Jordan Canonical Form - Revision history

Kfagerstrom at 16:03, 26 April 2023

2023-04-26T16:03:53Z

Kfagerstrom at 16:08, 14 April 2023

2023-04-14T16:08:12Z

Tboudreaux at 18:08, 4 April 2023

2023-04-04T18:08:09Z

Tboudreaux at 18:06, 4 April 2023

2023-04-04T18:06:36Z

Kfagerstrom at 18:43, 26 March 2023

2023-03-26T18:43:22Z

Kfagerstrom: Created page with "$F$ field, $V$ finite dim vector space, $V\xrightarrow{t} V$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so th..."

2023-03-25T22:26:55Z

Created page with "$F$ field, $V$ finite dim vector space, $V\xrightarrow{t} V$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so th..."

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$F$ field, $V$ finite dim vector space, $V\xrightarrow{t} V$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so things that look like $(x-r)^l$). Then there exists a basis $B$ such that $[t]_B^B=\begin{bmatrix} j_{e_1}(x_1)&\\ \; \ddots & \hspace{-12pt} j_{e_n}(x_n) \end{bmatrix}=j(t)$.

Each $r_i\in F$ is a root of $c_t$, the characteristic polynomial, and $e_i\geq 1$. The polynomials $(x-r_i)^{e_i}$ are the elementary divisors of $t$, and this jordan canonical form for $t$ is unique up to order of the blocks

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-$$ F $$ field, $$V$$ finite dim vector space, $$V\xrightarrow{t} V$$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so things that look like $(x-r)^l$). Then there exists a basis $$B$$ such that $$[t]_B^B=\begin{bmatrix} j_{e_1}(x_1)&\\ \; \ddots & \hspace{-12pt} j_{e_n}(x_n) \end{bmatrix}=j(t)$$.
+$$ F $$ field, $$V$$ finite dim vector space, $$V\xrightarrow{t} V$$ linear transformation. Assume that the characteristic polynomial of $$t$$ completely factor into linear forms (so things that look like $$(x-r)^l$$). Then there exists a basis $$B$$ such that $$[t]_B^B=\begin{bmatrix} j_{e_1}(x_1)&\\ \; \ddots & \hspace{-12pt} j_{e_n}(x_n) \end{bmatrix}=j(t)$$.
 Where each Jordan block $$J_n(r)$$ is an $$n\times n$$ matrix with entries $$a_{ij}=\begin{cases} r, & i=j \\ 1, & j=i+1 \\ 0 & \text{else}  \end{cases}$$
-Each $$r_i\in F$$ is a root of $$c_t$$, the characteristic polynomial, and $$e_i\geq 1$$. The polynomials $$(x-r_i)^{e_i}$$ are the elementary divisors of $$t$$, and this jordan canonical form for $t$ is unique up to order of the blocks
+Each $$r_i\in F$$ is a root of $$c_t$$, the characteristic polynomial, and $$e_i\geq 1$$. The polynomials $$(x-r_i)^{e_i}$$ are the elementary divisors of $$t$$, and this jordan canonical form for $$t$$ is unique up to order of the blocks

@@ Line 1: / Line 1: @@
-$$ F $$ field, $$V$$ finite dim vector space, $V\xrightarrow{t} V$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so things that look like $(x-r)^l$). Then there exists a basis $B$ such that $[t]_B^B=\begin{bmatrix} j_{e_1}(x_1)&\\ \; \ddots & \hspace{-12pt} j_{e_n}(x_n) \end{bmatrix}=j(t)$.
+$$ F $$ field, $$V$$ finite dim vector space, $$V\xrightarrow{t} V$$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so things that look like $(x-r)^l$). Then there exists a basis $$B$$ such that $$[t]_B^B=\begin{bmatrix} j_{e_1}(x_1)&\\ \; \ddots & \hspace{-12pt} j_{e_n}(x_n) \end{bmatrix}=j(t)$$.
-Where each Jordan block $J_n(r) is an $n\times n$ matrix with entries $a_{ij}=\begin{cases} r, & i=j \\ 1, & j=i+1 \\ 0 & \text{else}  \end{cases}
+Where each Jordan block $$J_n(r)$$ is an $$n\times n$$ matrix with entries $$a_{ij}=\begin{cases} r, & i=j \\ 1, & j=i+1 \\ 0 & \text{else}  \end{cases}$$
-Each $r_i\in F$ is a root of $c_t$, the characteristic polynomial, and $e_i\geq 1$. The polynomials $(x-r_i)^{e_i}$ are the elementary divisors of $t$, and this jordan canonical form for $t$ is unique up to order of the blocks
+Each $$r_i\in F$$ is a root of $$c_t$$, the characteristic polynomial, and $$e_i\geq 1$$. The polynomials $$(x-r_i)^{e_i}$$ are the elementary divisors of $$t$$, and this jordan canonical form for $t$ is unique up to order of the blocks

