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I read a theorem

If $A$ is an invertible square matrix, then $\operatorname{adj} A^T= (\operatorname{adj} A)^T$.

But after attempting to prove it myself and also reading the proof I am unable to tell why the function needs to be invertible for above to be true. As in the proof we no where use this condition.

Proof:

$$\begin{align} A \operatorname{adj}A & =|A|I_n \\ \implies (A \operatorname{adj} A)^T & = \{|A|I_n\}^T\\ \implies (\operatorname{adj} A)^T(A^T) & =|A|I_n \tag{1}\\ \text{Also,}\\ (\operatorname{adj} A^T)(A^T) & = |A^T|I_n \\ (\operatorname{adj} A^T)(A^T) & = |A|I_n \tag{2}\\ \text{from (1) and (2), we get} \\ (\operatorname{adj}A^T)(A^T)& =(\operatorname{adj} A)^T(A^T)\\ \implies \operatorname{adj} A^T&=(\operatorname{adj} A)^T \end{align}$$

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Pedro
    Mar 14, 2023 at 12:15

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You have already used $A$ invertible in the last step, where you removed $A^T$ on the right on both sides.

But you are right that this theorem does not need A to be invertible. By definition $$(\mathbb{adj}A)^T = \begin{bmatrix}A_{11} & A_{21} & \cdots & A_{n1}\\A_{12}&A_{22}&\cdots&A_{n2}\\\vdots&\vdots&\ddots&\vdots\\A_{1n}&A_{2n}&\cdots&A_{nn}\end{bmatrix}^T=\begin{bmatrix}A_{11} & A_{12} & \cdots & A_{1n}\\A_{21}&A_{22}&\cdots&A_{2n}\\\vdots&\vdots&\ddots&\vdots\\A_{n1}&A_{n2}&\cdots&A_{nn}\end{bmatrix}=\begin{bmatrix}{A^T}_{11} & {A^T}_{21} & \cdots & {A^T}_{n1}\\{A^T}_{12}&{A^T}_{22}&\cdots&{A^T}_{n2}\\\vdots&\vdots&\ddots&\vdots\\{A^T}_{1n}&{A^T}_{2n}&\cdots&{A^T}_{nn}\end{bmatrix}=\mathbb{adj}(A^T)$$

Here $A_{ij} = (-1)^{i+j}\mathbb{det}B_{ij},$ where $B_{ij}$ is the $(n-1)\times(n-1)$ matrix you get when you remove the i-th row and the j-th column of $A$. And ${A^T}_{ji} = (-1)^{i+j}\mathbb{det}C_{ji},$ where $C_{ji}$ is what you get when you remove the j-th row and the i-th column of $A^T$. You could directly check that $B_{ij}=(C_{ji})^T$, which implies thay have the same determinant, and hence $A_{ij}={A^T}_{ji}$.

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  • $\begingroup$ how $A_{11}$ is equal to $A_{11}^T$ $\endgroup$ Mar 14, 2023 at 7:43
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    $\begingroup$ @garudasupermemo Eliminating the first row and the first column of $A$ and $A^T$ will get matrices transposing to each other, and they have the same determinant. $\endgroup$
    – lxklp
    Mar 14, 2023 at 7:47
  • $\begingroup$ @lxklp I am unable to understand your last part of the solution can you elaborate further, or if you are willing to we can discuss on apps like discord. $\endgroup$ Mar 14, 2023 at 8:00
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    $\begingroup$ Vseh and Garuda, the entries of the adjugate matrix are defined e.g. here, and lxklp's explanation for $(A^T)_{ij}= A_{ji}$ in the case $i=j=1$ naturally adapts to the general case. Garuda, beware the flip on the indices in this equality. $\endgroup$ Mar 14, 2023 at 8:12
  • $\begingroup$ @garudasupermemo: Note that $A_{12}$ equals $A_{21}^T$, not $A_{12}^T$. $\endgroup$ Mar 14, 2023 at 8:13

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