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Hello guys please help me whit this problem. Hints is fine but if you can explain that would be great. For $1.$ I think is just the definition but I can't figure out how to combine the two conditions to pass from $i>j$ and $j>k$ to $i>k$. and for $2.$ I find out that the dimension is $n^2 - \frac{n(n-1)}{2}$.

A square matrix $A=(a_{ij})$$i,j \in \{1,2, ...., n\}$ is said to be upper triangular if $a_{ij}=0$ for $i>j$.

$1$. Show that the product of two upper triangular matrices is an upper triangular matrix.

$2$. Let $\xi=\{A$ $\in M_n(\mathbb{R})$: $A$ is upper triangular$\}$. Establish that $\xi$ is a subspace of $M_n(\mathbb{R})$. What is its dimension?

$3$. Let $E$ be a vector space over $\mathbb{R}$,$B=\{e_1,e_2, ...., e_n\}$ a basis of $E$ and $\phi \in L(E)$. Let $E_i=Span\{e_1,e_2, ...., e_i\}$ for $i=1,2, ...., n$. Show that $M(\phi,B)$ is upper triangular matrix iff $\phi(E_i)\subset E_i$ for every $i=1,2, ...., n$.

$4$.Let $A \in \xi$. We suppose that $A$ is reversible. Show that $A^{-1} \in \xi$

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What have you tried so far? Do you have any ideas? –  alexwlchan Jul 2 '13 at 15:03
    
@alexwlchan I told you what I have tried above conserning questions $1.$ and $2.$ and for the others Iam trying. –  Mohamez Jul 2 '13 at 15:07

1 Answer 1

up vote 1 down vote accepted

Part 1): Show that the product of two upper triangular matrices is upper triangular.

Let $A, B \in M_n(\mathbb{R})$ be upper triangular matrices (with respect to the standard basis). Then we can express them as

$$\left( \begin{array}{ccc} a_{11} & a_{12} & \cdots &a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & 0 & \cdots & \vdots \\ 0 & \cdots & 0 & a_{nn}\end{array} \right) \text{ and }\left( \begin{array}{ccc} b_{11} & b_{12} & \cdots &b_{1n} \\ 0 & b_{22} & \cdots & b_{2n} \\ \vdots & 0 & \cdots & \vdots \\ 0 & \cdots & 0 & b_{nn}\end{array} \right)$$

Hence their product is clearly

$$\left( \begin{array}{ccc} a_{11} & a_{12} & \cdots &a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & 0 & \cdots & \vdots \\ 0 & \cdots & 0 & a_{nn}\end{array} \right)\left( \begin{array}{ccc} b_{11} & b_{12} & \cdots &b_{1n} \\ 0 & b_{22} & \cdots & b_{2n} \\ \vdots & 0 & \cdots & \vdots \\ 0 & \cdots & 0 & b_{nn}\end{array} \right) = \left( \begin{array}{ccc} a_{11}b_{11} & a_{11}b_{12}+a_{12}b_{22} & \cdots &\sum_{i=1}^n a_{1i}b_{in} \\ 0 & a_{22}b_{22} & \cdots & \sum_{i=2}^n a_{2i}b_{in} \\ \vdots & 0 & \cdots & \vdots \\ 0 & \cdots & 0 & a_{nn}b_{nn}\end{array} \right)$$

I think I have given you enough information to describe the general pattern of entry $c_{jk}$ of the product matrix.

Part 2): To do this, show that triangular matrices are closed under linear combinations. In particular, show that (1) the zero matrix is technically upper triangular, (2) that any triangular matrix times a scalar is upper triangular, and (3) that the sum of any two upper triangular matrices is itself upper triangular. These should be very straightforward proofs. See here for a specific case of the dimension argument.

Part 3): If $M(\phi, B)$ is the matrix representation of $\phi$ and it is upper triangular, then this means that $\phi(e_k)$ is the first column and has all zero entries except for the first $k$. Stop and think about what this means in terms of the subspace of $E$ that contains $\phi(e_k)$. I am being intentionally vague because this is an important concept that you should reach on your own, but I can be more explicit and supply a reference need be.

Part 4): By reversible do you mean invertible? If so, just think about rearranging the basis vectors after inverting the map.

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