Change of Basis for dual space

Asked Nov 26 '15 at 00:55

Active Nov 26 '15 at 00:58

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If the set $B={\{e_1,...e_n\}}$ is a basis of a finite dimensional vector space $V$ then the set $D=\{\epsilon_1,...,\epsilon_n\}$ where $\epsilon^i(e_j)=\delta_j^i$ is a basis for the dual space $V^*$

If $B'={\{e'_1,...e'_n\}}$ is a different basis for $V$ and then we get a different basis for the dual space $D'=\{\epsilon'_1,...,\epsilon'_n\}$

I am trying to work out how to relate the change of basis matrix from $B$ to $B'$ to the change of basis matrix from $D$ to $D'$

That is, there is to be a matrix $M$ such that $e'_i=M_{ij} e_j$ and a matrix $N$ such that $\epsilon'_i= N_{ij} \epsilon_j$ but I feel like there should be a close relationship between the two things because $\epsilon_i$ depends on $e_i$.

Can anyone help me with this?

(Also, this is my first question so I apologise if I have made errors here and there!)

edited Nov 26 '15 at 00:58

asked Nov 26 '15 at 00:55

Bernard W

2,141

Fixed. Thank you Mr. Human. – Bernard W Nov 26 '15 at 00:59
1

When you have dual bases, the relationship is rather simple. Since, relative to dual bases, $A^=A^T$, one might expect the change of basis formula for the dual space to be $(BAB^{-1})^=(BAB^{-1})^T=(B^{-1})^TA^TB^T=(B^T)^{-1}A^*B^T$. See the above link for details. – amd Nov 26 '15 at 02:33
Can you clarify what you mean by $A^*$? – Bernard W Nov 26 '15 at 03:16
That's the conjugate transpose (or adjoint) of $A.$ – sunspots May 15 '21 at 07:00

Change of Basis for dual space

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