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12.17 Further properties of det

Claim

\(\det (AB)=\det A\cdot \det B\)

Proposition

If \(A\) is invertible, then \(\det A^{-1}=\left(\det A\right)^{-1}\)

Proof

\(1=\det(A\cdot A^{-1})=\left(\det A)\cdot\det\left(A^{-1}\right)\Rightarrow\left(\det A\right.\right)^{-1}=\det A^{-1}\)

True or False

  1. \(\det A=1\) iff \(A=I\)​

    False, counterexample: \(\det\begin{pmatrix}-1 & 0\\ 0 & -1\end{pmatrix}=1\)​

  2. \(\det A^{k}=0\Rightarrow\det A=0\)​

    True, since \(\det A^{k}=0\Rightarrow\left(\det A\right)^{k}=0\Rightarrow\det A=0\)

  3. \(A\in M_{n\times n}\left(\mathbb{C}\right)\) with \(A^3=I\Rightarrow\left|\det A\right|=1\)​

    True, since \(A^3=I\), then \(\det A^3=I\Rightarrow\left(\det A\right)^3=1\). In particular, \(|\det A|^3=1\). Then \(|\det A|=1\)​

Corollary

Similar matrices have equal determinant.

Proof

By definition \(B=P^{-1}AP\), then \(\det B=\det(P^{-1}AP)=\det(P^{-1})\det(A)\det(P)=\det A\)

(We can commute this two because we are in the field, but we cannot do that inside \(\det\))

True or False

  1. \(\begin{pmatrix}1 & 1\\ 1 & -1\end{pmatrix}\) is similar to \(\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\) or \(\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}\)

    False

    \(\det B-2\) \(\det =-1\) \(\det =1\)

  2. \(\begin{pmatrix}1 & 1\\ 1 & -1\end{pmatrix}\) is similar to \(\begin{pmatrix}2 & 0\\ 0 & -1\end{pmatrix}\)

    False

    Although the determinant is equal but the trace is not equal, recall this, then they are not similar

  3. If \(A\) and \(B\) have the same \(\det\) and the same \(\text{tr}\), then they are similar

    False

    Easy to know the similarity class of \(\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\) is itself. (\(I=P^{-1}\cdot I\cdot P=P^{-1}\cdot P=I\)​ )

    Then\(\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\cancel{\sim}\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}\)

  4. \(I\) is the only matrix whose similarity class has one element

    False

    \(I,-I,0,\lambda I\)​

    Hint: \(P^{-1}AP=A\;\iff\;AP=PA,\forall\text{ invertible P}\)​

The det of an operator

Given \(T\in \text{End}(V)\), let \(\det T=\det[T]_{\mathcal{B}}\), for some basis \(\mathcal{B}\) of \(V\)

We need to check well definition (check if we choose different basis, we always get the same result)

\(\det[T]_{\mathcal{B}}=\det[T]_{\mathcal{B}^{\prime}}\) since \([T]_{\mathcal{B}}\) is similar to \([T]_{\mathcal{B}^{\prime}}\) since \([T]_{\mathcal{B}^{\prime}}=P^{-1}[T]_{\mathcal{B}}P\)

\(\det A=\det A^{T}\)

Laplace row expansion

Given a row \(i\), it holds: \(\det A=\sum_{j=1}^{n}\left(-1\right)^{i+j}a_{ij}\det A\left(i\left|j\right.\right)\), where \(A(i|j)\) is the \((n-1)\times(m-1)\) matrix obtained from \(A\) by deleting it's i-th row and it's j-th column.

Example

  1. \(\det\begin{pmatrix}1 & 0 & -1\\ 2 & 2 & 1\\ -1 & 1 & 2\end{pmatrix}=?\)

    \(\text{row}~i=1:\left(-1\right)^2\times1\times\begin{vmatrix}2 & 1\\ 1 & 2\end{vmatrix}+\left(-1\right)^2\times0\times\begin{vmatrix}we\left(2\right) & do\left(1\right)\\ not\left(-1\right) & care\left(2\right)\end{vmatrix}+\left(-1\right)^4\times\left(-1\right)\times\begin{vmatrix}2 & 2\\ -1 & 1\end{vmatrix}=3-4=-1\)

    image​

    Choose the row which has the most \(0\) that can simplify the calculation

    If we choose \(\text{row }i=2:\left(-1\right)\times2\times\det\begin{vmatrix}0 & -1\\ 1 & 2\end{vmatrix}+\left(-1\right)^4\times2\times\begin{vmatrix}1 & -1\\ -1 & 2\end{vmatrix}+\left(-1\right)^5\times1\times\begin{vmatrix}1 & 0\\ -1 & 1\end{vmatrix}=-2+2-1=-1\)

    If we choose \(\text{row }i=3:\left(-1\right)^4\times\left(-1\right)\times\det\begin{vmatrix}0 & -1\\ 2 & 1\end{vmatrix}+\left(-1\right)^5\times1\times\begin{vmatrix}1 & -1\\ 2 & 1\end{vmatrix}+\left(-1\right)^6\times2\times\begin{vmatrix}1 & 0\\ 2 & 2\end{vmatrix}=-2-3+4=-1\)

