# Do elementary row operations change eigenvalues?

May 2, 2016

Yes. For a given matrix $\hat{A}$, elementary row operations do NOT retain the eigenvalues of $\hat{A}$.

For instance, take the following matrix:

$\textcolor{g r e e n}{\hat{A} = \left[\begin{matrix}2 & 2 \\ 0 & 1\end{matrix}\right]}$

The eigenvalues are determined by solving

$\setminus m a t h b f \left(\hat{A} \vec{v} = \lambda \vec{v}\right) ,$

such that $| \lambda I - \hat{A} | = 0$. Then, the eigenvectors $\vec{v}$ form a basis acquired from solving $\left[\lambda I - \hat{A}\right] \vec{v} = \vec{0}$ for $\vec{v}$.

$| \lambda I - \hat{A} | = 0$

$= | \left[\begin{matrix}\lambda & 0 \\ 0 & \lambda\end{matrix}\right] - \left[\begin{matrix}2 & 2 \\ 0 & 1\end{matrix}\right] |$

$= | \left(\lambda - 2 , - 2\right) , \left(0 , \lambda - 1\right) |$

$= \left(\lambda - 2\right) \left(\lambda - 1\right) - 0$

From this we acquire the characteristic equation:

$\implies \textcolor{g r e e n}{\left(\lambda - 2\right) \left(\lambda - 1\right) = 0} ,$

And we get the eigenvalues

$\implies \textcolor{b l u e}{\lambda = 1 , 2} ,$

whose eigenvectors are...

$\setminus m a t h b f \left(\lambda = 1\right)$:

$\left[\begin{matrix}\lambda - 2 & - 2 \\ 0 & \lambda - 1\end{matrix}\right] \left[\begin{matrix}{v}_{1} \\ {v}_{2}\end{matrix}\right] = \left[\begin{matrix}0 \\ 0\end{matrix}\right]$

$= \left[\begin{matrix}- 1 & - 2 \\ 0 & 0\end{matrix}\right] \left[\begin{matrix}{v}_{1} \\ {v}_{2}\end{matrix}\right] = \left[\begin{matrix}0 \\ 0\end{matrix}\right]$

$\to {v}_{1} = - 2 {v}_{2} \to \textcolor{b l u e}{\vec{v} = {v}_{1} \left[\begin{matrix}1 \\ - 2\end{matrix}\right]}$

$\setminus m a t h b f \left(\lambda = 2\right)$:

$\left[\begin{matrix}\lambda - 2 & - 2 \\ 0 & \lambda - 1\end{matrix}\right] \left[\begin{matrix}{v}_{1} \\ {v}_{2}\end{matrix}\right] = \left[\begin{matrix}0 \\ 0\end{matrix}\right]$

$= \left[\begin{matrix}0 & - 2 \\ 0 & 1\end{matrix}\right] \left[\begin{matrix}{v}_{1} \\ {v}_{2}\end{matrix}\right] = \left[\begin{matrix}0 \\ 0\end{matrix}\right]$

$\to {v}_{2} = 0 , {v}_{1} = \text{anything}$, :. let ${v}_{1} = 9000$. Then, $\textcolor{b l u e}{\vec{v} = {v}_{1} \left[\begin{matrix}9000 \\ 0\end{matrix}\right]}$

Of course, had you row-reduced $\hat{A}$, you would have gotten:

$\hat{A} = \left[\begin{matrix}2 & 2 \\ 0 & 1\end{matrix}\right]$

stackrel(1/2R_1; -R_2+R_1" ")(->)[(1,0),(0,1)],

where the notation $c {R}_{i} + {R}_{j}$ implies that $c$ times row $i$ is added to row $j$ and the result is stored into row $j$.

That would give you the characteristic equation $| \lambda I - \hat{A} | = {\left(\lambda - 1\right)}^{2} = 0$, and thus give you one eigenvalue $\lambda = 1$ of multiplicity $2$.

However, without row-reduction, we had gotten two distinct eigenvalues: $\lambda = 1 , 2$. Thus, the eigenvalues were not retained as a result of elementary row operations.