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Let $\mathcal{B}$ and $\mathcal{C}$ both be bases for $\mathbb{R}^n$. If $L:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is any linear operator, then prove that $[L]_\mathcal{B}$ and $[L]_\mathcal{C}$ are similar.

How should I attack this proof?

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Draw the appropriate commutative diagram. – copper.hat Mar 31 '13 at 21:49

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I think proofs of this statement can be found in most introductory textbooks on linear algebra. The standard approach is to show that $P[L]_\mathcal{C}=[L]_\mathcal{B}P$, where $P$ is the change-of-basis matrix from $C$ to $B$. That is, the $j$-th column of $P$ are the coefficients for the linear combination $\mathbf{c}_j=\sum_{i=1}^n p_{ij}\mathbf{b}_i$, where $\mathcal{B}=\{\mathbf{b}_1,\ldots,\mathbf{b}_n\}$ and $\mathcal{C}=\{\mathbf{c}_1,\ldots,\mathbf{c}_n\}$. Then argue that $P$ is invertible.

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We already proved $P$ is invertible and inverting it yields the cob matrix that goes the other way. – user54609 Apr 2 '13 at 19:25

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