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Find $S = \{X^{T}AX:X \in\mathbb{R}^n\}$ where $A \in \mathcal{M}_{n,n}(\mathbb{R})$ is nilpotent.

What I have done so far:

If $A$ is nilpotent then its only eigenvalue is $0$ and $A^n = 0$.

If $A = 0$ we have $S = \{0\}$.

If there exists $X \in \mathbb{R}^n$ such that $X^T AX>0$ then $\mathbb{R}_+\subset S$, likewise if $X^T AX<0$ then $\mathbb{R}_- \subset S$.


I have no idea how to use the fact that $A$ is nilpotent. Any help would highly be appreciated.

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    $\begingroup$ Hint: Consider the eigenvalues of $\frac 12 (A+A^T)$, which must be symmetric and have trace zero $\endgroup$ Jun 6 at 20:25
  • $\begingroup$ @BenGrossmann Thank you a lot for your always relevant comments and answers. $\endgroup$
    – Axel
    Jun 6 at 20:38
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Thanks to @BenGrossmann's hint I have shown that:

If we consider $B = \dfrac{1}{2}(A+A^{T})$, then $B$ is a real symmetric matrix hence triangularizable and the trace is the sum of the eigenvalues (counted with their multiplicity), here $\mathrm{Tr}(B) = 0$ by using linearity of the trace, the fact a matrix and its transpose have same trace and the nilpotence of $A$ which gives us $\mathrm{Tr}(A)=0$.

It is great to have such matrix because we can have much information about it than $A$, moreover we have,

$$\forall X \in \mathbb{R}^n, \, X^{T}BX = \dfrac{1}{2}(X^{T}AX+X^{T}A^{T}X )=\dfrac{1}{2}(X^{T}AX+X^{T}AX)=X^{T}AX$$

Therefore,

$$S = \{X^{T}AX : X \in \mathbb{R}^n \} = \{X^{T}BX : X \in \mathbb{R}^n \}$$


So if there exists $\lambda \neq 0$ an eigenvalue of $B$ then we can find another eigenvalue $\mu$ such that $\mu \lambda<0$ otherwise $\mathrm{Tr}(B)$ would be nonzero. Without loss of generality, I will take $\lambda>0$ and $\mu<0$. Moreover there exists, $X_{\mu}, X_{\lambda} \in \mathbb{R}^n\backslash\{0\}$ such that, $BX_{\mu} = \mu X_{\mu}$ and $BX_{\lambda} = \lambda X_{\lambda}$.

Hence, $X_{\mu}^TBX_{\mu} = \mu \Vert X_{\mu} \Vert^2<0 $ and $X_{\lambda}^TBX_{\lambda} = \lambda \Vert X_{\lambda} \Vert^2>0 $.

Therefore,

$$\{X^{T}BX : X \in \mathrm{Span}(X_{\mu})\} = \mathbb{R}_- \subset S$$

Likewise,

$$\{X^{T}BX : X \in \mathrm{Span}(X_{\lambda})\} = \mathbb{R}_+ \subset S$$

We deduce that $S = \mathbb{R}$.


If $0$ is the only eigenvalue of $B$ then as $B$ is diagonalizable ($B$ is real and symmetric) then $B$ is similar to the null matrix, therefore $B=0$. Hence $A = -A^T$, thus $A^2$ is symmetric and nilpotent so $A^2 = 0$. Hence $-A^2 = A A^{T}=0$ thus $\mathrm{Tr}(AA^T)=0$ and it implies that $A=0$ (using the definiteness of the inner product on $\mathcal{M}_{n,n}(\mathbb{R})$ defined by $\varphi(A,B) = \mathrm{Tr}(A^TB)$).


To conclude, for $A$ nilpotent, if $A = 0$ then $S = \{0\}$ else $S = \mathbb{R}$.

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    $\begingroup$ @Axel A key point that you forgot to mention (but that I suspect you noticed) is that $X^TBX = X^TAX$ for all $X \in \Bbb R^n$ $\endgroup$ Jun 7 at 1:04
  • $\begingroup$ @BenGrossmann I edited. Thank you again! $\endgroup$
    – Axel
    Jun 7 at 11:31

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