Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter

    Given a matrix A with complex entries, how do I find the invertible matrix P such that P^-1 * A * P is upper triangular? I have seen the proof of the fact that every matrix is similar to an upper triangular matrix, but the proof was by induction and so I don't think that it can be used to construct the matrix.
    • PS Helper

    PS Helper
    The idea is that if you can find an eigenvector then the image of that eigenvector is going to be a multiple of itself, so choose that to be your first basis vector; then you have a column that looks like \begin{matrix} \lambda \\ 0 \\ \vdots \\ 0 \end{matrix}. Then you need to proceed to find linearly independent vectors v_k such that for each k, when you apply your matrix to it what you get is some linear combination of v_1, \dots, v_k (i.e. and not v_{k+1} , \dots, v_n). Then the matrix must be upper-triangular with respect to the resulting basis.
Have you ever experienced racism/sexism at uni?
Useful resources

Make your revision easier


Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here


How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.