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By definition:

$$Ax = \lambda x$$

for $x \neq 0$

I was using this to calculate the eigenvectors for $A$:

$$A = \begin{bmatrix}2 & 0 & 0\\1 & 3 & 0\\ 2 & 3& 4\end{bmatrix}$$

I already found that the eigenvalues for this matrix are:

$$\lambda = 2,3,4$$

However, when I began evaluating the eigenvectors for these eigenvalues, I noticed that for $\lambda = 4$, the resulting eigenvector is:

$$x = \begin{bmatrix}0\\0\\0\end{bmatrix}$$

But how is this possible??


Please help me find my mistake.

To find the eigenvectors, we must solve the equation:

$$(\lambda I -A)x = 0$$

$$\begin{bmatrix} 4 & 0 & 0\\0 & 4& 0\\ 0 & 0 & 4 \end{bmatrix} - \begin{bmatrix}2 & 0 & 0\\1 & 3 & 0\\ 2 & 3& 4\end{bmatrix} = \begin{bmatrix}2 & 0 & 0\\-1 & 1 & 0\\ -2 & -3 & 0\end{bmatrix}$$

Row reducing, we see that

$$ \left[ \begin{array}{ccc|c} 1&0&0&0\\ 0&-1&0&0\\ 0&3&0&0 \end{array} \right] $$

...which tells us that $x_1=0, x_2=0, 3x_2=0$.

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    $\begingroup$ There is a nonzero eigenvector; you must have made a mistake somewhere. $\endgroup$
    – Javier
    May 1, 2016 at 19:27
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    $\begingroup$ Well, it is always true that $(A-\lambda I) 0 = 0$, but an eigenvector must be non zero. In this case, you can 'read off' an eigenvector corresponding to $\lambda =4$ as $(0,0,1)^T$. $\endgroup$
    – copper.hat
    May 1, 2016 at 19:27
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    $\begingroup$ The last line is wrong, $x_3$ is not $0$ from that equation system. $\endgroup$ May 1, 2016 at 20:00
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    $\begingroup$ @whatwhatwhat Au contraire: $\;z\;$ doesn't appear at all in the homogeneous system and it thus can take any value whatsoever! $\endgroup$
    – DonAntonio
    May 1, 2016 at 20:05
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    $\begingroup$ z is free choice because it has a 0 coefficient for each row. There is no equation that can restrict it if it is multiplied by 0 in the equations. $\endgroup$ May 1, 2016 at 20:05

3 Answers 3

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The resulting homogeneous system for $\lambda=4$ ( i.e., $\;(\lambda I-A)\vec x=\vec0\;$) is:

$$\begin{cases}2x=0\\-x+y=0\\-2x-3y=0\end{cases}\implies\;\;x=y=0\implies \begin{pmatrix}0\\0\\1\end{pmatrix}$$

is an eigenvector for $\;\lambda=4\;$ and, in fact, it is a basis for the corresponding eigenspace

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  • $\begingroup$ The difference between my work and yours is the $x+y$. I got $-x+y$. I got a negative value for $x$ because when doing $\lambda I -A$ we see that the 1st column 2nd row of $\lambda I$ is $0$. Subtracting the corresponding value from $A$, we get $0-1=-1$. $\endgroup$ May 1, 2016 at 19:43
  • $\begingroup$ So how did you get a positive value? $\endgroup$ May 1, 2016 at 19:43
  • $\begingroup$ @whatwhatwhat Thank you, you're completely right: it must be $\;-x\;$ there and, in fact, also $\;-2x-3y=0\;$ in the last equation. I did it in my mind and, of course, it makes no difference in this case. I shall edit now. $\endgroup$
    – DonAntonio
    May 1, 2016 at 19:52
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There seems to be some confusion here. The eigenvalues solve the characteristic equation $\det({\bf A}-\lambda {\bf I}) = 0$ , but there can be several eigenvalues for one eigenvector (multiple roots). Take for example the matrix $$A = \left[\begin{array}{ccc} 2&0&0\\1&2&0\\0&0&4 \end{array}\right] \hspace{1cm} A-2I = \left[\begin{array}{ccc} 0&0&0\\1&0&0\\0&0&2 \end{array}\right]$$

This only gives one vector ($[0,1,0]^T$), although there are two roots $\lambda = 2$.

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You made a mistake, the (normed) eigenvector for the eigenvalue $4$ is $e_z$.

>> A
A =

   2   0   0
   1   3   0
   2   3   4

>> [x,y]=eig(A)
x =

   0.00000   0.00000   0.66667
   0.00000   0.31623  -0.66667
   1.00000  -0.94868   0.33333

y =

Diagonal Matrix

   4   0   0
   0   3   0
   0   0   2

And this is easy to verify as $e_z = (0,0,1)^t$ picks the third column, $(0,0,4)^t$, of $A$ and this is $4 \cdot e_z$.

By hand you would solve $A_1 u = 0$ for $$ A_1 = \begin{pmatrix} 2 - 4 & 0 & 0 \\ 1 & 3-4 & 0 \\ 2 & 3 & 4-4 \end{pmatrix} = \begin{pmatrix} -2 & 0 & 0 \\ 1 & -1 & 0 \\ 2 & 3 & 0 \end{pmatrix} $$ This has the solutions $u = (x,y,z) = (0,0,z)$ for arbitrary $z \in \mathbb{R}$. Here is the Gauss elimination, using augmented matrices:

$$ A_1 u = 0 \iff \\ \left[ \begin{array}{rrr|c} -2 & 0 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 2 & 3 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|c} 1 & 0 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 2 & 3 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|c} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 3 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] $$

Update:

...which tells us that x1=0,x2=0,x3=0.

No it does not. It just tells you $x_1$ and $x_2$ are zero.

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  • $\begingroup$ I don't understand what this is supposed to tell me. $\endgroup$ May 1, 2016 at 19:44
  • $\begingroup$ $x$ is the matrix of eigenvectors, $y$ is the diagonal matrix with the corresponding eigenvalues. In other words I asked my machine for the results to compare with yours. $\endgroup$
    – mvw
    May 1, 2016 at 19:45

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