Find $\{X^{T}AX:X \in\mathbb{R}^n\}$ where $A$ is nilpotent.

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.

• Hint: Consider the eigenvalues of $\frac 12 (A+A^T)$, which must be symmetric and have trace zero Jun 6 at 20:25
– Axel
Jun 6 at 20:38

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 $$S = \mathbb{R}$$.

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