Proving the Change of Coordinate Matrix for Left-Multiplication Transformation

[iomtt6076]

Homework Statement

Prove: Let $$A \in \mathrm{M}_{n \times n}(\mathbb{F})$$ and let $$\gamma$$ be an ordered basis for $$\mathbb{F}^n$$. Then $$[\boldmath{L}_A]_{\gamma} = Q^{-1}AQ$$, where Q is the nxn matrix whose jth column is the jth vector of $$\gamma$$.

Homework Equations

$$\boldmath{L}_A$$ denotes the left-multiplication transformation.

The Attempt at a Solution

Let $$\beta$$ be the standard ordered basis for $$\mathbb{F}^n$$ and C the change of coordinate matrix from $$\beta$$-coordinates to $$\gamma$$-coordinates. Then $$[\boldmath{L}_A]_{\beta} = A$$ and we have $$[\boldmath{L}_A]_{\gamma} = C^{-1}AC$$. Where I'm stuck is showing that the jth column of C is the jth vector of $$\gamma$$. Any hints would be appreciated.

 

[Dick]

The matrix C whose jth column is the jth vector of gamma maps the standard basis (1,0...0), (0,1,...0)...(0,0...1) into gamma, doesn't it?

 

[iomtt6076]

The matrix C whose jth column is the jth vector of gamma maps the standard basis (1,0...0), (0,1,...0)...(0,0...1) into gamma, doesn't it?

Dick, thanks for the response. I think I got it now. C should be the change of coordinate matrix from gamma-coordinates to beta-coordinates, not beta to gamma as I had stated above at first. In other words, for $$[\boldmath{L}_A]_{\gamma} = C^{-1}AC$$ to hold true, $$C = [\boldmath{I}]_{\gamma}^{\beta}$$ where I is the identity transformation. Then everything makes sense.