ALG1_L16

Another formula for determinants

Let $M$ be an $n\times n$ matrix. Let $M_1,M_2,...,M_n$ be the rows of $M$.

We have $M_i^t=\left[\begin{array}{c}{m}_{i1}\\ m_{i2}\\ \cdots \\ m_{in}\end{array}\right]=\sum _{j=1}^n{m}_{ij}{e}_j$ where $e_j$ is the $j$th vector of the standard basis in canonical order.

Now, the determinant can be thought of as a function of the row vectors of $M$:

$$\det (M_1^t,M_2^t,\cdots ,M_n^t)=\det \left(\sum _{j=1}^n{m}_{1j}{e}_j,\sum _{j=1}^n{m}_{2j}{e}_j,\cdots ,\sum _{j=1}^n{m}_{nj}{e}_j\right).$$

By multilinearity of the determinant, we have

$$\det \left(\sum _{j=1}^n{m}_{1j}{e}_j,\sum _{j=1}^n{m}_{2j}{e}_j,\cdots ,\sum _{j=1}^n{m}_{nj}{e}_j\right)=\sum _{i_1,i_2,...,i_n}{m}_{1i_1}{m}_{2i_2}...m_{n{i}_n}\det (e_{i_1},e_{i_2},...,e_{i_n})$$

where $i_k\in \{1,..,n\}$.

The anti-symmetry property implies that $\det (e_{i_1},...,e_{i_n})=0$ if $i_k=i_l$ for $k\ne l$
Therefore we only need to consider the case when all the $i_k$ are pairwise distinct, i.e. a permutation of $\{1,..,n\}$.

Permutations

A permutation is defined to be a bijection of the set $\mathcal{N}=\{1,2,\dots ,n\}$ to itself. We define $S_n$ to be the set of all permutations of $\mathcal{N}$. The elements of $S_n$ are functions. For some $\sigma \in S$, $\sigma (i)$ is denoted by ${\sigma }_i$.

Define the length of a permutation $\sigma$ as

$$l(\sigma )\stackrel{\text{def}}{=}\text{card}\{(i,j)\mid i<j\text{ and }\sigma (i)>\sigma (j)\}$$

We claim that a permutation $\sigma$ can be reordered using $l(\sigma )$ swaps.
We define another quantity

$$\text{sign}(\sigma )={(-1)}^{l(\sigma )}$$

Observe that

$$\det (e_{{\sigma }_1},e_{{\sigma }_2},...,e_{{\sigma }_n})=(-1{)}^{l(\sigma )}.$$

Thus,

$$\det M=\sum _{\sigma \in S_n}\mathrm{sign}(\sigma )m_{1{\sigma }_1}{m}_{2{\sigma }_2}\dots m_{n{\sigma }_n}.$$

Note that we have arrived at this formula by using every single one of the defining properties of the determinant (contrast with what we have done previously, which is to present a formula and show that it satisfies the defining properties). This tells us that the determinant must indeed be uniquely determined, meaning that there is only one map from ${\mathbb{R}}^{n\times n}$ to $\mathbb{R}$ which satisfies the defining properties. The upshot of this is that all formulas covered in previous lectures for the determinant are equivalent.

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Properties of the determinant

Only singular matrices have zero determinant

Theorem 1(Theorem).

$\det A=0⟺A$ is not invertible.

Proof
$A$ is invertible
$⟹$ $\mathrm{RREF}(A)=id$
$⟹\det (A)\ne 0$

$A$ is not invertible
$⟹$ $\mathrm{RREF}(\mathrm{A})$ has a zero row
$⟹$ $\det (\mathrm{RREF}(A))=0$
$⟹\det (A)=0$
❏

Theorem 2(Corollary).

If $A$ is an $n\times n$ matrix and the rows of $A$ are linearly dependent, then $\det (A)=0$

Multiplicativity

Theorem 3(Theorem).

$\det AB=\det A\det B$.

Proof
Case 1: $A$ is not invertible.
In this case, the product $AB$ will not be invertible. Thus we have
$\det AB=0=0\cdot \det B=\det A\det B.$

Case 2: A is invertible.
Define the function $\mathrm{d}(B)=\frac{\det AB}{\det A}$.
This is well defined, since $\det A\ne 0$.
Observe that $\mathrm{d}$ satisfies the defining properties of the determinant. Therefore, $\mathrm{d}(B)=\det B$. ❏

Theorem 4(Corollary).

$\det A^n=(\det A{)}^n$.

Theorem 5(Corollary).

${\prod }_{i=1}^n{A}_i$ is invertible $⟺$ each $A_i$ is invertible.

Determinant of transpose

Theorem 6(Theorem).

$\det (A)=\det (A^T)$

Proof
Consider the function $f(A)\equiv \det A^T$. It is clear that $f$ satisfies the defining properties of the determinant. Thus, $f=\det$. ❏

Thus, all statements about row operations etc. can be made about column operations.

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Invariant Subspaces

Definition 7(Definition).

A subspace $W$ of $V$ is said to be invariant under a linear map $T$, or $T-$invariant, if $T(W)\subseteq W$, that is $w\in W⟹T(w)\in W$.

For such $T$-invariant subspaces, we can define a linear map $T{|}_W:W\to W$ called the restriction of $T$ to $W$.

Let $T:V\to V$ be a linear map. If $W$ is a $T-$invariant subspace, we can take a basis $(w_1,w_2,...,w_k)$ of $W$ and extend it to a basis ${\beta }_V=(w_1,w_2,...,w_k,v_1,v_2,...,v_{n-k})$ of $V$ where $n=\dim V$ and $k=\dim W$.

Matrix of T with respect to ${\beta }_V$

The first $k$ columns of $T$ will be the image of the $w_{i}s$.Since all $T(w_i)\in W$, they are expressible as linear combinations of only the vectors in $W$. Thus the matrix of $T$ with respect to the basis ${\beta }_V$ will look like:

$${\mathcal{M}}_{{\beta }_V,{\beta }_V}(T)=\left[\begin{array}{cc}A& B\\ 0& C\end{array}\right]$$

Also notice that $A={\mathcal{M}}_{{\beta }_W,{\beta }_W}(T{|}_W)$ where ${\beta }_W=(w_1,w_2,\dots ,w_n)$

Matrix of $T$ when $V=W_1\oplus W_2$

If it happens that $V=W_1\oplus W_2$ where $W_1$ and $W_2$ are $T$-invariant with bases ${\beta }_{W_1}=(w_1,...,w_k)$
${\beta }_{W_2}=(w_1^{\prime },...,w_{n-k}^{\prime })$
${\beta }_V=(w_1,\dots ,w_k,w_1^{\prime },\dots ,w_k^{\prime })$
then the matrix of $T$ will be

$${\mathcal{M}}_{{\beta }_V,{\beta }_V}(T)=\left[\begin{array}{cc}A& 0\\ 0& B\end{array}\right]$$

where $A={\mathcal{M}}_{{\beta }_{W_1},{\beta }_{W_1}}(T),B={\mathcal{M}}_{{\beta }_{W_2},{\beta }_{W_2}}(T)$.