Properties of determinants
Lecture 22
Recap
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Determinants
- For an \(n\times n\) matrix \(A\):
- \(\det(A)\) is defined by cofactor expansion along any row or any column; in practice, choose one with many zeros to simplify the computation.
- \(|\det(A)|\) equals the hypervolume of the hyperparallelogram formed by its \(n\) column vectors.
- The sign of \(\det(A)\) records the orientation of the column vectors.
- If \(\det(A)=0\), the columns are linearly dependent.
- If \(\det(A)\neq 0\), the columns are linearly independent and \(A\) is invertible (also called non-singular).
- Computations can be carried out far more efficiently by computer algebra systems (e.g., MATLAB or Mathematica) than by hand.
Determinants for Special Matrices
- Let \(D\) be an \(n\times n\) diagonal matrix with diagonal entries \(d_1,\dots,d_n\): \[\det(D)=d_1d_2\cdots d_n.\]
- More generally, if \(T\) is a triangular matrix (upper or lower) with diagonal entries \(d_1,\dots,d_n\): \[\det(T)=d_1d_2\cdots d_n.\]
- If \(E_{R_i\leftrightarrow R_j}\) is the elementary matrix corresponding to swapping two rows: \[\det\!\left(E_{R_i\leftrightarrow R_j}\right)=-1.\]
- Geometrically, swapping two rows reverses orientation, which explains the negative sign.
Determinant and Invertible Matrices
- Geometrically, \(A^{-1}\) undoes what \(A\) does to vectors in \(\mathbb R^n\).
- If the columns of \(A\) are linearly independent, the associated hyperparallelogram has nonzero hypervolume — nothing collapses to a lower dimension.
- Since no information is lost, the transformation can be reversed; hence \(A\) is invertible.
- From a row-reduction perspective: if the \(n\) columns are linearly independent, then \(\text{rank}(A)=n\), so the RREF of \(A\) has \(n\) pivots and equals \(I_n\).
- Therefore \(A\) can be transformed into \(I_n\) by a sequence of elementary matrices, and those same operations produce \(A^{-1}\).
- Consequently, \(\det(A)\neq 0 \quad \Longleftrightarrow \quad A \text{ is invertible.}\)
Summary for Invertibility
- Let \(A\) be an \(n\times n\) matrix. The following are equivalent:
- There exists a matrix \(B\) such that \(AB=BA=I_n\) (so \(A\) is invertible and \(B=A^{-1}\)).
- \(\det(A)\neq0\).
- \(\operatorname{RREF}(A)=I_n\).
- \(\mathop{\mathrm{rank}}(A)=n\) and \(\mathop{\mathrm{null}}(A)=\mathop{\mathrm{null}}(A^T)=0\).
- The columns of \(A\) are linearly independent.
- The rows of \(A\) are linearly independent.
- The homogeneous system \(A\vec{x}=\vec{0}\) has the unique solution \(\vec{x}=\vec{0}\).
- The linear equation \(A\vec{x}=\vec{b}\) has the unique solution \(\vec{x}=A^{-1} \vec{b}\).
Summary for Non-Invertibility
- Let \(A\) be an \(n\times n\) matrix. The following are equivalent:
- There is no matrix \(B\) such that \(AB=BA=I_n\) (so \(A\) is not invertible).
- \(\det(A)=0\).
- \(\operatorname{RREF}(A)\neq I_n\) and has some zero rows.
- \(\mathop{\mathrm{rank}}(A)<n\) and \(\mathop{\mathrm{null}}(A)=\mathop{\mathrm{null}}(A^T)>0\).
- The columns of \(A\) are linearly dependent.
- The rows of \(A\) are linearly dependent.
- The homogeneous system \(A\vec{x}=\vec{0}\) has infinitely many solutions.
Properties of Determinants
Multiplicativity for \(2\times2\) Matrices
- Let \(A,B:\mathbb{R}^2\to\mathbb{R}^2\) be linear transformations.
- Matrix multiplication \(AB\) represents function composition \(A\circ B\) (apply \(B\) first, then \(A\)).
- When we apply \(B\), the unit square is transformed into a parallelogram with signed area \(\det(B)\).
- Applying \(A\) afterward scales the signed area of any region by a factor of \(\det(A)\), since linear transformations scale area uniformly.
- Therefore, after applying \(B\) and then \(A\), the unit square becomes a parallelogram with signed area \(\det(A)\det(B)\).
- Since \(AB\) represents the combined transformation, and determinants measure area scaling, \[\det(AB)=\det(A)\det(B).\]
Multiplicativity in General
- The same geometric reasoning extends to \(n\times n\) matrices, where determinants represent signed hypervolume scaling in \(\mathbb{R}^n\).
- Let \(A,B\) be two \(n\times n\) matrices, viewed as linear transformations from \(\mathbb{R}^n\) to \(\mathbb{R}^n\).
- Applying \(B\) first scales hypervolume by \(\det(B)\), and then applying \(A\) scales it by \(\det(A)\).
- Therefore, \[\det(AB)=\det(A)\det(B).\]
- In other words, the determinant is multiplicative.
Determinant and Commutativity
- Let \(A\) and \(B\) be \(n\times n\) matrices. By multiplicativity, \[ \det(AB)=\det(A)\det(B)=\det(B)\det(A)=\det(BA). \]
- Caution: In general, \(AB\neq BA\) as matrices.
- Thus matrix multiplication is not commutative, but the determinant behaves as if it were: \[ \det(AB)=\det(BA). \] if \(A\) and \(B\) are square matrices of the same size.
- Caution: If \(A\) and \(B\) are not both \(n\times n\), then \(\det(A)\) or \(\det(B)\) may not even be defined, and multiplicativity does not apply.
