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Let $\{ v_1, \ldots, v_n \}$ be a basis for the $n$-dimensional vector space $V$. A family of linear maps $T : V \to V$ is given by $$ Tv_i = \sum_{j=1}^i a_{ij} v_j $$ with $a_{ii} = 1$ for $i = 1, \ldots, n$.

I want to show that $V_i = \langle v_1, \ldots, v_i \rangle$ is the only sequence of invariant subspaces with $1$-dimensional factor spaces for the set of all such linear maps (they correspond to the triangular matrices with ones on the diagonal).

We have to show that if $W_1, W_2, \ldots, W_n$ is another such sequence, then $V_i = W_i$ for $i = 1,\ldots, n$.

My Attempt: Suppose $W_1 = \langle w_1 \rangle$ with $TW_1 = W_1$, then $Tw_1 = \alpha w_1$, and by adjusting $\hat{w_1} := \alpha^{-1} w_1$ we can assume that $\alpha = 1$. We have $w_1 = \sum_{i=1}^n \beta_i v_i$, hence substituting into $w_1 = Tw_1$ we find \begin{align*} \sum_{i=1}^n \beta_i v_i & = \sum_{i=1}^n \beta_i \sum_{j=1}^i a_{ij} v_i \\ & = \beta_1(a_{11} v_1) + \beta_2(a_{21} v_1 + a_{22} v_2) + \ldots + \beta_n(a_{n1}v_1 + \ldots + a_{nn} v_n) \\ & = (\beta_1 a_{11} + \beta_2 a_{21} + \ldots + \beta_n a_{n1}) v_1 + (\beta_2 a_{22} + \beta_3 a_{23} + \ldots + \beta_n a_{n2}) v_2 + \ldots + \beta_n a_{nn} v_n. \end{align*} By comparing coefficients $$ \begin{array}{lllllll} \beta_1 & = & \beta_1 a_{11} & + & \beta_2 a_{21} + \ldots + & \beta_n a_{n1} \\ \beta_2 & = & & & \beta_2 a_{22} + \ldots + & \beta_n a_{n2} \\ & \vdots & & & \ddots & \vdots\\ \beta_n & = & & & & \beta_n a_{nn} \end{array} $$ But here as $a_{nn} = 1$, what I find is that $\beta_n$ could be arbitrary, but what I want is that $\beta_1$ could be arbitrary and $\beta_2 = \ldots = \beta_n = 0$, i.e. $W_1 = V_1$.

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If $w$ is an eigenvector for $T$ with eigenvalue $\lambda$ then $\lambda^{-1}w$ is also an eigenvector with the same eigenvalue, and not with eigenvalue 1.

Let $m$ be the largest index such that $\beta_m \neq 0$. Using your proof, you will get that $\beta_m = \lambda \beta_m a_{mm}=\lambda \beta_m$ since $a_{mm}=1$ and from here you get that $\lambda = 1$.

Once you have this, the equation for $m-1$ will give you that $\beta_{m-1}=\beta_{m-1}+a_{m,m-1}\beta_m$. As this should be true for all triangular matrices, you can assume that $a_{m,m-1}\neq 0$ and get a contradiction because then $a_{m,m-1}\beta_m \neq 0$. The only way you don't get a contradiction is if $m=1$ which means that $w=\beta_1 v_1$ which is exactly what you wanted.

The proof that I would use is that if $W$ is an invariant subspace under all the triangular matrices and $\sum_1^m \beta_i v_i\in W$ with $\beta_m\neq 0$ (and therefore in $V_m$), then you can show that every $v=\beta_m v_m+\sum_1^{m-1}\alpha_i v_i\in V_m$ can be written as $v=Tw$ for some triangular matrix $T$ with ones on the diagonal, and the span of these elements is all of $V_m$ (or in other words, $V_m$ is irreducible). Since $W$ is invariant, you get that $V_m\subseteq W$. It follows that $W=V_m$ for the largest $m$ such that $W$ contains an element of the form $\sum_1^m \beta_i v_i\in W$ with $\beta_m\neq 0$.

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Prove this by induction on $\dim V$. Let $W_1 = \left< w_1 \right>$ and consider the linear map defined by

$$ Tv_1 = v_1, Tv_2 = v_1 + v_2, Tv_3 = v_2 + v_3, \ldots, Tv_n = v_{n-1} + v_n. $$

Since $W_1$ is $T$-invariant, $w_1$ must be an eigenvector of $T$ but the only eigenvectors of $T$ are of the form $\alpha v_1$, hence $W_1 = V_1$. Then consider $V / W_1$.

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