ThmDex – An index of mathematical definitions, results, and conjectures.
Product of a real square matrix and its adjugate is a constant diagonal matrix
Formulation 0
Let $A \in \mathbb{R}^{N \times N}$ be a D6160: Real square matrix.
Then \begin{equation} A \text{Adj} (A) = \text{Adj}(A) A = \text{Det}(A) I_N \end{equation}
Subresults
R5061: Expression for a real matrix inverse in terms of adjugate
Proofs
Proof 0
Let $A \in \mathbb{R}^{N \times N}$ be a D6160: Real square matrix.
Denote by $C_{n, m}$ the $(n, m)$th D5941: Real square matrix cofactor of $A$. Since we have \begin{equation} \begin{split} A \text{Adj}(A) & = \begin{bmatrix} A_{1, 1} & \cdots & A_{1, N} \\ \vdots & \ddots & \vdots \\ A_{N, 1} & \cdots & A_{N, N} \end{bmatrix} \begin{bmatrix} C_{1, 1} & \cdots & C_{N, 1} \\ \vdots & \ddots & \vdots \\ C_{1, N} & \cdots & C_{N, N} \end{bmatrix} \\ & = \begin{bmatrix} \sum_{n = 1}^N A_{1, n} C_{1, n} & \cdots & \sum_{n = 1}^N A_{1, n} C_{N, n} \\ \vdots & \ddots & \vdots \\ \sum_{n = 1}^N A_{N, n} C_{1, n} & \cdots & \sum_{n = 1}^N A_{N, n} C_{N, n} \end{bmatrix} \end{split} \end{equation} and \begin{equation} \begin{split} \text{Adj}(A) A & = \begin{bmatrix} C_{1, 1} & \cdots & C_{N, 1} \\ \vdots & \ddots & \vdots \\ C_{1, N} & \cdots & C_{N, N} \end{bmatrix} \begin{bmatrix} A_{1, 1} & \cdots & A_{1, N} \\ \vdots & \ddots & \vdots \\ A_{N, 1} & \cdots & A_{N, N} \end{bmatrix} \\ & = \begin{bmatrix} \sum_{n = 1}^N C_{n, 1} A_{n, 1} & \cdots & \sum_{n = 1}^N C_{n, 1} A_{n, N} \\ \vdots & \ddots & \vdots \\ \sum_{n = 1}^N C_{n, N} A_{n, 1} & \cdots & \sum_{n = 1}^N C_{n, N} A_{n, N} \end{bmatrix} \end{split} \end{equation} then this result is a consequence of R5064: Cofactor partition for a real square matrix. $\square$