Example
- Let \[ A= \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}, \qquad B= \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix}. \]
- Compute \(BA\).
- Then compute \(\det(BA)\) and submit your answer on iClicker.
- Next, we will compute \(AB\) and \(\det(AB)\) and compare the results together.
Scan the QR code or go to join.iclicker.com/MBNJ.
Determinant of the Inverse
- If \(A\) is invertible, then by multiplicity \[ \det(A)\det(A^{-1}) = \det(AA^{-1}) = \det(I_n) = 1. \]
- Therefore, \[ \det(A^{-1})=\frac{1}{\det(A)}. \]
- In particular, \(A\) is invertible if and only if \(\det(A)\neq0\).
RREF and Determinants
- If \(\operatorname{RREF}(A)=R\), then \[ R=E_kE_{k-1}\cdots E_1 A, \] where each \(E_i\) is an elementary row matrix.
- Taking determinants, \[ \det(A) = \det(R)/\left(\det(E_k)\cdots\det(E_1)\right). \]
- Each elementary row operation has a simple determinant effect:
- Row swap \(\Rightarrow\) multiply by \(-1\),
- Row scaling by \(c\) \(\Rightarrow\) multiply by \(c\),
- Row replacement (adding a multiple of another row) \(\Rightarrow\) no change.
- Thus determinants can be computed efficiently via row reduction.
Example
Let \[ A= \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 0 \\ 1 & 1 & 1 \end{bmatrix}. \]
Row reduce to \(\operatorname{RREF}(A)=I_3\) while tracking determinant changes.
Step 1: \(R_2\leftarrow R_2-2R_1\), \(R_3\leftarrow R_3-R_1\)
(row replacement → determinant unchanged) \[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & -1 & -2 \\ 0 & -1 & 0 \end{bmatrix}. \]Step 2: \(R_3\leftarrow R_3-R_2\)
(row replacement → determinant unchanged) \[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & -1 & -2 \\ 0 & 0 & 2 \end{bmatrix}. \]Step 3: \(R_2\leftarrow -R_2\)
(row scaling by \(-1\) → determinant multiplies by \(-1\)) \[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{bmatrix}. \]Step 4: \(R_3\leftarrow \frac12 R_3\)
(row scaling by \(\frac12\) → determinant multiplies by \(\frac12\)) \[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}. \]Step 5: \(R_2\leftarrow R_2-2R_3\), \(R_1\leftarrow R_1-R_3\)
(row replacement → determinant unchanged) \[ \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. \]Step 6: \(R_1\leftarrow R_1-2R_2\)
(row replacement → determinant unchanged) \[ \operatorname{RREF}(A)= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=I_3. \]Let \(E\) be the product of elementary matrices for these row operations. Then \(EA=I_3\), so \(E=A^{-1}\).
Taking determinants: \[ \det(E)\det(A)=\det(I_3)=1 \quad\Rightarrow\quad \det(A)=\frac{1}{\det(E)}. \]
Only the scaling steps change \(\det(E)\): \[ \det(E)=(-1)\left(\frac12\right)=-\frac12. \]
Therefore, \[ \boxed{\det(A)=\frac{1}{-\frac12}=-2}. \]
Transpose of Multiplication
- Let \(A=[\vec{u}_1\ \dots\ \vec{u}_n]\) be a \(k\times n\) matrix and \(B=[\vec{v}_1\ \dots\ \vec{v}_m]\) a \(k\times m\) matrix.
- Then \(A^TB\) is an \(n\times m\) matrix with entries \[ (A^TB)_{ij}=\vec{u}_i\cdot \vec{v}_j. \]
- Similarly, \(B^TA\) is an \(m\times n\) matrix with \[ (B^TA)_{ij}=\vec{v}_i\cdot \vec{u}_j. \]
- Since the dot product is commutative, \(\vec{u}_i\cdot\vec{v}_j=\vec{v}_j\cdot\vec{u}_i\), we obtain \[ (B^TA)^T=A^TB. \]
Transpose of a Product
- Let \(C=B^T\). Then \(C^T=B\), and the previous identity \[ (B^TA)^T=A^TB \] becomes \[ (CA)^T=A^TC^T. \]
- Applying this rule repeatedly, for three matrices \(A,B,C\) (whose product is defined), \[ (ABC)^T = (AB\,C)^T = C^T(AB)^T = C^TB^TA^T. \]
- More generally, for matrices whose product is defined, \[ (A_1A_2\cdots A_\ell)^T = A_\ell^T\cdots A_2^T A_1^T. \]
Determinant and Transpose
- If \(\operatorname{RREF}(A)=R\), then \(A=E_kE_{k-1}\cdots E_1 R\) with some elementary row matrices \(E_i\).
- Taking transpose, \(A^T=R^T E_1^T\cdots E_k^T\).
- Taking determinants, by multiplicativity, \[ \det(A^T)=\det(R^T)\det(E_1^T)\cdots\det(E_k^T). \]
- Each elementary matrix and its transpose have the same determinant.
- Also, \(\det(R)=\det(R^T)\) (either \(1\) if \(R=I_n\), or \(0\) otherwise).
- Therefore \[\det(A^T)=\det(A).\]
- Consequently, cofactor expansion along rows gives the same determinant as expansion along columns.
Summary
- For \(n\times n\) matrices \(A\) and \(B\),
\[\det(AB)=\det(A)\det(B),\]
and therefore \(\det(AB)=\det(BA)\). - \(A\) is invertible if and only if \(\det(A)\neq0\); in that case, \[\det(A^{-1})=\frac{1}{\det(A)}.\]
- Determinants of elementary matrices are easy to compute, so \(\det(A)\) can be found efficiently using Gaussian or Gauss–Jordan elimination.
- The determinant is invariant under transpose: \(\det(A^T)=\det(A)\).