The transpose of a matrix

Simple properties of the transpose

• $(A^T)^T=A$; and
• $(A+B)^T=A^T+B^T$ if $A$ and $B$ are matrices of the same size; and
• $(cA)^T=c(A^T)$ for any scalar $c$.

Lemma: transposes and row-column multiplication

If $a$ is a $1\times m$ row vector and $b$ is an $m\times 1$ column vector, then \[ ab=b^Ta^T.\]

• Proof is a calculation (see Tutorial worksheet 4)

Observation: the transpose swaps rows with columns

Formally, for any matrix $A$ and any $i,j$, we have \begin{align*}\def\col#1{\text{col}_{#1}}\def\row#1{\text{row}_{#1}} \row i(A^T)&=\col i(A)^T\\\col j(A^T)&=\row j(A)^T .\end{align*}

Theorem: the transpose reverses the order of matrix multiplication

If $A$, $B$ are matrices and the matrix product $AB$ is defined, then $B^TA^T$ is also defined, and $(AB)^T=B^TA^T$.

Proof

• $A$ is $n\times m$ and $B$ is $m\times k$ for some $n,m,k$
  • $AB$ is an $n\times k$ matrix, so $(AB)^T$ is $k\times n$.
  • $B^T$ is $k\times m$ and $A^T$ is $m\times n$
  • So $B^TA^T$ is defined, and is $k\times n$.
• So $(AB)^T$ and $B^TA^T$ have the same sizes.
• To show that $(AB)^T=B^TA^T$, we have to check their entries all agree.

• $(i,j)$ entry of $(AB)^T$
  • =$(j,i)$ entry of $AB$
  • = $\row j(A)\cdot\col i(B)$
  • = $\col i(B)^T\cdot \row j(A)^T$ since $ab=b^Ta^T$ for row-col multiplication
  • = $\row i(B^T)\cdot \col j(A^T)$ by the Observation about transposes
  • = the $(i,j)$ entry of $B^TA^T$.
• Hence $(AB)^T$ and $B^TA^T$ have the same sizes and the same entries
• So $(AB)^T=B^TA^T$. ■

$n\times n$ determinants

For an $n\times n$ matrix $A$, we'll define a number $\det(A)$ so that \[ A\text{ is invertible} \iff \det(A)\ne0.\]

1. If $A=[a]$ is a $1\times 1$ matrix, then $\det[a]=a$.
2. If $A=\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\mat{a&b\\c&d}$ is a $2\times2$ matrix, then $\det(A)=ad-bc$.
3. If $A=\mat{a&b&c\\d&e&f\\g&h&i}$ is a $3\times 3$ matrix, then it turns out that $\det(A)=aei-afh+bfg-bdi+cdh-ceg$.
4. If $A$ is $4\times 4$, the formula for $\det(A)$ has $24$ terms.
5. If $A$ is $5\times 5$, the formula for $\det(A)$ has $120$ terms.
6. etc

• To understand $n\times n$ determinants, we'll use several steps:

1. Define the minors of a matrix
2. Use minors to define the cofactors of a matrix
3. Use cofactors to find the determinant of a $3\times 3$ matrix
4. Use cofactors to find the determinant of an $n\times n$ matrix

• After that: some other ways to calculate the determinant

Step 1: minors

• Short version: to find $M_{ij}$: delete the row and col containing $(i,j)$ entry, then take determinant.

Examples of minors

Short version: to find $M_{ij}$: delete the row and col containing $(i,j)$ entry, then take determinant.

1. If $A=\mat{3&5\\-4&7}$, then
   • $M_{11}=\det[7]=7$
   • $M_{12}=\det[-4]=-4$
   • $M_{21}=5$
   • $M_{22}=3$.
2. If $A=\mat{1&2&3\\7&8&9\\11&12&13}$, then
   • $M_{23}=\det\mat{1&2\\11&12}=1\cdot 12-2\cdot 11=-10$
   • $M_{32}=\det\mat{1&3\\7&9}=-12$.

Step 2: cofactors

Note that $(-1)^{i+j}$ is $+1$ or $-1$, and can looked up in the matrix of signs: $\mat{+&-&+&-&\dots\\-&+&-&+&\dots\\+&-&+&-&\dots\\\vdots&\vdots&\vdots&\vdots&\ddots}$. This matrix starts with a $+$ in the $(1,1)$ entry (corresponding to $(-1)^{1+1}=(-1)^2=+1$) and the signs then alternate.

• Short version: minors with some sign changes (according to matrix of signs).

Examples of cofactors

Short version: minors with sign changes $\mat{+&-&+&-&\dots\\-&+&-&+&\dots\\+&-&+&-&\dots\\\vdots&\vdots&\vdots&\vdots&\ddots}$

1. If $A=\mat{3&5\\-4&7}$, then
   • $C_{11}=+M_{11}=\det[7]=7$
   • $C_{12}=-M_{12}=-\det[-4]=4$
   • $C_{21}=-5$, and $C_{22}=3$.
2. If $A=\mat{1&2&3\\7&8&9\\11&12&13}$, then
   • $C_{23}=-M_{23}=-(-10)=10$
   • $C_{33}=+M_{33}=\det\mat{1&2\\7&8}=-6$.

Step 3: the determinant of a $3\times 3$ matrix

If $A=\mat{a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}}$ is a $3\times 3$ matrix, then \[\det A=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}.\] Here $C_{ij}$ are the cofactors of $A$.

• This formula is called Laplace expansion of $\det A$ along the first row
• Short version: sum along first row of (entries $\times$ cofactors)

Example

Short version: sum along first row of (entries $\times$ cofactors)

• $\det\mat{1&2&3\\7&8&9\\11&12&13}$
• = $1\cdot C_{11} + 2 C_{12} + 3 C_{13}$
• = $M_{11} - 2 M_{12} + 3 M_{13}$
• = $\det\mat{8&9\\12&13} -2\det\mat{7&9\\11&13} + 3\det\mat{7&8\\11&12}$
• = $-4 -2(-8)+3(-4)$
• = $0$.

Since it has zero determinant, $\mat{1&2&3\\7&8&9\\11&12&13}$ is not invertible.

Notation

• Writing $\det$ all the time is annoying
• Sometimes we write the entries of a matrix inside vertical bars $|\ |$ instead
• e.g.\begin{align*}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{1&2&3\\7&8&9\\11&12&13} &= 1\vm{8&9\\12&13} -2\vm{7&9\\11&13} + 3\vm{7&8\\11&12}\\ &=-4 -2(-8)+3(-4)\\ &=0.\end{align*}

Step 4: the determinant of an $n\times n$ matrix

Definition