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For any $m \times m$ matrix, I will get a characteristic polynomial of degree $m$ with $m$ eigenvalues. But for the matrix $$A = \pmatrix{2 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1}$$ I got the characteristic polynomial $$P(A)=t(t+1)(1-t)^3.$$ This means $5$ eigenvalues: $\{1,1,1,-1,0\}$.

Did I do some thing wrong?

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    Yup, you got the wrong characteristic polynomial.2012-08-02
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    If a matrix is $m\times m$ then the characteristic polynomial is of degree $m$, you must have a mistake since you should have a polynomial of degree $6$2012-08-02
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    http://www.wolframalpha.com/input/?i=[[2+%2C+1+%2C+1+%2C+1+%2C+1+%2C+1+]%2C[+1+%2C+1+%2C+0+%2C+1+%2C+0+%2C+1+]%2C[+1+%2C+0+%2C+1+%2C+0+%2C+0+%2C+1+]%2C[+1+%2C+0+%2C+0+%2C+1+%2C+0+%2C+0+]%2C[+1+%2C+0+%2C+0+%2C+0+%2C+1+%2C+0+]%2C[+1+%2C+0+%2C+0+%2C+0+%2C+0+%2C+1]]2012-08-02

2 Answers 2

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I calculated the same polynomial and I got

$$P(X)= X^2 (X-1)^3 (X-4) \,.$$

Note that $tr(A)=7$ has to be the sum of eigenvalues.


Just to get you started:

$$\det(A-tI)= \det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 1 & 0 & 0 & 1-t & 0 & 0 \\ 1 & 0 & 0 & 0 & 1-t & 0 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$

Subtract the 6th row from 4th and 5th: $$\det(A-tI)= \det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 0 & 0 & 0 & 1-t & 0 & t-1 \\ 0 & 0 & 0 & 0 & 1-t & t-1 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$

Now, $(1-t)$ common factor on rows 4 and 5.

$$\det(A-tI)= (t-1)^2\det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$

Next add Column 4 and column 5 to Column 6, and you can get a smaller $4 \times 4$ determinant....

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    $tr(A)=7$ and you must have a mistake since you have it that the sum is $8$2012-08-02
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    Ty fixed the typo.2012-08-02
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    ok, i notice but it correct to calculate the characteristic polynomial by finding the determinant of A-tI no short way??2012-08-02
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    @MissIndependent - in genrral no.2012-08-02
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    @celtschk it has been edited, other than the typo about the tr2012-08-02
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    @Belgi Fixed that too... Too many typos, I hope my students don't read this... I sometimes joke that it is good that I am teaching Linear Algebra, not taking it:)2012-08-02
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    @Belgi: I noticed after sending the comment; that's why I deleted the comment.2012-08-02
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    @Belgi what do you mean by in general no???? so how to calculate the characteristic polynomial if no??2012-08-02
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    Determinants are quite easily computable. Of course, computing by hand a $6 \times 6$ determinant is no fun, but elementary row operations allow you to compute a $n\times n$ determinant in $O(n^3)$ operations. Cf. https://en.wikipedia.org/wiki/Determinant#Decomposition_methods Finding the eigenvalues is another business...2012-08-02
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    @PseudoNeo thank you very much i needed the Jordan Block but don't we order the eigenvalue descending? meaning t=4 in the first Block???2012-08-02
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    @MissIndependent: that's really a matter of inessential convention. As far as I'm concerned, Jordan's form is well defined up to permutation of the blocks, so I don't care how these are ordered (but your course may have set a specific convention). BTW, how would you put the eigenvalues in “descending” order if some of them aren't real? [PS: unless you are analysing a chess game, there is no need for multiple question marks.]2012-08-02
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    yup , my course set a specific conversation and we can order them since any matrix A belonging to( m by m ,R) we don't deal with complex thanks for clarification2012-08-02
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    @MissIndependent, if you didn't catch it through the noise, in general it's not only correct but definitional that you'll calculate the characteristic polynomial by taking the determinant of the matrix you mention. In this case, it might be faster to use expansion by minors, if you've studied that technique, but it's a determinant no matter what.2012-08-02
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    @KevinCarlson , yup i already have studied and i use this technique to calculate determinants.2012-08-02
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In this particular case, it's not hard to find the eigenvalues and eigenvectors without computing the characteristic polynomial. Write out the equations for the entries of $A v - \lambda v = 0$: $$ \eqalign{2\,v_{{1}} &+v_{{2}}+v_{{3}}+v_{{4}}+v_{{5}}+v_{{6}}-\lambda\,v_{{1}}=0\cr v_{{1}} &+v_{{2}}+v_{{4}}+v_{{6}}-\lambda\,v_{{2}}=0\cr v_{{1}}&+v_{{3}}+v_ {{6}}-\lambda\,v_{{3}}=0\cr v_{{1}}&+v_{{4}}-\lambda\,v_{{4}}=0\cr v_{{1}}&+v_ {{5}}-\lambda\,v_{{5}}=0\cr v_{{1}}&+v_{{6}}-\lambda\,v_{{6}}=0\cr}$$ From the last three equations we see that (unless $\lambda = 1$) $v_4 = v_5 = v_6 = v_1/(\lambda-1)$. The second and third equations then become $$ \eqalign{\frac{1+\lambda}{\lambda-1} v_1 - (\lambda - 1) v_2 &= 0\cr \frac{\lambda}{\lambda-1} v_1 - (\lambda - 1) v_3 &= 0\cr}$$ so $v_2 = (1+\lambda)/(\lambda-1)^2$ and $v_3 = \lambda v_1/(\lambda-1)^2$. The first equation then becomes $$ \frac{\lambda^2 (4-\lambda)}{(\lambda-1)^2} v_1 = 0$$ so (since $v_1 = 0$ would make all $v_i = 0$, which we don't want), $\lambda = 0$ or $4$, and taking $v_1 = 1$ gives us the eigenvectors $$ \pmatrix{1\cr 5/9 \cr 4/9 \cr 1/3 \cr 1/3 \cr 1/3\cr} \ \text{for} \ \lambda = 4, \pmatrix{1 \cr 1\cr 0 \cr -1\cr -1\cr -1\cr} \ \text{for} \ \lambda=0 $$ We still need to consider the case $\lambda = -1$, in which the equations become $$ \eqalign{ v_1 &+ v_2 + v_3 + v_4 + v_5 + v_6 = 0\cr v_1 &+ v_4 + v_6 = 0\cr v_1 &+ v_6 = 0\cr v_1 &= 0\cr}$$ so $v_1 = v_4 = v_6 = 0$ and $v_5 = -v_2 - v_3$ with $v_2, v_3$ arbitrary. This gives us eigenvectors $$ \pmatrix{0 \cr 1 \cr 0 \cr 0\cr -1 \cr 0\cr}, \ \pmatrix{0 \cr 0 \cr 1 \cr 0\cr -1 \cr 0\cr} \ \text{for}\ \lambda=1 $$

Note, by the way, that there are only four linearly independent eigenvectors: eigenvalue $0$ has algebraic multiplicity $2$ and geometric multiplicity $1$, while eigenvalue $1$ has algebraic multiplicity $3$ and geometric multiplicity $2$.