Question

Find the determinants of the linear transformations in Exercises 17 through 28.

23.$\mathit{L}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}{\mathit{A}}^{\mathbf{T}}\mathbf{}\mathit{f}\mathit{r}\mathit{o}\mathit{m}\mathbf{}{\mathbf{ℝ}}^{\mathbf{n}\mathbf{\times }\mathbf{n}}\mathbf{}\mathit{t}\mathit{o}\mathbf{}{\mathbf{ℝ}}^{\mathbf{n}\mathbf{\times }\mathbf{n}}$

Short answer

Therefore, the determinant of the linear transformations is given by,

$\det T=\det B={(-1)}^{\frac{n(n-1)}{2}}$

Step-by-step solution

Definition. 

A determinant is a unique number associatedwith a square matrix.

A determinant is a scalar value that is a function of the entries of a square matrix.

It is the signed factor by which areas are scaled by this matrix. If the sign is negative the matrix reverses orientation.

Given. 

Given linear transformation,

$L\left(A\right)=A^{\mathrm{T}}from{\mathrm{ℝ}}^{\mathrm{n}\times \mathrm{n}}to{\mathrm{ℝ}}^{\mathrm{n}\times \mathrm{n}}$

To find determinant.

Let . $A={\left(a_{i,j}\right)}_{i,j}\in {\mathrm{ℝ}}^{n\times n}$

We have $T{\left(A_j\right)}_{i,j}={\left(A^T\right)}_{ij}=a_{ij}$. If is the canonical basis for ${\mathrm{ℝ}}^{n\times n}$, then the matrix $B={\left(b_{ij}\right)}_{ij}\in {\mathbb{ℝ}}^{n\times n}$of T corresponding to has 1 on places

localid="1659521205647" $$\begin{gathered} (1,1),(2,n+1),(3,2n+1),...,(n-1,n^2-2n+1),(n,n^2-n+1), \\ (n+1,2),(n+2,n+2),(n+3,2n+2),....,(2n-1,n^2-2n),(2n,n^2-n+2), \\ (2n+1,3)(2n+2,n+3),(2n+3,2n+3),...,(3n-1,n^2-2n+3),(3n,n^2-n+3) \end{gathered}$$ .

.

.

$$\begin{gathered} (n^2-2n+1,n-1),(n^2-2n+2,2n-1),(n^2-2n+3,3n-1),..., \\ (n^2-n-1,n^2-n-1)({(n-1)}^2,n^2-1)(n^2-n+1,n)(n^2-n+2,2n),(n^2-n+3+3n),.., \\ (n^2-1,{(n-1)}^2),(n^2,n^2) \end{gathered}$$

and 0 everywhere else. To obtain the identity matrix $l_{n\times n}$, we clearly need

$(n-1)+(n-2)+(n+3)+...+1=\frac{n(n-1)}{2}$row interchanges.

Therefore,

$\det T=\det B={(-1)}^{\frac{n(n-1)}{2}}$