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--- lecture_14_slides [2017/03/08 16:51] – [Proof by induction on $n$] rupert
+++ lecture_14_slides [2017/03/09 10:45] (current) – [Corollary: the determinant of an upper triangular matrix] rupert
@@ Line 76: / Line 76: @@
 \end{align*}
-==== Definition: upper triangular matrices ====
+==== Corollary: a matrix with a zero row or column isn't invertible ====
-An $n\times n$ matrix $A$ is **upper triangular** if all the entries below the main diagonal are zero.
+If an $n\times n$ matrix $A$ has a zero row or a zero column, then $\det(A)=0$ and $A$ isn't invertible.
+=== Proof ===
-==== Definition: diagonal matrices ====
+Expand $\det(A)$ along the zero row or column
+  * get a sum of terms $0\times C_{ij}$
+  * so $\det(A)=0$
+  * so $A$ isn't invertible.■
-An $n\times n$ matrix $A$ is **diagonal** if the only non-zero entries are on its main diagonal.
+==== Definition: "upper triangular" and "diagonal" ====
+An $n\times n$ matrix is **upper triangular** if all the entries below the main diagonal are zero.
+An $n\times n$ matrix is **diagonal** if the only non-zero entries are on its main diagonal.
 ==== Corollary: the determinant of an upper triangular matrix ====
-The determinant of an upper triangular $n\times n$ matrix is the product of its diagonal entries: \[\det(A)=a_{11}a_{22}\dots a_{nn}.\]
+If $A$ is an upper triangular matrix or a diagonal matrix, then the determinant of $A$ is the product of its diagonal entries: \[\det(A)=a_{11}a_{22}\dots a_{nn}.\]
+  * Every diagonal matrix is upper triangular!
+  * So it suffices just to prove it for upper triangular matrices.
 ==== Proof by induction on $n$ ====
@@ Line 93: / Line 105: @@
   * This is true for $n=1$, trivially.
   * For $n>1$, assume inductively that it is true for $n-1$
-  * Let $A$ be $n\times n$. Laplace expansion of $\det(A)$ along column 1:
+  * Let $A$ be $n\times n$. Expand $\det(A)$ along column 1:
   * $\det(A)=a_{11}C_{11}+0+\dots+0=a_{11}C_{11}$.
-  * $C_{11}=\det(B)$ where $B=(A$ with first row and column removed$)$
+  * $C_{11}=\det(B)$ for $B=(A$ without row 1 and column 1$)$
   * $B$ is $(n-1)\times (n-1)$, upper triangular, diagonal entries $a_{22},a_{33},\dots,a_{nn}$.
-  * By inductive assumption, $C_{11}=\det(B)=a_{22}a_{33}\dots a_{nn}$.
+  * By assumption, $C_{11}=\det(B)=a_{22}a_{33}\dots a_{nn}$.
   * So $\det(A)=a_{11}C_{11}=a_{11}a_{22}a_{33}\dots a_{nn}$.■
-==== Corollary: the determinant of a diagonal matrix ====
-The determinant of an $n\times n$ diagonal matrix is the product of its diagonal entries: \[\det(A)=a_{11}a_{22}\dots a_{nn}.\]
-=== Proof ===
-  * Any diagonal matrix is upper triangular, so this is a special case of the last Corollary (about upper triangular matrices).■
@@ Line 112: / Line 119: @@
   - For any $n$, we have $\det(I_n)=1\cdot 1\cdots 1 = 1$.
   - For any $n$, we have $\det(5I_n)=5^n$.
+    * Careful: $\det(5A)\ne 5\det(A)$!
+    * Actually, $\det(5A)=5^n\det(A)$ for any  $n\times n$ A (exercise)
   - $\vm{1&9&43&23434&4&132\\0&3&43&2&-1423&-12\\0&0&7&19&23&132\\0&0&0&2&0&0\\0&0&0&0&-1&-903\\0&0&0&0&0&6}=1\cdot3\cdot7\cdot2\cdot(-1)\cdot6 = 252$.
@@ Line 123: / Line 132: @@
   - If $B$ is another $n\times n$ matrix, then $\det(AB)=\det(A)\det(B)$.
+=== Corollary on invertibility ===
+  - $A^T$ is invertible if and only if $A$ is invertible
+  - $AB$ is invertible if and only if **both** $A$ and $B$ are invertible
+  * Warning: it's not true that $A+B$ is invertible if and only if $A$ and $B$ are invertible!
