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12.16 The determinant

Determinant

It is a function \(\det:M_{n\times n}\left(\mathbb{F}\right)\rightarrow\mathbb{F}\). It is not linear.

It serves to determine whether a matrix is invertible or not.

Namely \(\det(A)\neq 0\) iff \(A\) is invertible

Example

\(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\), \(\det A=ad-bc\)

Recall that we know that \(A\) is invertible iff \(ad-bc\neq 0\)

The defining properties of det

  1. \(\det\) is a multilinear function of the rows.

    Namely, \(\det\begin{pmatrix}--a_1--\\ -a_{i}+\lambda a_{i}^{\prime}-\\ --a_{n}--\end{pmatrix}=\det\begin{pmatrix}-a_1-\\ -a_{i}-\\ -a_{n}-\end{pmatrix}+\lambda\det\begin{pmatrix}-a_1-\\ -a_{i}^{\prime}-\\ -a_{n}-\end{pmatrix}\)​

  2. \(\det\) is an alternating function of the rows

    Namely, \(\det\begin{pmatrix}-a_1-\\ -a_{i}-\\ -a_{j}-\\ -a_{n}-\end{pmatrix}=-\det\begin{pmatrix}-a_1-\\ -a_{j}-\\ -a_{i}-\\ -a_{n}-\end{pmatrix}\)

  3. \(\det (I)=1\)

Consequence

  • If \(A\) has a zero row, the \(\det A=0\)​

Since \(\det\begin{pmatrix}-a_1-\\ -0-\\ -a_{n}-\end{pmatrix}=\det\begin{pmatrix}-a_1-\\ -0+0-\\ -a_{n}-\end{pmatrix}=\det\begin{pmatrix}-a_1-\\ -0-\\ -a_{n}-\end{pmatrix}+\det\begin{pmatrix}-a_1-\\ -0-\\ -a_{n}-\end{pmatrix}\)

Then \(\det\begin{pmatrix}-a_1-\\ -0-\\ -a_{n}-\end{pmatrix}=0\)

The matrix which has zero rows or columns can never be invertible since it has null space, then the rank is not full * If \(A\) has two identical rows, then \(\det A=0\)​

Since \(\det\begin{pmatrix}-a_1-\\ -a_{i}-\\ -a_{i}-\\ -a_{n}-\end{pmatrix}=-\det\begin{pmatrix}-a_1-\\ -a_{i}-\\ -a_{i}-\\ -a_{n}-\end{pmatrix}=0\), then \(\det A=0\)​ * If \(A=\begin{pmatrix}\lambda_1 & \cdots & 0\\ \vdots & \ddots & \vdots\\ 0 & \cdots & \lambda_{n}\end{pmatrix}\), then \(\det A=\lambda_1\cdot ...\cdot \lambda_n\)​

Since multilinear, then \(\det A=\begin{pmatrix}\lambda_1 & \cdots & 0\\ \vdots & \ddots & \vdots\\ 0 & \cdots & \lambda_{n}\end{pmatrix}=\lambda_1\det A=\begin{pmatrix}1 & \cdots & 0\\ \vdots & \ddots & \vdots\\ 0 & \cdots & \lambda_{n}\end{pmatrix}=\lambda_1\cdot\ldots\cdot\lambda_{n}\det A=\begin{pmatrix}1 & \cdots & 0\\ \vdots & \ddots & \vdots\\ 0 & \cdots & 1\end{pmatrix}\)​

Det and row operations

  1. Multiply the i-th row by \(\lambda\), then \(A=\begin{pmatrix}-a_1-\\ -a_{i}-\\ -a_{n}-\end{pmatrix}\rightarrow\tilde{A}=\begin{pmatrix}-a_1-\\ -\lambda a_{i}-\\ -a_{n}-\end{pmatrix}\).

    Thus by multilinear, we get \(\det A=\lambda \det\tilde{A}\)​

  2. Replace the i-th row by itself plus a multilinear of the j-th row, where \(i\neq j\), then \(A=\begin{pmatrix}-a_1-\\ -a_{i}-\\ -a_{n}-\end{pmatrix}\rightarrow\tilde{A}=\begin{pmatrix}--a_1--\\ -a_{i}+\lambda a_{j}-\\ --a_{n}--\end{pmatrix}\)

    Thus by multilinear, we get \(\det\tilde{A}=\det A+\lambda\det\begin{pmatrix}-a_1-\\ -a_{j}-\\-a_{j}-\\ -a_{n}-\end{pmatrix}\). Since we have two identical rows, then it \(\det\) is \(0\)

    Thus \(\det \tilde{A}=\det A\)

