Linear Algebra II HT23, Determinants
Determinants
When building up determinants as a determinantal mapping $D: M _ n(\mathbb{R}) \to \mathbb{R}$, what are the three conditions a determinantal map must satsify?
- Multilinear in the columns \(D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]\)
- Alternating \(D[\ldots,\pmb{a}\ _ i,\pmb{a}\ _ {i+1},\ldots] = 0 \text{ when } \pmb{a}\ _ i = \pmb{a}\ _ {i+1}\)
- $D(I _ n) = 1$ for the $n \times n$ identity matrix
What’s the geometric interpretation behind a determinantal map being multilinear in the columns, i.e. $D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]$?
It represents the fact that stretching a face with scale the area/volume accordingly.
What’s the geometric interpretation behind a determinantal map being alternating, i.e. $D[\ldots,\pmb{a} _ i,\pmb{a} _ {i+1},\ldots] = 0 \text{ when } \pmb{a} _ i = \pmb{a} _ {i+1}$ ?
A shape with thickness has no volume.
If $\lambda = \det [\ldots, \pmb{a}, \ldots, \pmb{b}, \ldots]$, then what is $\det [\ldots, \pmb{b},\ldots, \pmb{a}, \ldots]$?
When proving that there exists a map $D _ n$ that satisfies the properties of being a determinantal map on $n\times n$ matrices
Multilinear in the columns \(D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]\)
Alternating \(D[\ldots,\pmb{a}\ _ i,\pmb{a}\ _ {i+1},\ldots] = 0 \text{ when } \pmb{a}\ _ i = \pmb{a}\ _ {i+1}\)
$D(I _ n) = 1$ for the $n \times n$ identity matrix
In the inductive step, what do you define $D _ n$ as, in terms of $D _ {n-1}$?
Multilinear in the columns \(D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]\) Alternating \(D[\ldots,\pmb{a}\ _ i,\pmb{a}\ _ {i+1},\ldots] = 0 \text{ when } \pmb{a}\ _ i = \pmb{a}\ _ {i+1}\) $D(I _ n) = 1$ for the $n \times n$ identity matrix
Can you give the Laplace expansion formula for $\det A$ along row $i$?
^laplace-expansion-row
Can you give the Laplace expansion formula for $\det A$ along column $j$?
Can you give the permutation formula for $\det A$?
Why is $\det A$ multilinear in rows as well as columns, despite this not being in the definition?
Because $\det A = \det A^\intercal$.
Using the Rule of Sarrus (cool name) mneumonic, what is
\(\left \vert \begin{matrix} a \& b \& c \\\\ d \& e \& f \\\\ g \& h \& i \end{matrix}\right \vert\)?
If $A$ is a upper or lower triangular matrix, what is the determinant?
To prove the more general $\det AB = \det A \det B$, what do you show first?
\(\det EA = \det E\det A\) where $E$ is an elementary matrix.
How does the determinant change when you swap two rows of a matrix?
It switches sign.
How does the determinant change when you add a scalar multiple of one row to another in a matrix?
It does not change.
How does the determinant change when you multiply one row of a matrix by a non-zero scalar $\lambda$?
The determinant is multiplied by $\lambda$.
Proofs
Important ones
Prove that there exists a map $D _ n$ that satisfies the properties of being a determinantal map on $n\times n$ matrices.
Multilinear in the columns \(D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]\)
Alternating \(D[\ldots,\pmb{a} _ i,\pmb{a} _ {i+1},\ldots] = 0 \text{ when } \pmb{a} _ i = \pmb{a} _ {i+1}\)
$D(I _ n) = 1$ for the $n \times n$ identity matrix
Multilinear in the columns \(D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]\) Alternating \(D[\ldots,\pmb{a} _ i,\pmb{a} _ {i+1},\ldots] = 0 \text{ when } \pmb{a} _ i = \pmb{a} _ {i+1}\) $D(I _ n) = 1$ for the $n \times n$ identity matrix
Todo?
Prove if there exists a unique map $D _ n$ that satisfies the properties of being a determinantal map on $n\times n$ matrices:
Multilinear in the columns \(D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]\)
Alternating \(D[\ldots,\pmb{a} _ i,\pmb{a} _ {i+1},\ldots] = 0 \text{ when } \pmb{a} _ i = \pmb{a} _ {i+1}\)
$D(I _ n) = 1$ for the $n \times n$ identity matrix
Hint: This involves showing the following formula is true for the determinant:
\[\det A = \sum _ {\sigma \text{ perm. of } \\{1,\ldots,n\\} \space} \text{sgn}(\sigma)a _ {\sigma(1)1}a _ {\sigma(2)2}\ldots a _ {\sigma(n)n}\]
Multilinear in the columns \(D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]\) Alternating \(D[\ldots,\pmb{a} _ i,\pmb{a} _ {i+1},\ldots] = 0 \text{ when } \pmb{a} _ i = \pmb{a} _ {i+1}\) $D(I _ n) = 1$ for the $n \times n$ identity matrix
Todo?
Not-so-important ones
Prove that if a map $D _ n$ satisfies the properties of being a determinantal map on $n \times n$ matrices
Multilinear in the columns \(D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]\)
Alternating \(D[\ldots,\pmb{a} _ i,\pmb{a} _ {i+1},\ldots] = 0 \text{ when } \pmb{a} _ i = \pmb{a} _ {i+1}\)
$D(I _ n) = 1$ for the $n \times n$ identity matrix
Then the alternating condition can be strengthened to
\[\det [\ldots,\pmb{a},\ldots,\pmb{a},\ldots] = 0\]
Multilinear in the columns \(D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]\) Alternating \(D[\ldots,\pmb{a} _ i,\pmb{a} _ {i+1},\ldots] = 0 \text{ when } \pmb{a} _ i = \pmb{a} _ {i+1}\) $D(I _ n) = 1$ for the $n \times n$ identity matrix
Todo?
Prove that if a map $D _ n$ satisfies the properties of being a determinantal map on $n \times n$ matrices
Multilinear in the columns \(D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]\)
Alternating \(D[\ldots,\pmb{a} _ i,\pmb{a} _ {i+1},\ldots] = 0 \text{ when } \pmb{a} _ i = \pmb{a} _ {i+1}\)
$D(I _ n) = 1$ for the $n \times n$ identity matrix
Then it is also true that
\[\det [\ldots,\pmb{a},\ldots,\pmb{b},\ldots] = -\det [\ldots,\pmb{b},\ldots,\pmb{a},\ldots]\]
Multilinear in the columns \(D[\ldots,\lambda \pmb{b} _ i + \mu\pmb{c} _ i, \ldots] = \lambda D[\ldots,\pmb{b} _ i,\ldots] + \mu D[\ldots,\pmb{a} _ i,\ldots]\) Alternating \(D[\ldots,\pmb{a} _ i,\pmb{a} _ {i+1},\ldots] = 0 \text{ when } \pmb{a} _ i = \pmb{a} _ {i+1}\) $D(I _ n) = 1$ for the $n \times n$ identity matrix
Todo?
Prove that if $V$ is a vector space and $T : V \to V$ is linear then $\det T$ defined by $\det T = \det M^\mathcal{B}$ where $M^\mathcal{B}$ is the transformation matrix of $T$ with respect to a basis $\mathcal{B}$ does not depend on the choice of basis.
Todo.