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${\bf Problem }$ If $A$ is a square matrix, we define the trace of A, $tr(A)$, to be the sum of the diagonal entries of $A$. Let V be a finite dimensional vector space over $\Bbb C$ , with $\dim(V ) = n \geq 1$, and let $T \in L(V )$. We define the trace of T by $tr(T) = tr([T]_{\alpha}^{\alpha})$ where $\alpha $ is any basis for V .

${\bf Part (a)}$

Prove that $tr(T)$ is well-defined. That is, prove that if $\alpha $ and $\beta$ are bases for V , then $tr([T]_\alpha^\alpha) = tr([T]_\beta^\beta) $

Note that a basis transformation of a transformation $T$ is $ATA^{-1}$ where $A$ is an invertible matrix, we also know that the trace has the property $\operatorname{tr}(AB)=\operatorname{tr}(BA)$. This implies that $$\operatorname{tr}((AT)A^{-1}) = \operatorname{tr}(A^{-1}(AT))= \operatorname{tr}(A^{-1}AT) = \operatorname{tr}T$$ the result follows.

${\bf Part (b)}$

Prove that T is nilpotent iff $tr(T) = tr ( T^2) = \cdots = tr(T^n)=0 $

We start by showing that If $T\in L(V) $ is nilpotent then we prove that the only eigenvalue of T is $0$. Then there is a positive integer $n$ s.t $T^n =0$

This implies that T is not injective hence $0$ is an eigenvalue of T.

Conversely, suppose $\lambda $ is an eigenvalue of T.

Then there exists a non-zero vector $v\in V $ such that $\lambda v = Tv $ we apply T $n$ times and we get that $\lambda^n v = T^nv $ but $ T^n v=0 $ it follows that $ \lambda^n v =0 $ but v is non-zero so $ \lambda =0$

Secondly we have that the characteristic polynomial of an $n\times n$ matrix $A$ is equivalent to $$p(t) = \det(A-tI) = (-1)^n \big(t^n - (\text{tr} A) \,t^{n-1} + \dots + (-1)^n \det A\big)\,.$$ On the other hand, $p(t) = (-1)^n(t-\lambda_1)\dots (t-\lambda_n)$, where the $\lambda_j$ are the eigenvalues of $A$. So, comparing coefficients, we have $\text{tr}A = \lambda_1 + \dots + \lambda_n$.

Now $( \Rightarrow ) $ assume that T is nilpotent we know by above that the only eigenvalues of T are zero and also by above that the $\operatorname{tr}(T) =0 $ now we wish to show that if all the eigenvalues are $0$ of T that all the eigenvalues of $T^k $ are zero for any $k \in \Bbb N$ well we know by part (a) that the trace is independent of the basis and we also know we can always find a basis so that the matrix T is upper triangular with all of the diagonal entries also equal to zero by above. Now consider $\operatorname{tr} T^k$ where k is any integer we know that $\operatorname{tr} (T^k)=\operatorname{tr} (ATA^{-1})^k =\operatorname{tr} (U^k) $ where A is a change of basis matrix to make T upper triangular which we denote as U. we notice that $\operatorname{tr} (U^k)=\operatorname{tr} (U (U^{k-1}) $ Note that $U^{k-1} $ is the product of $k-1$ upper triangular matrices so it is upper triangular let us denote it $R $ let $C=UR $ but all the diagonal entries of $U $ are zero we have that the $C_{ij} = \sum_{k =1}^n a_{ik}b_{ki} = \sum_{k =1}^{i-1}a_{ik}b_{kj} + \sum_{k=i}^na_{ki}b_{kj} = 0 + 0 = 0, $

Since $k \leq i \implies a_{ik} =0 $ and $k \geq i \implies b_{kj} =0 $ for this particular combination of $(i,j)$. We conclude that $C_{ij} = 0$ if $i \geq j$ hence $C$ is upper triangular and all of its diagonal entries are zero. it follows that $\operatorname {tr}(C) =0 = \operatorname{tr} (U^k) = \operatorname{tr} (T^k)$ for any value of k so the result follows.

Now for the hard part:

$( \Leftarrow ) $ Assume that $tr(T) = tr ( T^2) = \cdots = tr(T^n)=0 $ want to show that T is nilpotent.

