# The upper triangular matrix problem.

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})$$i,j \in \{1,2, ...., n\} is said to be upper triangular if a_{ij}=0 for i>j. 1. Show that the product of two upper triangular matrices is an upper triangular matrix. 2. Let \xi=\{A \in M_n(\mathbb{R}): A is upper triangular\}. Establish that \xi is a subspace of M_n(\mathbb{R}). What is its dimension? 3. Let E be a vector space over \mathbb{R},B=\{e_1,e_2, ...., e_n\} a basis of E and \phi \in L(E). Let E_i=Span\{e_1,e_2, ...., e_i\} for i=1,2, ...., n. Show that M(\phi,B) is upper triangular matrix iff \phi(E_i)\subset E_i for every i=1,2, ...., n. 4.Let A \in \xi. We suppose that A is reversible. Show that A^{-1} \in \xi - What have you tried so far? Do you have any ideas? – alexwlchan Jul 2 '13 at 15:03 @alexwlchan I told you what I have tried above conserning questions 1. and 2. and for the others Iam trying. – Mohamez Jul 2 '13 at 15:07 add comment ## 1 Answer Part 1): Show that the product of two upper triangular matrices is upper triangular. Let A, B \in M_n(\mathbb{R}) be upper triangular matrices (with respect to the standard basis). Then we can express them as$$\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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