# Invertible Matrix Question

## Homework Statement

Find all n x n-matrices A such that Q-1 A Q = A for all invertible n x n matrices Q.

None

## The Attempt at a Solution

Proof by induction
Assumption: True for A = k(In) where k is a real number

Base case: k = 1
When k =1
A = 1 * In = In
Q-1 * In * Q = In
(Q-1 * In) * Q = In
(Q-1) * Q =In
In = In.

When k = m
A = m * In
Q-1 * (m *In) * Q = m * In
m * (Q-1 * In) * Q = m * In
m * (Q-1) * Q = m * In
m * In = m * In.

m * In + In = In * (m + 1) =

When k = m+1
A = m+1 * In
Q-1 * ((m + 1) *In) * Q = (m + 1) * In
(m + 1) * (Q-1 * In) * Q = (m + 1) * In
(m + 1) * (Q-1) * Q = (m + 1) * In
(m + 1) * In = (m + 1) * In.

Since they equal this is true by induction.

This is one of 6 bonus problems the linear algebra class were given. It's the first three people to turn in each of the problems gets the two points. However you only get one submission. So I thought I'd have someone else check my work to see if it is correct or not before I turn it in.

## The Attempt at a Solution

Related Calculus and Beyond Homework Help News on Phys.org
lanedance
Homework Helper
i'm not too sure why you are using induction?

you should be able to show A=kIn is a solution to the equation for any k in the reals using just matrix multiplication. However you may need to do some work to show it is teh only solution

HallsofIvy
You say that the problem was "find all n x n-matrices A such that $Q^{-1} A Q = A$ for all invertible n x n matrices Q" but then you ask about proving, by induction, that this is true for A= kIn. Do you understand that if you do prove that, you will NOT have done what you were asked to? In orfder to do that you would have to also prove that any other kind of matrix, this is NOT true.
Note that $$\displaystyle Q^{-1}AQ= A$$ is the same (multiply on both sides, on the right, by Q) AQ= QA. That is, A must commute with all invertible matrices.