1. ## General Solution Eigenvale

I'm reading my textbook, in the Diagonalization and Eigenvalues section, and am wondering how they get this...

"λI-A = $\left(\begin{array}{ccc} -5 & -5\\ -1 & -1\end{array}\right)$

so the general solution to (λI-A)X = 0 is X = t $\left(\begin{array}{ccc} -1\\ 1\end{array}\right)$ where t is an arbitrary real number."

The book makes it sound as if the general solution of X = t $\left(\begin{array}{ccc} -1\\ 1\end{array}\right)$ is very obvious, but I'm not seeing how it's done.

Could someone enlighten me?

2. Originally Posted by Thomas
I'm reading my textbook, in the Diagonalization and Eigenvalues section, and am wondering how they get this...

"λI-A = $\left(\begin{array}{ccc} -5 & -5\\ -1 & -1\end{array}\right)$

so the general solution to (λI-A)X = 0 is X = t $\left(\begin{array}{ccc} -1\\ 1\end{array}\right)$ where t is an arbitrary real number."

The book makes it sound as if the general solution of X = t $\left(\begin{array}{ccc} -1\\ 1\end{array}\right)$ is very obvious, but I'm not seeing how it's done.

Could someone enlighten me?
the reduced row echelon form of $\lambda I - A$ is $\left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$

thus we have that $x_1 + x_2 = 0$ if we set $\lambda I - A = 0$. letting $x_2 = t$ we have that $x_1 = -t$. so it follows $\bold{x} = {x_1 \choose x_2} = t {-1 \choose 1}$

text books make things seem obvious sometimes, when they are not always obvious. don't get discouraged if you don't see it right away

3. Thank you. I knew it wasn't too hard - but sometimes text books like to skip over small details like that.

4. Originally Posted by Thomas
Thank you. I knew it wasn't too hard - but sometimes text books like to skip over small details like that.
hehe, sometimes they skip over big details too. most of the time they want the reader to figure out what happened, sometimes though, they're just trying to save page space