  2. \(\begin{vmatrix}i & i & 0 & 0\\ 1 & 0 & -1 & 0\\ 0 & 0 & i & i\\ 1 & 0 & 1 & 0\end{vmatrix}=i\times\begin{vmatrix}0 & -1 & 0\\ 0 & i & i\\ 0 & 1 & 0\end{vmatrix}+\left(-i\right)\times\begin{vmatrix}1 & -1 & 0\\ 0 & i & i\\ 1 & 1 & 0\end{vmatrix}=0+\left(-i\right)\times\left(-i-i\right)=-2\)​

    Choose \(\text{row }1\)

    \(\begin{vmatrix}i & i & 0 & 0\\ 1 & 0 & -1 & 0\\ 0 & 0 & i & i\\ 1 & 0 & 1 & 0\end{vmatrix}=\left(-i\right)\times\begin{vmatrix}1 & -1 & 0\\ 0 & i & i\\ 1 & 1 & 0\end{vmatrix}=\left(-i\right)\times\left(-i-i\right)=-2\)

    Choose \(\text{column }2\)

    Hint \(\begin{pmatrix}a & b & c\\ d & e & f\\ g & h & i\end{pmatrix}=\left(aei-ceg\right)-afh-bdi+bfg+cdh\) image​

    We need to add two mirror triangular

Remark

Since \(\det A=\det A^t\), we mat expand \(\det A\) by a column.

Also, if there is a zero column, \(\det\) also equal to zero since same reason

Cofactors and the adjoint

Let \(c_{ij}=\left(-1\right)^{i+j}\det A\left(i\left|j\right.\right)\) be called the \((i,j)-\)cofactor of \(A\). The matrix \(C\) with \(C_{ij}=c_{ij}\) is the cofactors matrix associated to \(A\)

We know that \(\det A=\sum_{j=1}^{n}a_{ij}c_{ij}\)(i-th row expansion) and also \(\det A=\sum_{i=1}^{n}a_{ij}c_{ij}\)(j-th column expansion)

Now, let \(B\) be obtained form \(A\) by replacing it's j-th column by it's k-th column (\(k\neq j\))

[\(B\) has two identical columns: j-th and k-th].

Then \(\det B=0=\sum_{i=1}^{n}b_{ij}\left(-1\right)^{i+j}\det B\left(i\left|j\right.\right)=\sum_{i=1}^{n}a_{ik}\left(-1\right)^{i+j}\det A\left(i\left|j\right.\right),k\neq j\)

Now, \(\sum_{i=1}^{n}a_{ik}\left(-1\right)^{i+j}\det A\left(i\left|j\right.\right)=\begin{cases}{}0,\text{ if }k\neq j\\ \det A,\text{ if }k=j\end{cases}=\delta_{kj}\det A\) where \(c_{ij}= \left(-1\right)^{i+j}\det A\left(i\left|j\right.\right)\)​

Let \(\text{adj}A=C^t\), \((\text{adj}A)_{ji}=c_{ji}\). Hence \(\sum_{i=1}^{n}\left(\text{adj}_{ji}\right)a_{ik}=\delta_{kj}\det A\) and on the other hand, \(\sum_{i=1}^{n}\left(\text{adj}_{ji}\right)a_{ik}=\left(\left(\text{adj}A\right)\cdot A\right)_{jk}\)

Therefore, \(\left(\left.\text{adj}A\right)\cdot A\right.=\begin{pmatrix}\det A & \cdots & 0\\ \vdots & \ddots & \vdots\\ 0 & \cdots & \det A\end{pmatrix}=\det A\cdot I\)

Conclusion: If \(\det A\neq0\left(\;\iff\;A\text{ is invertible }\right)\), then \(A^{-1}=\frac{1}{\det A}\cdot\text{adj}A\)

Example

  1. \(A = \begin{pmatrix}1 & 0 & 1 \\0 & 1 & 0 \\-1 & 1 & 1\end{pmatrix}\)​\(\text{adj} A = \begin{pmatrix}+1 & -(-1) & +(-1) \\-0 & +2 & -0 \\+1 & -1 & +1\end{pmatrix}=\begin{pmatrix}1 & 1 & -1 \\0 & 2 & 0 \\1 & -1 & 1\end{pmatrix}\)

    \(\begin{pmatrix}1 & 0 & 1\\ 0 & 1 & 0\\ -1 & 1 & 1\end{pmatrix}\begin{pmatrix}1 & 1 & -1\\ 0 & 2 & 0\\ 1 & -1 & 1\end{pmatrix}=\begin{pmatrix}2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{pmatrix}=\det A\cdot I\Rightarrow A^{-1} =\begin{pmatrix}\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\0 & 1 & 0 \\\frac{1}{2} & -\frac{1}{2} & \frac{1}{2}\end{pmatrix}\)

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