Let $ m_i$ denote the multiplicity of the eigenvalue and let $\lambda_1, \cdots , \lambda_r $ denote all the distinct nonzero eigenvalues. using an above result and the assumption we have that $$ m_1 \lambda_1 + \cdots + m_r \lambda_r =0 $$ $$ m_1 (\lambda_1)^2 + \cdots + m_r (\lambda_r)^2 =0 $$ $$ \vdots $$

$$ m_1 (\lambda_1)^n + \cdots + m_r (\lambda_r)^n =0 $$

We have that $$\begin{pmatrix} \lambda_1 & \cdots & \lambda_r \\ (\lambda_1)^2 & \cdots & (\lambda_r)^2 \\ \vdots & \ddots & \vdots \\ (\lambda_1)^{n} & \cdots & (\lambda_r)^{n} \end{pmatrix} \begin{pmatrix} m_1 \\ \vdots \\ m_r \end{pmatrix} = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix}$$

If we can conclude that each $ m_i=0 $ then the result will follow and it is sort of clear That this sucker is non-invertable (i think but not sure on the justification) so we know the determinant is zero. We are also given that a Vandermonde-matrix is a matrix of this form:

$$\begin{pmatrix} 1 & \cdots & 1 \\ a_1 & \cdots & a_r \\ \vdots & \ddots & \vdots \\ a_1^{r-1} & \cdots & a_r^{r-1} \end{pmatrix} $$ where each $a_i \in \Bbb C$ and $\det(V)= \Pi_{1\leq i j \leq r} (a_j - a_i)$ we are given this as a true fact to use w.o proof and it seem to be the key to finishing off the problem. ( i did try and prove the required result with induction my only suggestion on that avenue is don't.)

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I propose another method:

  • If $A$ is nilpotent, then $A$ is triangulable over $\mathbb C$ with $0$ as unique eigenvalue: $A = P^{-1}TP$. Hence $A^k = P^{-1}T^kP$ with $T^k$, as $T$, a triangular matrix with $0$ on diagonal. Hence $Tr(A^k) = 0$.

  • If $Tr(A) = ... = Tr(A^n) = 0$ then for all polynomial $P$ with degree $≤ n$ and such that $P(0) = 0$, $Tr(P(A)) = 0$ (by linearity). $A$ is triangulable over $\mathbb C$: $T = Q^{-1}AQ$.

    Take $P$ with degree $≤ n$ such that $P(0) = 0$ and for all eigenvalue $\lambda \neq 0$ of $A$, $P(\lambda) = 1$ (Lagrange polynomial). Then $Tr(P(T)) = Tr(P(A)) = 0$ implies all eigenvalue is $0$ ; so $A$ is nilpotent.

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  • $\begingroup$ Sorry i really dont understand your second argument you claim: then for all polynomial $P$ with degree $≤ n$ and such that $P(0) = 0$, $Tr(P(A)) = 0$ (by linearity). i am not sure why this is true. secondly Take $P$ with degree $≤ n$ such that $P(0) = 0$ and for all eigenvalue $\lambda \neq 0$ of $A$, $P(\lambda) = 1$. Then $Tr(P(T)) = Tr(P(A)) = 0$ implies all eigenvalue is $0$ ; so $A$ is nilpotent. I am also not sure why $Tr(P(T)) = Tr(P(A)) = 0$ ? i think i believe that it implies all eigenvalues are $0$ though We have never learned about any properties of trace in class $\endgroup$ – Faust Feb 28 '18 at 15:25
  • $\begingroup$ The first point is because $Tr$ is linear i.e. $Tr(aA) = aTr(A)$ and $Tr(A+B) = Tr(A) + Tr(B)$. $\endgroup$ – user371663 Feb 28 '18 at 15:31
  • $\begingroup$ Sorry i understand that its linear i just see no correlation between it being linear and $P(A)=0 \implies Tr (P(A))=0 $ infact i am not sure it even makes sense to talk about $Tr(P(A)) $ is $P(A) $ a matrix? i thought it was a polynomial equation. $\endgroup$ – Faust Feb 28 '18 at 15:35
  • $\begingroup$ Ok. $P(A) = 0$ ? I suppose you meant $P(0) = 0$. It is just equivalent to $P$ has no constant coefficient because $Tr(P(A))=0$ would be wrong if this constant coefficient is not $0$. So $P = a_1X + a_2X^2 + ... + a_nX^n$ so $P(A) = a_1A + a_2A^2 + ... + a_nA^n$ and $Tr(P(A)) = a_1Tr(A) + a_2Tr(A^2) + ... + a_nTr(A^n) = 0 + 0 + ... + 0$. $\endgroup$ – user371663 Feb 28 '18 at 15:43
  • $\begingroup$ Thats very clever! i understand what your trying to do now thank you. i'll try and write out a complete solution in this way and will accept once i feel i understand it. $\endgroup$ – Faust Feb 28 '18 at 15:47
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$( \Leftarrow ) $ Assume that $tr(T) = tr ( T^2) = \cdots = tr(T^n)=0 $ and that T is not nilpotent.

