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Please help me calculate the Eigen vectors of this matrix.
$$\begin{pmatrix} 3 & 0 & 1\\ 1 & 3 & 0\\ 0 & 1 & 3 \end{pmatrix}$$

The first vector comes out to be null, no clue how to find out the other two.

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    By null do you mean the zero vector? In that case you must've made a mistake somewhere.2012-12-21
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    I dont get you, the first Eigen vector comes out to be a zero vector. Its quite visible from the question.2012-12-21
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    http://www.wolframalpha.com/input/?i=eigenvectors+of+%7B%7B3%2C0%2C1%7D%2C%7B1%2C3%2C0%7D%2C%7B0%2C1%2C3%7D%7D2012-12-21
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    This matrix has three distinct eigenvectors (two complex). It has a complete set of non-zero eigenvectors.2012-12-21
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    Did you compute the characteristic polynomial of the matrix?2012-12-21
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    The eigenvector that is obvious is $(1,1,1)$, with eigenvalue $4$, since each row has the same sum ($4$).2012-12-21
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    After observing that, calculate the characteristic polynomial, $\text{det}(M-\lambda I)$, divide it by $\lambda - 4$, and find the remaining two roots.2012-12-21
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    @AdnanZahid Eigenvectors by definition cannot be the zero vector, after all the main interest in eigenvectors is for forming eigenbasis in which zero vectors are a big no. Also, eigenvalues themselves are defined as the values for which $\det(A-\lambda I) = 0$ so there must be a non-trivial solution.2012-12-21
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    Thanks alot, I often get confused with Eigen-value problems. I have a few other questions too, if you guys dont mind :)2012-12-21

1 Answers 1

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Try $v_1=(1,1,1)^T$, $v_2=(1-i\sqrt{3},-2,1+i\sqrt{3})^T$, $v_3 = \overline{v_2}$.

$A v_i = \lambda_i v_i$, where $A$ is the matrix above and $\lambda_i$ can be found by solving $\lambda^3-9\lambda^2+27 \lambda -28 = 0$. (By inspection, $4$ is a solution, and synthetic division results in $x^2-5x+7=0$.)

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    Can you suggest how to evaluate its complex Eigen values?2012-12-21
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    Yes, solve $x^2-5x+7=0$.2012-12-21
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    Okay, thank you so much.2012-12-21
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    You are very welcome.2012-12-21
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    Also suggest how to compute eigen vectors for identity matrix.2012-12-21
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    Since $I v = 1 v$ for any $v$, **any** non-zero vector is an eigenvalue of $1$ corresponding to the eigenvalue $1$. The characteristic polynomial is $\det (xI-I)= \det((x-1)I) = (x-1)^n$, hence all eigenvalues are $1$.2012-12-21
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    I'm sorry my question was regarding the Eigen 'vectors'.2012-12-21
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    Oops, my mistake the first word 'eigenvalue' should have been 'eigenvector'. Every non-zero vector is an eigenvector of $I$.2012-12-21
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    You left out the $i$'s.2012-12-21
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    @RobertIsrael: Fixed. Thanks!2012-12-21