@@ Line 1: / Line 1: @@
-$$ F $$ field, $V$ finite dim vector space, $V\xrightarrow{t} V$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so things that look like $(x-r)^l$). Then there exists a basis $B$ such that $[t]_B^B=\begin{bmatrix} j_{e_1}(x_1)&\\ \; \ddots & \hspace{-12pt} j_{e_n}(x_n) \end{bmatrix}=j(t)$.
+$$ F $$ field, $$V$$ finite dim vector space, $V\xrightarrow{t} V$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so things that look like $(x-r)^l$). Then there exists a basis $B$ such that $[t]_B^B=\begin{bmatrix} j_{e_1}(x_1)&\\ \; \ddots & \hspace{-12pt} j_{e_n}(x_n) \end{bmatrix}=j(t)$.
 Where each Jordan block $J_n(r) is an $n\times n$ matrix with entries $a_{ij}=\begin{cases} r, & i=j \\ 1, & j=i+1 \\ 0 & \text{else}  \end{cases}
 Each $r_i\in F$ is a root of $c_t$, the characteristic polynomial, and $e_i\geq 1$. The polynomials $(x-r_i)^{e_i}$ are the elementary divisors of $t$, and this jordan canonical form for $t$ is unique up to order of the blocks

@@ Line 1: / Line 1: @@
-$F$ field, $V$ finite dim vector space, $V\xrightarrow{t} V$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so things that look like $(x-r)^l$). Then there exists a basis $B$ such that $[t]_B^B=\begin{bmatrix} j_{e_1}(x_1)&\\ \; \ddots & \hspace{-12pt} j_{e_n}(x_n) \end{bmatrix}=j(t)$.
+$$ F $$ field, $V$ finite dim vector space, $V\xrightarrow{t} V$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so things that look like $(x-r)^l$). Then there exists a basis $B$ such that $[t]_B^B=\begin{bmatrix} j_{e_1}(x_1)&\\ \; \ddots & \hspace{-12pt} j_{e_n}(x_n) \end{bmatrix}=j(t)$.
 Where each Jordan block $J_n(r) is an $n\times n$ matrix with entries $a_{ij}=\begin{cases} r, & i=j \\ 1, & j=i+1 \\ 0 & \text{else}  \end{cases}
 Each $r_i\in F$ is a root of $c_t$, the characteristic polynomial, and $e_i\geq 1$. The polynomials $(x-r_i)^{e_i}$ are the elementary divisors of $t$, and this jordan canonical form for $t$ is unique up to order of the blocks

← Older revision		Revision as of 18:43, 26 March 2023
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	$F$ field, $V$ finite dim vector space, $V\xrightarrow{t} V$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so things that look like $(x-r)^l$). Then there exists a basis $B$ such that $[t]_B^B=\begin{bmatrix} j_{e_1}(x_1)&\\ \; \ddots & \hspace{-12pt} j_{e_n}(x_n) \end{bmatrix}=j(t)$.		$F$ field, $V$ finite dim vector space, $V\xrightarrow{t} V$ linear transforamtion. Assume that the charateristic polynomial of $t$ compleately factor into linear forms (so things that look like $(x-r)^l$). Then there exists a basis $B$ such that $[t]_B^B=\begin{bmatrix} j_{e_1}(x_1)&\\ \; \ddots & \hspace{-12pt} j_{e_n}(x_n) \end{bmatrix}=j(t)$.
		+
		+	Where each Jordan block $J_n(r) is an $n\times n$ matrix with entries $a_{ij}=\begin{cases} r, & i=j \\ 1, & j=i+1 \\ 0 & \text{else} \end{cases}

	Each $r_i\in F$ is a root of $c_t$, the characteristic polynomial, and $e_i\geq 1$. The polynomials $(x-r_i)^{e_i}$ are the elementary divisors of $t$, and this jordan canonical form for $t$ is unique up to order of the blocks		Each $r_i\in F$ is a root of $c_t$, the characteristic polynomial, and $e_i\geq 1$. The polynomials $(x-r_i)^{e_i}$ are the elementary divisors of $t$, and this jordan canonical form for $t$ is unique up to order of the blocks