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Let $A\ne 0$ be a $3\times 3$ matrix with real entries such that $A^3+A=0$. We need to show $\mathrm{rank}(A)=2$.

$\det A(A^2+I)=0\Rightarrow\det A=0\Rightarrow \mathrm{rank}(A)<3$, Suppose $\mathrm{rank}(A)=1$, Then I showed one matrix with rank $1$ which do not satisfies the given relation, is my answer is ok? Thank you for help and discussion

Myshkin
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    Strange coincidence (deleted question from 10 hours ago asking the exact same thing)? At any rate, the statement is false: if $A$ is the zero matrix, then $A^3+A=0$, but $\mathrm{rank}(A)=0$. – Zev Chonoles May 13 '13 at 03:37
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    $A$ was not $0$ so far I remember – Myshkin May 13 '13 at 03:47
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    If it is assumed that $A \ne 0$, edit the problem please. – Eric Jablow May 13 '13 at 03:49
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    Presumably, it's meant to be $\text{rank}(A)\leq 2$. – Glen O May 13 '13 at 03:57
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    Your implication $\det A(A^2+I)=0 \implies \det A=0$ is incorrect (you have it backwards). For example, the matrix \begin{pmatrix}0&1\-1&0\end{pmatrix} satisfies $\det A(A^2+I)=0$ but $\det A=1$. You need to use the fact that $A$ is $3\times 3$ to rule out the possibility that $A$ is invertible. – Greg Martin May 13 '13 at 04:00
  • @GregMartin: presumably, you meant $$\pmatrix{0&1\-1&0}$$as with the matrix you mentioned, the square of the matrix is the identity. – Glen O May 13 '13 at 04:01
  • @GlenO thanks, fixed. – Greg Martin May 13 '13 at 04:02
  • what is true is that \begin{align} \det(A^2+I)&= \det((A+iI)(A-iI))\ &=\det(A+iI)\det(A-iI)\ &=\det(A+iI)\det(\overline{A+iI})\ &=\det(A+iI)\overline{\det({A+iI})} \geq 0 \end{align} – clark May 14 '13 at 21:04

2 Answers2

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The "standard" way to solve this kind of problems is to examine the minimal polynomial of $A$. Yet, for this particular problem, there are other approaches as well. Here is one of them:

  1. By considering the determinants of both sides of $A^3 = -A$, argue that $\operatorname{rank}(A)\ne3$.
  2. If $\operatorname{rank}(A)=1$, then $A=uv^T$ for some nonzero vectors $u,v\in\mathbb{R}^3$. Show that $A^3+A$ is a nonzero scalar multiple of $A$ and hence it cannot be zero.
  3. As $\operatorname{rank}(A)\ne3,1$ and $A\ne0$ by assumption, the only possibility is that $\operatorname{rank}(A)=2$.
user1551
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By the Cayley Hamilton theorem, the minimal polynomial of a square matrix divides its characteristic polynomial. Therefore, since $A \neq 0$ by assumption, $(x^2 + 1) | f_A(x)$ which means that $A$ has at least the eigenvalues $\pm i$. Since these are distinct eigenvalues, the corresponding eigenvectors are linearly independent. Since they lie in the column space of $A$, the rank of $A$ is at least $2$. As you mentioned, a determinant argument can be used to show that it is less than $3$.

Rookatu
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    This is not quite correct. First, you say the minimal polynomial divides the characteristic polynomial; what is relevant here is that the minimal polynomial divides any polynomial that vanishes at $A$ - in this case, $x^3+x$. Second, the minimal polynomial might be a proper divisor - in this case, $x$ might be the minimal polynomial. This is impossible if $A$ is assumed to not be the zero matrix, but it shows that that hypothesis is necessary. – Greg Martin May 13 '13 at 03:57
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    To make this srgument work, you have to think of $A$ as a linear map on $\mathbb C^3$, which may be beyond the scope of OP's course. More directly, by Cayley-Hamilton, if $A\ne 0$, the characteristic polynomial must be $p(t)=t^3+t$, which has $0$ as a single root. Therefore, $\dim E(0)=\dim \ker A =1$, so $\text{rank }A= 2$. – Ted Shifrin May 13 '13 at 03:59
  • @TedShifrin could you please tell me how did you calculate $\dim ker A=1$? – Myshkin May 13 '13 at 07:53
  • Since the characteristic polynomial must be $p(t)=t^3+t=t(t^2+1)$, $0$ is a root with multiplicity $1$, and so the geometric multiplicity of the eigenvalue $0$ must be exactly $1$. But $E(0) = \ker A$. – Ted Shifrin May 19 '13 at 18:32