1

In diagonalisation and, I think, generally speaking, why are we able to change lines of $A-\lambda I$ and not those of $A$?

Is it because when the search for the eigen values of $A$ we assume that we are solving a system of linear eauations where $\lambda$ is the unknown?

Furthermore, why doing such an operation is not correct and gives 5 eigen-values: \begin{bmatrix} 1 & -\lambda & 1 & \\ 0 & 1-\lambda^2 & \lambda \\ 0 & 1 & -\lambda \\ \end{bmatrix}

$$\Leftrightarrow \begin{bmatrix} 1 & -\lambda & 1 & \\ 0 & 1-\lambda^2 & \lambda \\ 0 & 0 & -\lambda(1-\lambda^2)-\lambda \\ \end{bmatrix}$$

Why do we have to do the following one, which allow us to have 3 eigen values:

\begin{bmatrix} 1 & -\lambda & 1 & \\ 0 & 1 & -\lambda \\ 0 & 1-\lambda^2 & \lambda \\ \end{bmatrix}

Revolucion for Monica
  • 1,511

1 Answers1

0

When you find eigenvalues by row reducing $A - \lambda I$, you're just manipulating coefficients of a system of linear equations by conventional elementary row operations. If you operate on $A$ instead of on $A - \lambda I$, you're solving a different set of equations (which happen to be those giving the null space of $A$).

Not sure what your proposed operation is that gives five solutions, but it's not a sequence of elementary row operations.

Andrew D. Hwang
  • 78,195
  • So If I operate on A instead of on A−λI, and then applying −λI on A I'm solving a different set of equations and this one is giving the null space of A, $\ker(A)$. Am I right? – Revolucion for Monica Nov 16 '15 at 13:49
  • An why does what I did with the matrix I gave as an exemple is not a sequence of elementary row operations? I only multiiplied the third line by $(1-\lambda)^2$ and subtracted $L_2$ from it to remove the $A_{32}$ coefficient of the matrix. – Revolucion for Monica Nov 16 '15 at 13:51
  • 1.(a) If you row reduce $A$, you're finding the null space, yes. 1.(b) If you row reduce $A$ and then subtract $\lambda I$, you're solving a system with no obvious relation to $A$. 2. Multiplying by $(1 - \lambda^{2})$ is not an elementary row operation if $\lambda = \pm1$. Those are the two extra "solutions" you pick up. – Andrew D. Hwang Nov 16 '15 at 23:09