dec 1st 2009, 06:14 PMnoles2188
According to my book:

E(sub 7) = ker(A - 7I(sub 2)) = {**v**[IMG]file:///C:/Users/Geoffrey/AppData/Local/Temp/moz-screenshot.png[/IMG]in**R**^2 : A**v**= 7**v**}

7 is an eigenvalue and, to me, the question is asking to find matrices A such that A**v**= 7**v**. How would I go about this? - Dec 1st 2009, 11:12 PMlepton
$\displaystyle E_7$ denotes the eigenspace associated with the eigenvalue 7.

- Dec 2nd 2009, 02:40 AMtonio

Ok then, let $\displaystyle A=\left(\begin{array}{cc}a&b\\c&d\end{array}\right )$ be such a matrix, then $\displaystyle \left(\begin{array}{cc}a&b\\c&d\end{array}\right)\ left(\begin{array}{c}x\\y\end{array}\right)=\left( \begin{array}{c}7x\\7y\end{array}\right)\,\,\foral l\,x\,,\,y\in \mathbb{R}$ , since they want $\displaystyle E_7=\mathbb{R}^2$.

Well, now choose wisely $\displaystyle x\,,\,y\in\mathbb{R}$ to find the coefficients of the matrix. For example, choosing $\displaystyle x=1\,,\,y=0$ gives us at once that $\displaystyle a=7\,,\,c=0$...

Tonio