Characteristic polynomial - using rank?

Question

Q: Let $A$ be an $n\times n$ matrix defined by $A_{ij}=1$ for all $i,j$. Find the characteristic polynomial of $A$.

There is probably a way to calculate the characteristic polynomial $(\det(A-tI))$ directly but I've spent a while not getting anywhere and it seems cumbersome. Something tells me there is a more intelligent and elegant way. The rank of $A$ is only 2. Is there a way to use this?

Related: https://math.stackexchange.com/questions/2853981/find-the-eigenvalues-and-their-multiplicities-of-a-special-matrix?noredirect=1&lq=1 — user557, Jul 17 '18 at 00:58

score 1 · Answer 1 · answered Nov 01 '12 at 21:52

1

Notice that $n$ is an eigenvalue with a certain eigenvector.
Notice that the rank is 1, so the kernel has dimension $n-1$.

This gives you a basis of eigenvectors ($n-1$ of them have eigenvalue 0).

answered Nov 01 '12 at 21:52

wj32

4,236

score 1 · Answer 2 · answered Nov 01 '12 at 21:53

1

Actually the rank is $1$, not $2$. Yes, there is a much better way: find the nonzero eigenvalue. Hint: the eigenvector is $(1,\ldots,1)$.

answered Nov 01 '12 at 21:53

Robert Israel

448,999

Characteristic polynomial - using rank?

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