Problems

Problem 1. \(\ \) Let

\[\begin{split}\boldsymbol{C}\ \ =\ \ \left[\begin{array}{cc} \boldsymbol{A} & \boldsymbol{0}\, \\[4pt] \cdots & \boldsymbol{B}\, \end{array}\right] \ \ =\ \ \left[\begin{array}{ccc|ccc} a_{11} & \ldots & a_{1p} & 0 & \ldots & 0 \\[4pt] \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\[4pt] a_{p1} & \ldots & a_{pp} & 0 & \ldots & 0 \\[4pt] \hline \ldots & \ldots & \ldots & b_{11} & \ldots & b_{1q} \\[4pt] \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\[4pt] \ldots & \ldots & \ldots & b_{q1} & \ldots & b_{qq} \end{array}\right]\end{split}\]

where the lower left rectangle is filled with any elements. Prove that

\[\det{\boldsymbol{C}}\ \,=\ \, \det{\boldsymbol{A}}\,\cdot\,\det{\boldsymbol{B}}\,.\]

Proof. \(\ \) Induction with respect to \(\ q.\)

I. \(\ \) We verify the statement for \(\ q = 1.\ \) Then \(\ \boldsymbol{B}\ =\ [b_{11}]_{1 \times 1}:\)

\[\begin{split}\boldsymbol{C}\ \ =\ \ \left[\begin{array}{ccc|c} a_{11} & \ldots & a_{1p} & 0 \\[4pt] \ldots & \ldots & \ldots & \ldots \\[4pt] a_{p1} & \ldots & a_{pp} & 0 \\[4pt] \hline \ldots & \ldots & \ldots & b_{11} \end{array}\right]\end{split}\]

By the Laplace expansion along the last column we get

\[\det{\boldsymbol{C}}\ \,=\ \, b_{11}\,\cdot\:(-1)^{2\,(p+1)}\ \det{\boldsymbol{A}}\ \,=\ \, \det{\boldsymbol{A}}\,\cdot\:b_{11}\ \,=\ \, \det{\boldsymbol{A}}\,\cdot\,\det{\boldsymbol{B}}\,.\]

II. \(\ \) We assume that the statement is true for some \(\ q-1,\ \\\) and have to prove that the statemant is then true for \(\ q.\)

We shall use the following denotements (\(i=1,2,\ldots,q\)):

\(M_{iq}\ -\ \) the minor of matrix \(\ \boldsymbol{B}\ \) obtained by removing the \(\ i\)-th row and the \(\ q\)-th column;

\(B_{iq}\ -\ \) the cofactor of the element \(\ b_{iq}\ \) in the matrix \(\ \boldsymbol{B};\)

\(C_{iq}\ -\ \) the cofactor of the element \(\ b_{iq}\ \) in the matrix \(\ \boldsymbol{C}.\)

In virtue of the induction hypothesis:

\[\begin{split}\begin{array}{rl} C_{iq} & =\ \ (-1)^{(p+i)+(p+q)}\ \cdot\ \det{\boldsymbol{A}}\ \cdot\ M_{iq}\ \ = \\ & =\ \ (-1)^{2p+(i+q)}\ \cdot\ \det{\boldsymbol{A}}\ \cdot\ M_{iq}\ \ = \\ & =\ \ \det{\boldsymbol{A}}\ \cdot\ (-1)^{i+q}\ \cdot\ M_{iq}\ \ =\ \ \det{\boldsymbol{A}}\ \cdot\ B_{iq}\,. \end{array}\end{split}\]

The Laplace expansion along the last column of matrix \(\ \boldsymbol{C}\ \) yields

\[\det{\boldsymbol{C}}\ \,=\ \, \displaystyle\sum_{i=1}^q\ b_{iq}\,C_{iq}\ =\ \det{\boldsymbol{A}}\ \cdot\ \displaystyle\sum_{i=1}^q\,b_{iq}\ B_{iq}\ =\ \det{\boldsymbol{A}}\ \cdot\ \det{\boldsymbol{B}}\,. \quad\bullet\]

