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I am trying to calculate eigenvalues and eigenvectors of this matrix

$$\frac12\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1&-1&1&-1\\ 1&1&-1&-1\\ 1&-1&-1&1 \end{pmatrix}?$$

Determinat is $\frac{1}{16}(\lambda^2-4)^2$ and it should have two eigenvalues $2$ and $-2$. I also need eigenvectors. I decided to check it by wolframalpha. However, there are 4 eigenvectors, I have only two. Where I made mistake?

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    $\begingroup$ You also have $4$ eigenvalues: $2,2,-2-2$. $\endgroup$ – Dietrich Burde May 8 '18 at 15:05
  • $\begingroup$ One eigenvalue can give you more than one eigenvector $\endgroup$ – joseabp91 May 8 '18 at 15:07
  • $\begingroup$ You must operate to realize in your example $\endgroup$ – joseabp91 May 8 '18 at 15:08
  • $\begingroup$ The eigenvalues are $\pm 1$, not $\pm 2$ - note the factor of $1/2$. $\endgroup$ – Math1000 May 8 '18 at 15:12
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    $\begingroup$ The spectrum is $\pm 1$. Note that $A A^T = A^T A$, so the matrix is normal and has a full set of eigenvectors. $\endgroup$ – copper.hat May 8 '18 at 15:21
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Note that also the identity $I$ has $1$ eigenvalues but $4$ independent eigenvectors.

In your case we have 2 eigenvalues with algebraic multiplicity equal to 2 but the key point for eigenvectors is the geometric multiplicity of each eigenvalue, that is $n-r$ with $r=$rank of $(A-\lambda I)$.

In this case, if $rank(A-\lambda I)=2$ for each eigenvalue we can find $2 $ corresponding eigenvectors by the solution of $(A-\lambda I)x=0$.

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  • $\begingroup$ @amd Yes of course! I used a really bad term, Thanks $\endgroup$ – gimusi May 8 '18 at 18:03
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The system $$(A-\lambda I)V =0 $$ has more than one solution for $\lambda =2$ or$\lambda =-2$ .

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  • $\begingroup$ The system has exactly one solution for $\lambda = \pm 2$ :-). $\endgroup$ – copper.hat May 8 '18 at 15:44

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