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lecture_14
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| lecture_14 [2015/03/05 11:22] – [Step 3: the determinant of a $3\times 3$ matrix using Laplace expansion along the first row] rupert | lecture_14 [2017/03/09 10:49] (current) – [Corollary on invertibility] rupert | ||
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| - | ==== Observation: | + | === Example |
| - | Formally, for any matrix $A$ and any $i,j$, we have | + | \begin{align*}\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\det\mat{1& |
| - | \begin{align*}\def\col#1{\text{col}_{#1}}\def\row#1{\text{row}_{#1}} | + | &= 1 \cdot (+M_{11}) + 2 \cdot (-M_{12}) + 3 \cdot(+M_{13})\\ |
| - | \row i(A^T)&=\col i(A)^T\\\col j(A^T)&=\row j(A)^T | + | & |
| - | .\end{align*} | + | &= \det\mat{8& |
| + | & | ||
| + | &=-4 -2(-8)+3(-4)\\ | ||
| + | & | ||
| + | &=0.\end{align*} | ||
| - | ==== Theorem: the transpose reverses the order of matrix multiplication ==== | + | From this, we can conclude that $\mat{1& |
| - | If $A$ and $B$ are matrices and the [[matrix product]] $AB$ is defined, then $B^TA^T$ is also defined. Moreover, in this case we have | + | === Notation === |
| - | \[ (AB)^T=B^TA^T.\] | + | |
| - | === Proof === | + | To save having to write $\det$ all the time, we sometimes write the entries of a matrix inside vertical bars $|\ |$ to mean the determinant of that matrix. Using this notation (and doing a few steps in our heads), we can rewrite the previous example as: |
| + | |||
| + | \begin{align*}\def\vm# | ||
| + | &=-4 -2(-8)+3(-4)\\ | ||
| + | &=0.\end{align*} | ||
| + | |||
| + | ==== Step 4: the determinant of an $n\times n$ matrix ==== | ||
| + | |||
| + | |||
| + | ===Definition=== | ||
| + | |||
| + | {{page> | ||
| + | |||
| + | ===Example=== | ||
| - | If $AB$ is defined, then $A$ is $n\times m$ and $B$ is $m\times k$ for some $n,m,k$, so $B^T$ is $k\times m$ and $A^T$ is $m\times n$, so $B^TA^T$ is defined. Moreover, in this case $B^TA^T$ is an $k\times n$ matrix, and $AB$ is an $n\times k$ matrix, so $(AB)^T$ is a $k\times n$ matrix. Hence $B^TA^T$ has the same size as $(AB)^T$. To show that they are equal, we calculate, using the fact that the transpose swaps rows with columns: | ||
| \begin{align*} | \begin{align*} | ||
| - | \text{the }(i,j)\text{ entry of }(AB)^T&= \text{the }(j,i)\text{ entry of }AB | + | \def\vm# |
| - | \\&= \row j(A)\cdot\col i(B) | + | \vm{\color{red}1&\color{red}0&\color{red}2&\color{red}3\\0& |
| - | \\&=\col i(B)^T\cdot \row j(A)^T | + | & |
| - | \\&=\row i(B^T)\cdot \col j(A^T) | + | & |
| - | \\&=\text{the }(i,j)\text{ entry of }B^TA^T | + | & |
| + | &=36. | ||
| \end{align*} | \end{align*} | ||
| - | Hence $(AB)^T=B^TA^T$. ■ | ||
| - | ===== Determinants of $n\times n$ matrices ===== | + | ==== Theorem: Laplace expansion along any row or column gives the determinant |
| - | Given any $n\times n$ matrix $A$, it is possible to define a number | + | - For any fixed $i$: $\det(A)=a_{i1}C_{i1}+a_{i2}C_{i2}+\dots+a_{in}C_{in}$ (Laplace expansion along row $i$) |
| - | \[ A\text{ is invertible} \iff \det(A)\ne0.\] | + | - For any fixed $j$: $\det(A)=a_{1j}C_{1j}+a_{2j}C_{2j}+\dots+a_{nj}C_{nj}$ |
| - | - If $A$ is a $1\times 1$ matrix, then $A=[a]$ for some number $a$, and we just define $\det[a]=a$. | + | === Example |
| - | - If $A$ is a $2\times 2$ matrix, then $A=\def\mat# | + | |
| - | - If $A$ is a $3\times 3$ matrix, then $A=\mat{a& | + | |
| - | - If $A$ is a $4\times 4$ matrix, then the formula for $\det(A)$ is more complicated still, with $24$ terms. | + | |
| - | - If $A$ is a $5\times 5$ matrix, then the formula for $\det(A)$ has $120$ terms. | + | |
| - | Trying to memorise a formula in every case (or even in the $3\times 3$ case!) isn't convenient unless we understand it somehow. We will approach | + | We can make life easier by choosing expansion rows or columns with lots of zeros, if possible. Let's redo the previous example with this in mind: |
| - | ==== Step 1: minors ==== | + | \begin{align*} |
