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Yes, I have searched for this question. I didn't see another question that was asking the same thing, and the answers seemed to gloss over what I'm missing. I know this is really simple but I can't seem to find where I'm going wrong.

I want to find the eigenvectors of: $$A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}$$

Since $A$ is triangular I know the eigenvalues are $1,4,6$.

Using $Av-\lambda I_3 v = 0$, and factoring out $v$ to get $(A-\lambda I_3)v = 0$, I get that:

\begin{align} E_6 = \begin{bmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 1 \end{bmatrix}\vec{v} = \vec{0} \end{align}

Solving for $v_1,v_2,v_3$ I come up with the following equations:

\begin{align*} -5v_1 + 2v_2 +3v_3 = 0 \\ -2v_2+5v_3 = 0 \\ v_3 = 0 \end{align*}

Clearly $v_3 = 0$, and now we can substitute $v_3$ in the remaining equations. This leads to: \begin{align*} -5v_1+2v_2 = 0\\ -2v_2= 0 \end{align*}

But now it's clear that $v_2 = 0$. This is repeated again for $v_1$ and I've just found that $v_1 = v_2 = v_3 = 0$. Putting $E_6$ in RREF right away makes this clear at a glance. I know this is wrong because eigenvectors are defined to be not all zeros.

I am interested in where I went wrong, and particularly what theoretical hole in my understanding lead me to this wrong answer.

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    $\begingroup$ ..what is $6-6$? $\endgroup$ – user127.0.0.1 Jan 31 '14 at 6:47
  • $\begingroup$ @user127.0.0.1 Wow. Thanks $\endgroup$ – Alan Jan 31 '14 at 6:49
  • $\begingroup$ You're welcome. Btw: The matrix (that you have called $E_6$ in this special case) has to be singular by construction, no matter what $A$ and corresponding $\lambda$ look like. Thus it was not hard to spot the error :) $\endgroup$ – user127.0.0.1 Jan 31 '14 at 6:56
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$A e_1 = e_1$.

$A(\mu,1,0)^T = (2+\mu, 4 ,0)^T$, hence if we choose $4 \mu = (2+\mu)$, we get $\mu = {2 \over 3}$ and $A({2 \over 3},1,0)^T = 4({2 \over 3}, 1 ,0)^T$.

A similar analysis on $A(x_1,x_2,1)^T = 6 (x_1,x_2,1)^T$ yields $x_1 = {8 \over 5}, x_2 = {5 \over 2}$.

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Not sure this needs a full answer, but I've decided to add some details to this:

As pointed out in the comments, this is a matter of a simple subtraction mistake ($6-6\ne 1)$, however, you could have stopped and realized you've made a mistake at two points already:

1) When you found out $v_2$ is zero: We have a 3x3 matrix and three distinct eigenvalues. Since there has to be at least one eigenvector per eigenvalue (why?), we cannot end up with one having two.

2) When you found out matrix $E_6$ is regular: The matrix must be singular by its construction. (Try to prove that!)

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