Notes - 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 $\det [\ldots, \pmb{b},\ldots, \pmb{a}, \ldots]$?


\[-\lambda\]

What is $\det [\ldots,\pmb{a},\ldots,\pmb{a},\ldots]$?


\[0\]

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 $D _ {n-1}$?


\[D_n(A) = a_{11}D_{n-1}(A_{11}) - a_{12}D_{n-2}(A_{12}) + \ldots + (-1)^{n-1}a_{1n}D_{n-1}(A_{1n})\]

Can you give the Laplace expansion formula for $\det A$ along row $i$?


\[\det A = \sum_{j=1}^{n} a_{ij} (-1)^{i+j} \det A_{ij}\]

Can you give the Laplace expansion formula for $\det A$ along column $j$?


\[\det A = \sum^n_{i=1} a_{ij}(-1)^{i+j} \det A_{ij}\]

Can you give the permutation formula for $\det A$?


\[\det A = \sum_{\sigma \text{ perm. of } \\{1,\ldots,n\\} \space} \text{sgn}(\sigma)a_{\sigma(1)1}a_{\sigma(2)2}\ldots a_{\sigma(n)n}\]

What is $\det A^\intercal$?


\[\det A\]

What is $\det BA$?


\[\det A \det B\]

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

\[\left|\begin{matrix} a \& b \& c \\\\ d \& e \& f \\\\ g \& h \& i \end{matrix}\right|\]

?


\[\begin{aligned} &aei + dhc + gbf\\\\ -&ceg-fha-ibd \end{aligned}\]

If $A$ is a upper or lower triangular matrix, what is the determinant?


\[\prod^n_{i=1} a_{ii}\]

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


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:

\[\det A = \sum_{\sigma \text{ perm. of } \\{1,\ldots,n\\} \space} \text{sgn}(\sigma)a_{\sigma(1)1}a_{\sigma(2)2}\ldots a_{\sigma(n)n}\]

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

\[\det [\ldots,\pmb{a},\ldots,\pmb{a},\ldots] = 0\]

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

\[\det [\ldots,\pmb{a},\ldots,\pmb{b},\ldots] = -\det [\ldots,\pmb{b},\ldots,\pmb{a},\ldots]\]

Todo?

Prove that $\det A = \det A^\intercal$.


Todo.

Prove

\[\det A = 0 \iff A \text{ singular}\]

Todo.

Prove

\[\det E A = \det E \det A\]

where $E$ is an elementary matrix.


Todo

Prove

\[\det A B = \det A \det B\]

by appealing to a lemma.


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.




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