Corollary. \(\\\) Suppose that \(\ \boldsymbol{A}_1,\,\boldsymbol{A}_2,\,\ldots,\,\boldsymbol{A}_k\ \) are square matrices, possibly of different sizes. Once again by induction one may easily prove the formula for the determinant of the block-diagonal matrix:

\[\begin{split}\left|\begin{array}{cccc} \boldsymbol{A}_1 & \boldsymbol{0} & \ldots & \boldsymbol{0} \\[4pt] \boldsymbol{0} & \boldsymbol{A}_2 & \ldots & \boldsymbol{0} \\[4pt] \cdots & \cdots & \cdots & \cdots \\[4pt] \boldsymbol{0} & \boldsymbol{0} & \ldots & \boldsymbol{A}_k \end{array}\right|\ \ =\ \ \det{\boldsymbol{A}_1}\ \cdot\ \det{\boldsymbol{A}_2} \ \cdot\ \ldots\ \cdot\ \det{\boldsymbol{A}_k}\,.\end{split}\]

In particular, if \(\ \boldsymbol{I}_n\in M_n(K)\ \) is the identity matrix, \(\ \boldsymbol{A}\in M_m(K),\ \) then

\[\begin{split}\det{(\boldsymbol{I}_n\otimes\boldsymbol{A})}\ \ =\ \ \underbrace{ \left|\begin{array}{cccc} \boldsymbol{A} & \boldsymbol{0} & \cdots & \boldsymbol{0} \\[3pt] \boldsymbol{0} & \boldsymbol{A} & \cdots & \boldsymbol{0} \\[3pt] \cdots & \cdots & \cdots & \cdots \\[3pt] \boldsymbol{0} & \boldsymbol{0} & \cdots & \boldsymbol{A} \end{array}\right|}_{n\ \text{blocks}}\ \ =\ \ \left(\det{\boldsymbol{A}}\right)^n.\end{split}\]

Problem 2. \(\ \)

Prove that if \(\ \boldsymbol{P}_\sigma\in M_n(K)\ \) is the matrix of a permutation \(\,\sigma\in S_n\,,\ \) then \(\ \det\boldsymbol{P}_\sigma\ =\ \text{sgn}\,\sigma\,.\)

Proof.

Any permutation \(\,\sigma\in S_n\ \) may be represented as a product of transpositions:

\[\sigma\ =\ \tau_k\,\dots\,\tau_2\ \tau_1\,.\]

Accordingly, the matrix \(\ \boldsymbol{P}_\sigma\,\) is a product of matrices representing transpositions:

\[\boldsymbol{P}_\sigma\ =\ \boldsymbol{P}_{\tau_k\,\dots\,\tau_2\ \tau_1}\ =\ \boldsymbol{P}_{\tau_1}\ \boldsymbol{P}_{\tau_2}\ \ldots\ \boldsymbol{P}_{\tau_k}\,.\]

Transpositions are represented by elementary matrices of the first kind \(\\\) (obtained from the identity matrix by a transposition of two rows). So

\[\boldsymbol{P}_\sigma\ =\ \boldsymbol{E}_1^{(1)}\,\boldsymbol{E}_1^{(2)}\,\ldots\ \boldsymbol{E}_1^{(k)}\,.\]

Every elementary matrix of the first kind has determinant \(-1\):

\[\det{\boldsymbol{E}_1^{(i)}}\ =\ -1\,,\qquad i=1,2,\ldots\,k.\]

Determinant of a product of matrices being equal to the product of determinants:

\[\det{\left(\boldsymbol{E}_1^{(1)}\,\boldsymbol{E}_1^{(2)}\,\ldots\, \boldsymbol{E}_1^{(k)}\right)}\ \ =\ \ \det{\boldsymbol{E}_1^{(1)}}\cdot\ \det{\boldsymbol{E}_1^{(2)}}\, \cdot\ \ldots\ \cdot\ \det{\boldsymbol{E}_1^{(k)}}\,,\]

we obtain (\(\,k\,\) is the number of factors in any representation of \(\,\sigma\,\) as a product of transpositions):

\[\det{\boldsymbol{P}_{\sigma}}\ =\ (-1)^k\ =\ \text{sgn}\,\sigma\,. \quad\bullet\]