Let $ m_i$ denote the multiplicity of the eigenvalue and let $\lambda_1, \cdots , \lambda_r $ denote all the distinct nonzero eigenvalues. using an above result and the assumption we have that $$ m_1 \lambda_1 + \cdots + m_r \lambda_r =0 $$ $$ m_1 (\lambda_1)^2 + \cdots + m_r (\lambda_r)^2 =0 $$ $$ \vdots $$

$$ m_1 (\lambda_1)^n + \cdots + m_r (\lambda_r)^r =0 $$

We have that $$\begin{pmatrix} \lambda_1 & \cdots & \lambda_r \\ (\lambda_1)^2 & \cdots & (\lambda_r)^2 \\ \vdots & \ddots & \vdots \\ (\lambda_1)^{r} & \cdots & (\lambda_r)^{r} \end{pmatrix} \begin{pmatrix} m_1 \\ \vdots \\ m_r \end{pmatrix} = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix}$$

It follows that $$ \begin{pmatrix} 1 & \cdots & 1 \\ (\lambda_1) & \cdots & (\lambda_r) \\ \vdots & \ddots & \vdots \\ (\lambda_1)^{r-1} & \cdots & (\lambda_r)^{r-1} \end{pmatrix} C_1 C_2 \cdots C_r \begin{pmatrix} m_1 \\ \vdots \\ m_r \end{pmatrix} = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix}$$

it follows from $$\begin{pmatrix} 1 & \cdots & 1 \\ a_1 & \cdots & a_r \\ \vdots & \ddots & \vdots \\ a_1^{r-1} & \cdots & a_r^{r-1} \end{pmatrix} $$ where each $a_i \in \Bbb C$ and $\det(V)= \Pi_{1\leq i j \leq r} (a_j - a_i)$ we are given this as a true fact to use w.o proof and it seem to be the key to finishing off the problem.

so we have that

$$\frac{1}{\lambda_1 \lambda_2 \cdots \lambda_r} \det \begin{pmatrix} 1 & \cdots & 1 \\ (\lambda_1) & \cdots & (\lambda_r) \\ \vdots & \ddots & \vdots \\ (\lambda_1)^{r-1} & \cdots & (\lambda_r)^{r-1} \end{pmatrix} = \frac{1}{\lambda_1 \lambda_2 \cdots \lambda_r} \Pi_{1\leq i<j \leq r} (\lambda_j - \lambda_i)=0$$ But we have that $(\lambda_j - \lambda_i) $ where $ i \neq j $ so this number is always non zero as each eigenvalue is distinct so the determinate is non-zero but then we have that $\frac{1}{\lambda_1 \lambda_2 \cdots \lambda_r} \neq 0 $ as well so the only soultion to $$ \begin{pmatrix} 1 & \cdots & 1 \\ (\lambda_1) & \cdots & (\lambda_r) \\ \vdots & \ddots & \vdots \\ (\lambda_1)^{r-1} & \cdots & (\lambda_r)^{r-1} \end{pmatrix} C_1 C_2 \cdots C_r \begin{pmatrix} m_1 \\ \vdots \\ m_r \end{pmatrix} = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix}$$

it follows that $m_1=m_2=\cdots=m_r=0 $ but the eigenvalues were picked in a such a way that this implies that all the eigenvalues of T are $0$ but then T is nilpotent!

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  • $\begingroup$ Actually, I did not mention it but we only have to consider the associated Lagrange polynomial as $P$ (in my answer). There is at most $n+1$ different value (the case in which all eigenvalues are distinct and non-zero) to fix so we know the degree of $P$ is $≤ n$. $\endgroup$ – user371663 Mar 1 '18 at 16:53
  • $\begingroup$ i am not saying your answer is not correct i actually prefer it but it turns out in the proof of what you stated and used you end up with a Vandermonde-matrix that you have to solve in a similar way as i have done so above. so i thought it would be worth adding the direct solution. $\endgroup$ – Faust Mar 1 '18 at 18:10
  • $\begingroup$ Yes I understood; actually, there is no Vandermonde in my answer, you just need to use Lagrange polynomial, what I have not precised. That's why it is shorter, but in return it is a bit tricky. The usual method is to use Vandermonde as you have done. $\endgroup$ – user371663 Mar 1 '18 at 18:15

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