  3. Interchanger rows \(i\) and \(j\)​

    We have \(A=\begin{pmatrix}-a_1-\\ -a_{i}-\\ -a_{j}-\\ -a_{n}-\end{pmatrix}\rightarrow\tilde{A}=\begin{pmatrix}-a_1-\\ -a_{j}-\\ -a_{i}-\\ -a_{n}-\end{pmatrix}\), then by alternating, we get \(\det\tilde{A}=-\det A\)

Example

\(A=\begin{pmatrix}1 & 2 & -1\\ 0 & 1 & 3\\ 1 & -1 & 0\end{pmatrix},\det A=?\)

We reduce it \(A\rightarrow A_1=\begin{pmatrix}1 & 2 & -1\\ 0 & 1 & 3\\ 0 & -3 & 1\end{pmatrix}\rightarrow A_2=\begin{pmatrix}1 & 0 & -7\\ 0 & 1 & 3\\ 0 & 0 & 10\end{pmatrix}\rightarrow A_3=\begin{pmatrix}1 & 0 & -7\\ 0 & 1 & 3\\ 0 & 0 & 1\end{pmatrix}\rightarrow A_4=\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}\)

In this process, \(\det A_1=\det A,\det A_2=\det A_1,\det A_3=\frac{1}{10}\det A_2,\det A_4=\det A_3\)​

Then \(\det A=\det A_1=\det A_2=10\det A_3=10\det A_4=10\)

Remark

To compute \(\det A\) we may reduce \(A\) to its row-reduced echelon form \(\tilde{A}\) recording all row operation of type (1) (3). Type (2) operations do not affect \(\det\)

Conclusion

\(\det A=\left(-1\right)^{k}\lambda_1^{-1}\cdot...\cdot\lambda_{m}^{-1}\det(\tilde{A})=\begin{cases}\left(-1\right)^{k}\lambda_1^{-1}\cdot...\cdot\lambda_{m}^{-1},\text{ if }\tilde{A}=I\\ 0,\text{ Otherwise}\end{cases}\) where \(k\) is the number of time we interchanged rows and \(\lambda_1,...,\lambda_m\) are the non-zero scalars that we used to multilinear rows

Example

\(A = \begin{pmatrix} i & 1 & 0 \\ 0 & -i & 1 \\ -1 & 0 & -1 \end{pmatrix}\), find \(\det A\)?

\(A_{k=1}\sim\begin{pmatrix}-1 & 0 & -1\\ 0 & -i & 1\\ i & 1 & 0\end{pmatrix}_{\lambda=-1}\sim\begin{pmatrix}1 & 0 & 1\\ 0 & -i & 1\\ i & 1 & 0\end{pmatrix}\sim\begin{pmatrix}1 & 0 & 1\\ 0 & -i & 1\\ 0 & 1 & -i\end{pmatrix}_{k=2}\sim\begin{pmatrix}1 & 0 & 1\\ 0 & 1 & -i\\ 0 & -i & 1\end{pmatrix}\sim\begin{pmatrix}1 & 0 & 1\\ 0 & 1 & -i\\ 0 & 0 & 2\end{pmatrix}_{\lambda=\frac12}\sim\begin{pmatrix}1 & 0 & 1\\ 0 & 1 & -i\\ 0 & 0 & 1\end{pmatrix}\sim\begin{pmatrix}1\\ 1\\ 1\end{pmatrix}\)

\(\det A=(-1)^2\lambda_1^{-1}\lambda_2^{-1}=(-1)^2\cdot-2=-2\)

Proposition

Let \(A\in M_{n\times n}\left(\mathbb{F}\right)\)

  1. If \(A\) has a zero row or two identical rows, then \(\det A=0\)​

  2. If \(A\) is upper triangular \(A=\begin{pmatrix}\lambda_1 & \cdots & \star\\ \vdots & \ddots & \vdots\\ 0 & \cdots & \lambda_{n}\end{pmatrix}\) \([A_{ij}=0,i>j]\), then \(\det A=\lambda_1\cdot ... \cdot \lambda_n\)​

    Proof

    If \(\lambda_1,\lambda_2,...,\lambda_n\) all are not equal to \(0\), then \(\det A=\lambda_1\cdot\lambda_2\cdot...\cdot\lambda_{n}\det\begin{pmatrix}1 & \cdots & \star\\ \vdots & \ddots & \vdots\\ 0 & \cdots & 1\end{pmatrix}=\lambda_1\cdot\lambda_2\cdot...\cdot\lambda_{n}\det I\). (After reducing to identity matrix)

    If some \(\lambda_i=0\), then \(\tilde{A}\) has a zero row. Then \(\det \tilde{A}=0\)

    Since \(\det A=c\det \tilde{A}\), then \(\det A=0=\lambda_1\cdot\lambda_2\cdot...\cdot\lambda_{n}\)​

