Question

nick1337 · Accepted Answer

To represent the transpose homomorphism from the vector space of 2x2 matrices over the field of real numbers, denoted as

M_{2 \times 2} (ℝ)

, to itself, we need to consider the standard bases for both the domain and the codomain.
The standard basis for

M_{2 \times 2} (ℝ)

is given by:

B_{V} = {[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]}

.
Let's denote the transpose homomorphism as

A

. For a matrix

M

in

M_{2 \times 2} (ℝ)

, the transpose of

M

is denoted as

M^{T}

. The transpose homomorphism

A

can be defined as

A (M) = M^{T}

.
Now, let's consider a matrix

M

in

M_{2 \times 2} (ℝ)

represented in the standard basis

B_{V}

:

M = a [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] + b [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] + c [\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}] + d [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]

,
where

a, b, c, d

are real numbers.
The representation of

M

with respect to the standard basis

B_{W}

can be obtained by taking the transpose of

M

and representing it as a linear combination of the basis vectors in

B_{W}

.
Since

B_{W}

is the same as

B_{V}

in this case, the representation of

M

with respect to

B_{W}

is the same as the original representation of

M

.
The representation of

M

with respect to

B_{W}

is given as:

[M]_{B_{W}} = [\begin{matrix} a & b \\ c & d \end{matrix}]

.
Therefore, the transpose homomorphism

A

maps a matrix

M

in

M_{2 \times 2} (ℝ)

to its transpose

M^{T}

without changing the representation of

M

with respect to the standard basis

B_{V}

.
The transpose homomorphism can be represented as:

A : M_{2 \times 2} (ℝ) \to M_{2 \times 2} (ℝ)

,

A (M) = M^{T}

.
And the representation of

M

with respect to

B_{W}

can be represented as:

[M]_{B_{W}} = [\begin{matrix} a & b \\ c & d \end{matrix}]

.

Answered question