Change of basis: why is Q an identity matrix?

Question

Let $\beta$ and $\beta'$ be two ordered bases for a finite-dimensional vector space V, and let Q= $[I_V]^{\beta}_{\beta'}$. Then

a). Q is invertible

b). for any $v \in V$, $[v]_\beta=Q[v]_{\beta'}$

My question is why is Q an identity matrix?

Answer 1

The key is to understand what exactly is going on with that bracket notation. For a transformation $T:V \to W$ and bases $\beta,\beta'$ of $V$ and $W$, $[T]^\beta_{\beta'}$ is defined to be the matrix for which $$ [T]^\beta_{\beta'}[v]_{\beta} = [T(v)]_{\beta'}. $$ Here, $[v]_{\beta}$ denotes the coordinate vector of $v$ relative to the basis $\beta$. On the other hand, the change of basis matrix $Q$ is defined so that we just get the input with respect to the new basis. In other words, we want $Q$ to satisfy $$ Q[v]_\beta = [v]_{\beta'}. $$ If we want to replace the $Q$ in this equation with $[T]^\beta_{\beta'}$ for some linear transformation $T$ on $V$, then the only transformation that makes sense is the identity map $I_V$ defined by $I_V(v) = v$. So, $Q = [I_V]^\beta_{\beta'}$.