# finding the basis of an eigenspace

"Given the matrix $A=\left[\begin{array}{cc}6 & 6\\-2 & -1\end{array}\right].$

"The roots of the characteristic polynomial of $A$ are: \begin{align}\det(A-tI)=0 &\Leftrightarrow \det\left[\begin{array}{cc}6-t & 6\\-2 & -1-t\end{array}\right]=0\\ &\Leftrightarrow 6-5t+t^2=0\\ &\Leftrightarrow (t-2)(t-3)=0\\ &\Leftrightarrow t=2,3\end{align} "So, the eigenvalues of $A$ are $\fbox{2,3}.$ "

Then I subtract the diagonals by the eigenvalue $2$ and I guess we are then supposed to find the rref but I get a different answer than the solution manual of my textbook:

"Now by using the reduced row echelon form, we have $$\left[\begin{array}{cc}2 & 3\\0 & 0\end{array}\right]\left[\begin{array}{c}x_1\\x_2\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]\Rightarrow 2x_1+3x_2=0.$$ "Therefore $x_1=3$ $x_2=-2$ satisfying the above equations.

"Therefore, $\left[\begin{array}{c}3\\-2\end{array}\right]$ is the eigenvector corresponding to the eigenvalue $\lambda=2.$

"So $\fbox{$\left\{\left[\begin{array}{c}3\\-2\end{array}\right]\right\}$}$ is a basis for the eigenspace of $A$ corresponding to eigenvalue $\lambda=2.$"

Shouldnt the RREF be

$$\left[\begin{array}{cc}1 & 3/2\\0 & 0\end{array}\right]$$

so then the basis vector would be

$$\left[\begin{array}{c}-3/2\\1\end{array}\right]?$$

Am I doing something wrong?

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I have edited your post (and fixed the matrix at the beginning). Please let me know if any of my changes are not in the spirit of your original post. – Cameron Buie Jul 21 '13 at 15:31

What you've done is not wrong. (I do note, though, that your matrix $A$ at the beginning doesn't seem to match the rest of your post.) In fact, your sole basis vector is a non-zero multiple of their sole basis vector, so they span the same space (as they should). Note that the book simply asked for a basis. There are infinitely-many such, and yours is one of them.