= Solution
\b[Nilpotence and eigenvalues.] If $Av=\lambda v$ with $v\ne0$, then $A^nv=\lambda^nv$. Thus $A^n=0$ forces $\lambda=0$ for every <eigenvalue>. Conversely, if all $n$ complex <eigenvalues>, counted with <algebraic multiplicity>, vanish, the monic <characteristic polynomial> is $\chi_A(t)=t^n$. The <Cayley-Hamilton theorem> then gives $A^n=0$. Hence
$$
\boxed{A^n=0\quad\Longleftrightarrow\quad\operatorname{spec}(A)=\{0\}.}
$$
This is the criterion for a <nilpotent matrix>.
\b[Symmetric case.] Here a <symmetric matrix> is understood to be real. The <spectral theorem> gives $A=QDQ^T$ with $Q$ orthogonal and $D$ real diagonal. The previous result makes every diagonal entry zero, so $A=0$. Alternatively, choose $2^r\geq n$, so $A^{2^r}=0$. For every real <vector> $x$,
$$
x^TA^{2^r}x=\left\|A^{2^{r-1}}x\right\|^2=0.
$$
It follows that $A^{2^{r-1}}=0$. Repeat until $A=0$. This uses positivity of the <quadratic form> $x^TM^TMx$. The restriction to real symmetric, or complex <Hermitian matrices>, matters: a complex matrix satisfying only $A^T=A$ need not have this positivity.
\b[The parameters and powers.] The first two columns of $T$ coincide, so $\det T=0$ for all $\alpha,\beta$. The sum of its principal two-by-two determinants, each a <minor of a matrix>, is
$$
0+(\beta-1)+(\beta-\alpha)=2\beta-\alpha-1.
$$
Together with $\operatorname{tr}T=2+\beta$, this gives
$$
\chi_T(t)=t^3-(2+\beta)t^2+(2\beta-\alpha-1)t.
$$
For every <eigenvalue> to vanish both remaining coefficients must be zero. Therefore
$$
\boxed{\beta=-2,\qquad\alpha=-5.}
$$
Direct multiplication at these values gives
$$
T=\begin{pmatrix}1&1&1\\1&1&-5\\1&1&-2\end{pmatrix},\qquad
T^2=\begin{pmatrix}3&3&-6\\-3&-3&6\\0&0&0\end{pmatrix},\qquad\boxed{T^3=0.}
$$
Indeed the first row of $T^2$ multiplied by $T$ is $3\operatorname{row}_1(T)+3\operatorname{row}_2(T)-6\operatorname{row}_3(T)=0$, and the second row is its negative.
\b[Geometry of repeated application.] For any parameters the <image of a linear map> lies in the <span> of $a=(1,1,1)^T$ and $b=(1,\alpha,\beta)^T$, since $Tx=(x_1+x_2)a+x_3b$. Thus it lies in a fixed <plane> through the origin, independently of the input. When $a,b$ are independent its normal is $a\times b=(\beta-\alpha,1-\beta,\alpha-1)$; if they coincide, the image is a line and is still contained in a plane.
For $\alpha=-5,\beta=-2$, put $s=x_1+x_2$ and $t=x_3$. Then
$$
Tx=(s+t,s-5t,s-2t)^T,\qquad x'_1+x'_2-2x'_3=0.
$$
The columns span this entire <plane>, so $T$ has <rank> two. The next two steps are
$$
\boxed{T^2x=3(x_1+x_2-2x_3)(1,-1,0)^T,\qquad T^3x=0.}
$$
Thus the <image of a linear map> contracts from a plane to a line and then to the origin under successive applications. The line from $T^2$ is precisely the <kernel> of $T$.
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