 ==== Theorem: row/column operations and determinants ====
@@ Line 154: / Line 170: @@
   * $\det(A_{Ri\leftrightarrow Rj})=-\det(A)$, so $\vm{0&0&2\\0&3&0\\4&0&0} = -\vm{4&0&0\\0&3&0\\0&0&2}=-4\cdot 3\cdot 2 = -24$.
+==== ====
   * $\det(A_{Ri\to c Ri})=c\det(A)$, and the same for columns. So \begin{align*}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\vm{ 2&4&6&10\\5&0&0&-10\\9&0&81&99\\1&2&3&4} &= 2\vm{ 1&2&3&5\\5&0&0&-10\\9&0&81&99\\1&2&3&4} = 2\cdot 5\vm{ 1&2&3&5\\1&0&0&-2\\9&0&81&99\\1&2&3&4}\\&=  2\cdot 5\cdot 9 \vm{ 1&2&3&5\\1&0&0&-2\\1&0&9&11\\1&2&3&4}=2\cdot 5\cdot 9\cdot 2 \vm{ 1&1&3&5\\1&0&0&-2\\1&0&9&11\\1&1&3&4}\\&=2\cdot 5\cdot 9\cdot 2\cdot 3 \vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4}.\end{align*}
@@ Line 162: / Line 180: @@
 ==== Corollary ===
-If $\def\row{\text{row}}\row_i(A)=c\cdot \row_j(A)$ for some $i\ne j$ and some $c\in \mathbb{R}$, then $\det(A)=0$.
+If $\def\row{\text{row}}\row_i(A)=c\cdot \row_j(A)$ for some $i\ne j$ and some $c\in \mathbb{R}$, then $\det(A)=0$ (and so $A$ isn't invertible).
 === Proof ===
@@ Line 168: / Line 186: @@
   * $\row_i(A)-c \cdot\row_j(A)=0$
   * So $A_{Ri\to Ri-c\,Rj}$ has a zero row
-  * By Laplace expansion along this row: $\det(A_{Ri\to Ri-c\,Rj})=0$
+  * So $\det(A_{Ri\to Ri-c\,Rj})=0$
   * So $\det(A)=\det(A_{Ri\to Ri-c\,Rj})=0$.■
@@ Line 190: / Line 208: @@
 \begin{align*}\vm{1&3&1&3\\\color{red}4&\color{red}8&\color{red}0&\color{red}{12}\\0&1&3&6\\2&2&1&6}&= \color{red}{4}\vm{1&3&1&\color{blue}3\\1&2&0&\color{blue}3\\0&1&3&\color{blue}6\\2&2&1&\color{blue}6}=4\cdot \color{blue}3\vm{\color{green}1&3&1&1\\\color{red}1&2&0&1\\\color{red}0&1&3&2\\\color{red}2&2&1&2}
-=12\vm{1&3&1&1\\\color{blue}0&\color{blue}{-1}&\color{blue}{-1}&\color{blue}{0}\\\color{blue}0&\color{blue}1&\color{blue}3&\color{blue}2\\0&-4&-1&-0}
+\\&=12\vm{1&3&1&1\\\color{blue}0&\color{blue}{-1}&\color{blue}{-1}&\color{blue}{0}\\\color{blue}0&\color{blue}1&\color{blue}3&\color{blue}2\\0&-4&-1&-0}
-\\&=\color{blue}{-}12\vm{1&3&1&1\\0&\color{green}1&3&2\\0&\color{red}{-1}&{-1}&{0}\\0&\color{red}{-4}&-1&0}
+=\color{blue}{-}12\vm{1&3&1&1\\0&\color{green}1&3&2\\0&\color{red}{-1}&{-1}&{0}\\0&\color{red}{-4}&-1&0}
-=-12\vm{1&3&1&1\\0&1&3&2\\0&0&\color{green}2&2\\0&0&\color{red}{11}&8}
+\\&=-12\vm{1&3&1&1\\0&1&3&2\\0&0&\color{green}2&2\\0&0&\color{red}{11}&8}
 =-12\vm{1&3&1&1\\0&1&3&2\\0&0&2&2\\0&0&0&-3}
 \\&=-12(1)(1)(2)(-3)=72.
 \end{align*}