0
$\begingroup$

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 $A^{-1} \in \xi$

$\endgroup$
  • $\begingroup$ @alexwlchan I told you what I have tried above conserning questions $1.$ and $2.$ and for the others Iam trying. $\endgroup$ – Mohamez Jul 2 '13 at 15:07
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.