# Linear Algebra, eigenvalues and eigenvectors

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. – N. S. Aug 2 '12 at 19:44
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 $6$ – Belgi Aug 2 '12 at 19:45
wolframalpha.com/input/… – PseudoNeo Aug 2 '12 at 19:54

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 $8$ – Belgi Aug 2 '12 at 19:51
Ty fixed the typo. – N. S. Aug 2 '12 at 19:53
ok, i notice but it correct to calculate the characteristic polynomial by finding the determinant of A-tI no short way?? – Miss Independent Aug 2 '12 at 19:55
@MissIndependent - in genrral no. – Belgi Aug 2 '12 at 19:56
@celtschk it has been edited, other than the typo about the tr – Belgi Aug 2 '12 at 19:56

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 $2$.

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