Question is from Intro to Linear Algebra (5th Ed) by Gilbert Strang, Chapter 6-39.

Without writing down any calculations, can you find the eigenvalues of this matrix? Also find $A^{2017}$.

$$ A = \begin{Bmatrix} 110 & 55 & -164\\ 42 & 21 & -62\\ 88 & 44 & -131 \end{Bmatrix} $$

Obviously one of the eigenvalues is $0$. Not sure how to find the rest without calculation.


I'm going to actually give a couple of hints that should help you solve this.

First, you already have seen that 0 is an eigenvalue, probably by noticing that your columns are not linearly independent. Try to see if you can find the rank of the matrix by observation. That will let you find the multiplicity of 0 as an eigenvalue.

Your matrix has constant row sums, this row sum will be an eigenvalue (with the all-ones vector as an eigenvalue).

Next, recall that the trace of a matrix is the sum of the eigenvalues (including multiplicity). That should help you narrow things down as well.

Once you have the eigenvalues, you have that $A = P^{-1} D P$, where $D$ is a diagonal matrix consisting of the eigenvalues on the diagonal. So $A^{2017} = P^{-1} D^{2017} P$, you will only be able to do this in your head if $D$ is reaaaaly really nice looking (it will be in this case).

  • $\begingroup$ The rank is 1, so does this mean that the multiplicity of 0 is 2? $\endgroup$ – helios321 Feb 4 '18 at 5:15
  • $\begingroup$ @helios321 are you sure that the rank is 1? $\endgroup$ – user328442 Feb 4 '18 at 5:22
  • $\begingroup$ I don't think the rank is 1. Also, I think you can get another eigenvalue by inspection. You can obtain another by noticing that all row sums are constant, so $(1,1,1)^{T}$ is an eigenvector (with eigenvalue = to the row sum). $\endgroup$ – Morgan Rodgers Feb 4 '18 at 5:27
  • $\begingroup$ @MorganRodgers I discovered using calculator that $A^n$ repeats itself for every $n$ odd. So the answer is A. Can you figure it out? $\endgroup$ – helios321 Feb 4 '18 at 5:56
  • 4
    $\begingroup$ The trace is zero, so its eigenvalues are $0$, $1$ and $-1$. As each of these raised to the power $2017$ remains the same, so does $A$. $\endgroup$ – Lord Shark the Unknown Feb 4 '18 at 7:12

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