1

I would like to calculate the eigenvalues of the following matrix $A$, but the factorization of the characteristic polynomial does not seem to be easy to compute.

$$A=\pmatrix{ a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \\ },\ a\neq1,\ a\neq-2$$

$f(\lambda) = \operatorname{Char}(A,\lambda) = (a-\lambda)^3-3(a-\lambda)+2 = -\lambda^3 + 3a\lambda^2 + 3\lambda(1-3a^2) + (a-1)^2(a+2)$

I have thought about using the Rational-Root Theorem (R.R.T.), so possible roots of $f(\lambda)$ are $(a-1)$, $(-a+1)$, $(a+2)$, $(-a-2)$, and much more, as for example in the case $a=2$ we should also test whether $f(\pm2)=0$ or not, am I wrong?

The eigenvalues of $A$ are $a-1$ and $a+2$ (computed with Wolfram Alpha). This result can be obtained using R.R.T., computing $f(a-1)$ and $f(a+2)$ and realizing that both are equal to zero but, is there an easier and ‘elegant’ way to find these eigenvalues?

Rócherz
  • 3,976
Gibbs
  • 544
  • Can you do it for $a=1$? – Angina Seng Apr 18 '19 at 16:09
  • Well, but that's a particular case. Can I assume from that case that eigenvalues are $a-1$ and $a+2$ ($0$ and $-2$ if $a=1$)? – Gibbs Apr 18 '19 at 16:11
  • Can you go from one particular case to the general case? – Angina Seng Apr 18 '19 at 16:21
  • 1
    Related: https://math.stackexchange.com/questions/904926/determinant-of-a-rank-1-update-of-a-scalar-matrix-or-characteristic-polynomia, https://math.stackexchange.com/questions/689111/find-the-eigenvalues-of-a-matrix-with-ones-in-the-diagonal-and-all-the-other-el. – StubbornAtom Apr 18 '19 at 16:45
  • Sometimes it’s easier to find eigenvectors first. Try simple linear combinations of the columns, such as summing all of them or taking two at a time. You can also take advantage of the symmetry of $A$ to narrow the search. – amd Apr 18 '19 at 17:19
  • Effectively a duplicate of https://math.stackexchange.com/q/86644/265466. – amd Apr 18 '19 at 17:19

4 Answers4

3

One way to see $a+2$ is an eigenvalue is that $$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}a+2\\a+2\\a+2\end{bmatrix}.$$ Then you can use the fact that $x-(a+2)$ divides the characteristic polynomial.

More generally: if all the rows of $A$ add up to $\lambda$, then $\lambda$ is an eigenvalue.

Dave
  • 13,568
3

Basically, you need to solve $(a-\lambda)^3-3(a-\lambda)+2 =0$ for $\lambda$. Don't expand the brackets, instead denote: $t=a-\lambda$. Then: $$t^3-3t+2=0 \Rightarrow (t-1)^2(t+2)=0 \Rightarrow \\ t_1=1 \Rightarrow a-\lambda =1 \Rightarrow \lambda_1 =a-1\\ t_2=-2\Rightarrow a-\lambda =-2 \Rightarrow \lambda_2=a+2.$$

farruhota
  • 31,482
1

Hint: if $I$ denotes the identity matrix, then the eigenvalues of $A+cI$ are easily obtained from the eigenvalues of $A$: $$ (A+cI)v=\lambda v \iff Av=(\lambda-c)v $$ What if you take $c=1-a$?

egreg
  • 238,574
  • If $A=I$, $\lambda_1-c=0$ and $\lambda_2-c=3$, then $(A+(a-1))v=\lambda v$, so $\lambda_1 - (a-1) = 0 \to \lambda_1=a-1$, and $\lambda_2 - (a-1) = 3 \to \lambda_2 = a+2$. Good, thanks for your answer!! – Gibbs Apr 18 '19 at 16:42
0

I think so. The matrix in question is called the rank-one perturbation of the identity matrix. That is $A = (a-1)I +uu^\top$ where $u$ is called the unit vector with all entries one. It is know that $uu^\top$ has eigenvalues $uu^\top$ and zero with multiplicity dimension of $u$ minus 1 and the associated eigenvectors $u$ and $u^\perp$. The eigenvalues of $(a-1)I + uu^\top$ are $a-1+u^\top u = a+2$ and two times $a-1$, the determinant is then the product of these eigenvalues. That is $(a-1)^2(a+2)$.

Rócherz
  • 3,976
abel
  • 29,170