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  3. If \(A\) is block diagonal(square matrix put in matrix) \(A=\begin{pmatrix}A_1 & \cdots & 0\\ \vdots & A_2 & \vdots\\ 0 & \cdots & A_{k}\end{pmatrix}\), then \(\det A=\left(\det A_1\right)\cdot\left(\det A_2\right)\cdot\ldots\cdot\left(\det A_{k}\right)\)

Example

  1. \(A=\begin{pmatrix}1 & 2 & 3 & 1\\ -1 & 1 & 2 & 0\\ 0 & 0 & 2 & 1\\ 0 & 0 & 1 & 0\end{pmatrix}\rightarrow\begin{pmatrix}1 & 2 & 0 & 0\\ -1 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{pmatrix}\rightarrow\begin{pmatrix}1 & 2 & 0 & 0\\ 0 & 3 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}\)​

    \(k=1\)​

    \(\det(A)=(-1)^1\cdot 3=-3\)​

  2. \(A = \begin{pmatrix} i+1 & \sqrt{2} & 2 & 2^{-1} & \beta \\ 0&i & 1 & 0 & 2^2 \\ 0&0&-i & 0 & i+1 \\ 0&0&0&i-1 & 1 \\0&0&0&0&1\end{pmatrix} \Rightarrow \det A = -2\). The multiplication of diagonal line (upper triangular)

  3. For matrix \(A=\begin{pmatrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{pmatrix}\)

    \(\det A=\begin{pmatrix}1 & 2 & 3\\ 1+3 & 2+3 & 3+3\\ 1+6 & 2+6 & 3+6\end{pmatrix}\rightarrow\det A=\det\begin{pmatrix}1 & 2 & 3\\ 1 & 2 & 3\\ 7 & 8 & 9\end{pmatrix}+\det\begin{pmatrix}1&2&3\\3 & 3 & 3\\ 1+6 & 2+6 & 3+6\end{pmatrix}\)

    \(=0+\det\begin{pmatrix}1 & 2 & 3\\ 3 & 3 & 3\\ 1 & 2 & 3\end{pmatrix}+\det\begin{pmatrix}1 & 2 & 3\\ 3 & 3 & 3\\ 6 & 6 & 6\end{pmatrix}=0+0+2\det\begin{pmatrix}3 & 3 & 3\\ 3 & 3 & 3\\ 3 & 3 & 3\end{pmatrix}=0\)

  4. Vandermonde matrix \((x_{i}\text{ different})\)​

    \(\det\begin{pmatrix}1 & x_0 & x_0^2\\ 1 & x_1 & x_1^2\\ 1 & x_2 & x_2^2\end{pmatrix}=\det\begin{pmatrix}1 & x_0 & x_0^2\\ 0 & x_1-x_0 & x_1^2-x_0^2\\ 0 & x_2-x_0 & x_2^2-x_0^2\end{pmatrix}=(x_1-x_0)(x_2-x_0)\det\begin{pmatrix}1 & x_0 & x_0^2\\ 0 & 1 & x_1+x_0\\ 0 & 1 & x_2+x_0\end{pmatrix}\)

    \(=(x_1-x_0)(x_2-x_0)\det\begin{pmatrix}1 & x_0 & x_0^2\\ 0 & 1 & x_1+x_0\\ 0 & 0 & x_2-x_1\end{pmatrix}= (x_1-x_0)(x_2-x_0)(x_2-x_1)\det\begin{pmatrix}1 & x_0 & x_0^2\\0 & 1 & x_1+x_0\\0 & 0 & 1\end{pmatrix}\)

    \(= (x_1-x_0)(x_2-x_0)(x_2-x_1)\)

T/F

  1. \(\det(-A) = -\det A\) (F)

    Counterexample: \(A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}, -A=\begin{pmatrix} -a & -b \\ -c & -d \end{pmatrix}\), then \(\det(-A) = (-1)\det\begin{pmatrix} a & b \\ -c & -d \end{pmatrix} = (-1)(-1)\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \det A\)

    Thus \(\det(-A) = \begin{cases} \det A, & \text{if }n\text{ is even} \\ -\det A, & \text{if }n\text{ is odd} \end{cases}\)

  2. \(\det(A+B) = \det A + \det B\) (F)

    \(A=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, B=-A\)

    \(\det(A+B) = \det(0) = 0\)

    \(\det(A) + \det(B) = 1 + 1 = 2\)

Notice that \(\det:M_{1\times1}\left(\mathbb{F}\right)\rightarrow\mathbb{F}\) is the identity

Theorem

\(A\) is invertible iff \(\det A\neq 0\)

Proof

\(\det A=c\det(\tilde{A})\) where \(\tilde{A}\) is the row-reduced echelon form of \(A\)

\(\tilde{A}=\begin{cases}I\\\text{has a zero row}\end{cases}\) \(A\) is invertible iff \(\tilde{A}=I\) iff \(\det \tilde{A}=1\) iff \(\det A=c\neq 0\)

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