4

Prove that $A$, a matrix of rank $3$, can't have characteristic polynomial of $p(x) = x^7 - x^5 + x^3$

My attempt to contradict:

Because of that characteristic polynomial, the matrix must be a $7 \times 7$ matrix. Also, $-\mathrm{tr}(A) = 0$, because $x^6 = 0$. $A$ is a matrix with rank of $3$ so the determinant is $0$ and therefore it isn't invertible.

I am unable to progress from this point onwards.

Fakemistake
  • 2,718
Din
  • 43

1 Answers1

5

Hint: The geometric multiplicity of an eigenvalue is always at most the algebraic multiplicity.
What is the eigenspace corresponding to the eigenvalue $0$?

leoli1
  • 6,979
  • 1
  • 12
  • 36
  • the algebraic multiplicity of 0 is 3, but how do I calculate eigenspace without knowing A? – Din Jan 09 '21 at 10:53
  • You don't need to calculate the eigenspace, you just need to relate the eigenspace for $0$ with some other subspace associated to the matrix which you can in turn relate to the rank (Hint: Rank-nullity-theorem) – leoli1 Jan 09 '21 at 10:56
  • 1
    @user3123 You don't need to find the eigenspace, just the geometric multiplicity. $v\ne0$ is an eigenvector with eigenvalue $0$ iff $v\in\ker A$. Thus geometric multiplicity of $0$ is equal to $\dim\ker A=7-3=4$. – Shubham Johri Jan 09 '21 at 10:56
  • @ShubhamJohri It should be $\dim\ker A\leq 3$ since the geometric multiplicity might be strictly smaller than the algebraic multiplicity which is $3$. – leoli1 Jan 09 '21 at 10:59
  • When did I contradict that? – Shubham Johri Jan 09 '21 at 11:03
  • @ShubhamJohri You wrote $\dim \ker A=7-3=4$ – leoli1 Jan 09 '21 at 11:03
  • 1
    But $\dim\ker A$ is indeed $4$. It follows from the rank-nullity theorem. $\dim\ker A$ needed to be $\le3$ that is why it causes a contradiction. – Shubham Johri Jan 09 '21 at 11:04
  • @ShubhamJohri Oh, I misunderstood you. I was refering to what we can deduce from the characteristic polynomial. You deduced $\dim\ker A$ from the rank that was given. Apologies – leoli1 Jan 09 '21 at 11:06
  • Nevermind. +1... – Shubham Johri Jan 09 '21 at 11:07
  • Why does $dim ker A = 7 - 3$? What does 3 stand for? – Din Jan 09 '21 at 11:14
  • @user3123 the $3$ comes from the assumption that the rank of $A$ is $3$ and the formula for $\dim\ker A$ comes from the rank-nullity theorem. – leoli1 Jan 09 '21 at 11:23