| + | \def\vm# | ||
| + | \vm{1& | ||
| + | -\color{red}0+\color{red}2\vm{1& | ||
| + | &=2\left(-\color{purple}2\vm{2& | ||
| + | &=2(-2(-8)-(-2))\\ | ||
| + | &=36. | ||
| + | \end{align*} | ||
| - | === Definition === | + | ==== Definition: upper triangular matrices ==== |
| - | {{page> | + | |
| - | === Examples === | + | An $n\times n$ matrix $A$ is **upper triangular** if all the entries below the main diagonal are zero. |
| - | * If $A=\mat{3& | ||
| - | * If $A=\mat{1& | ||
| - | ==== Step 2: cofactors | + | ==== Definition: diagonal matrices |
| - | ===Definition=== | + | An $n\times n$ matrix $A$ is **diagonal** if the only non-zero entries are on its main diagonal. |
| - | {{page> | + | |
| - | Note that $(-1)^{i+j}$ is $+1$ or $-1$, and can looked up in the matrix of signs: | + | ==== Corollary: the determinant of upper triangular matrices and diagonal matrices ==== |
| - | $\mat{+& | + | |
| - | This matrix starts with a $+$ in the $(1,1)$ entry (corresponding to $(-1)^{1+1}=(-1)^2=+1$) and the signs then alternate. | + | |
| - | ===Examples=== | + | - The determinant of an upper triangular $n\times n$ matrix is the product of its diagonal entries: $\det(A)=a_{11}a_{22}\dots a_{nn}$. |
| + | - The determinant of an $n\times n$ diagonal matrix is the product of its diagonal entries: $\det(A)=a_{11}a_{22}\dots a_{nn}$. | ||
| - | * If $A=\mat{3& | + | === Proof === |
| - | * If $A=\mat{1& | + | |
| - | ==== Step 3: the determinant of a $3\times 3$ matrix using Laplace expansion along the first row ==== | + | |
| - | ===Definition=== | + | - This is true for $n=1$, trivially. For $n>1$, assume inductively that it is true for $(n-1)\times (n-1)$ matrices and use the Laplace expansion of an upper triangular $n\times n$ matrix $A$ along the first column of $A$ to see that $\det(A)=a_{11}C_{11}+0+\dots+0=a_{11}C_{11}$. Now $C_{11}$ is the determinant of the $(n-1)\times (n-1)$ matrix formed by removing the first row and and column of $A$, and this matrix is upper triangular with diagonal entries $a_{22}, |
| + | - Any diagonal matrix is upper triangular, so this is a special case of statement 1. ■ | ||
| - | {{page> | + | ==== Examples ==== |
| + | - For any $n$, we have $\det(I_n)=1\cdot 1\cdots 1 = 1$. | ||
| + | - For any $n$, we have $\det(5I_n)=5^n$. | ||
| + | - $\def\vm# | ||
| - | === Example | + | ==== Theorem: important properties of the determinant ==== |
| - | \begin{align*}\det\mat{1& | + | Let $A$ be an $n\times n$ matrix. |
| - | &= 1 \cdot (+M_{11}) + 2 \cdot (-M_{12}) + 3 \cdot(+M_{13})\\ | + | |
| - | &= M_{11}-2M_{12}+3M_{13}\\ | + | |
| - | &= \det\mat{8& | + | |
| - | &= (8\cdot 13-9\cdot 12) -2(7\cdot 13-9\cdot 11)+3(7\cdot 12-8\cdot 11)\\ | + | |
| - | &=-4 -2(-8)+3(-4)\\ | + | |
| - | & | + | |
| - | &=0.\end{align*} | + | |
| - | From this, we can conclude that $\mat{1& | + | - $A$ is invertible if and only if $\det(A)\ne0$. |
| + | - $\det(A^T)=\det(A)$ | ||
| + | - If $B$ is another $n\times n$ matrix, then $\det(AB)=\det(A)\det(B)$ | ||
| - | === Notation | + | ==== Corollary on invertibility ==== |
| - | To save having to write $\det$ all the time, we sometimes write the entries of a matrix inside vertical bars $|\ |$ to mean the determinant of that matrix. Using this notation (and doing a few steps in our heads), we can rewrite the previous example as: | + | - $A^T$ is invertible if and only if $A$ is invertible |
| + | - $AB$ is invertible if and only if **both** $A$ and $B$ are invertible | ||
| - | \begin{align*}\def\vm#1{\begin{vmatrix}# | + | === Proof === |
| - | &=-4 -2(-8)+3(-4)\\ | + | |
| - | &=0.\end{align*} | + | - We have $\det(A^T)=\det(A)$. So $A^T$ is invertible $\iff$ $\det(A^T)\ne0$ $\iff$ $\det(A)\ne 0$ $\iff$ $A$ is invertible. |
| + | - We have $\det(AB)=\det(A)\det(B)$. So $AB$ is invertible $\iff$ $\det(AB)\ne 0$ $\iff$ $\det(A)\det(B)\ne0$ $\iff$ $\det(A)\ne0$ and $\det(B)\ne 0$ $ \iff$ $A$ is invertible and $B$ is invertible. ■ | ||
lecture_14.1425554530.txt.gz · Last modified: by rupert