Prove that $(U^{t})^{-1}=(U^{-1})^{t}$.

Question

Note that $f_{i}$ and $f^{'}_{i}$ are dual bases of $\mathfrak{B}=\{\alpha_{1},\ldots,\alpha_{n} \}$ and $\mathfrak{B^{'}}=\{\alpha^{'}_{1},\ldots,\alpha^{'}_{n} \}$:

Let U be the invertible linear operator such that $U\alpha_{j}=\alpha^{'}_{j}$. Then the transpose of $U$ is given by $U^{t}f^{'}_{i}=f_{i}$. It is easy to verify that since $U$ is invertible, so is $U^{t}$ and $(U^{t})^{-1}=(U^{-1})^{t}$.

The above statement suggests that

$U^{t}$ is invertible,
$(U^{t})^{-1}=(U^{-1})^{t}.$

It is also trivial to me that the former one is true for sure. However, in order to prove the latter (without using matrix notation), I have no idea where to start from. Can you give me some hints?

I'm probably being naive; but, $(x^z)^y = x^{zy}$ - therefore - $(x^z)^y = (x^y)^z $ — Tobi, Jun 12 '16 at 05:23
@Tobi Here $\cdot^t$ denotes the transpose, it's not an exponent. — M. Vinay, Jun 12 '16 at 05:44

score 2 · Accepted Answer · answered Jun 12 '16 at 05:44

$$ U^{-1} U = I \implies U^{-1} \alpha_j' = \alpha_j \implies (U^{-1})^t f_i = f_i' $$

$$ (U^t)^{-1} U^t = I \implies (U^t)^{-1} f_i = f_i' $$

Equating expressions for $f_i'$ gives $(U^{-1})^t = (U^t)^{-1}$ on the basis $f_i$, so those operators are equal by linearity.

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

2

Transpose of composition of functions property tells us that $(U^{-1})^tU^t=(UU^{-1})^t=(I)^t=I$. That is, $(U^{-1})^tU^t=I$ and therefore $(U^t)^{-1}=(U^{-1})^t$.

edited Apr 13 '17 at 12:21

Community

1

answered Jun 12 '16 at 06:31

rych

4,205

Prove that $(U^{t})^{-1}=(U^{-1})^{t}$.

2 Answers2