Question

Linear transformation matrix representation with differentiation answer confirmation

score 604 · Answer 1 · 2014-06-25 04:04:51

604

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Linear transformation matrix representation with differentiation answer confirmation

Published on 25 Jun 2014 - 4:04

score 111 · Answer 2 · 2026-03-28 05:00:49

111

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Diagonalizable operator of a finite vector space

Published on 28 Mar 2026 - 5:00

score 81 · Answer 3 · 2014-06-26 06:07:49

$V$ and $W$ finite vector spaces with dimension $n$ and $r$ with $\{ v_1,v_2,\ldots,v_n \} \subset \ker T$ and $\{ u_1,u_2,...,u_s \} \subset V$

score 285 · Answer 4 · 2014-06-26 08:28:34

285

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$\{ v_1,v_2,...,v_n\}$ is basis of $V$ if and only if $\{ v_1,v_1 + v_2,...,v_1 + v_2+...+v_n,\}$ is a basis of $V$

Published on 26 Jun 2014 - 8:28

score 61 · Answer 5 · 2014-06-26 09:43:41

61

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Showing a set is a basis

Published on 26 Jun 2014 - 9:43

score 456 · Answer 6 · 2026-03-26 01:09:10

456

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How does the z transform work in practice?

Published on 26 Mar 2026 - 1:09

score 80 · Answer 7 · 2014-06-26 18:58:55

80

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Matrix transformations

Published on 26 Jun 2014 - 18:58

score 101 · Answer 8 · 2014-06-27 08:26:38

101

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Matrices as linear transformations

Published on 27 Jun 2014 - 8:26

score 190 · Answer 9 · 2014-06-27 08:39:03

190

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If $f\in V$ of degree $n$ then for every $g \in P_n(\Bbb R)$ there exist scalars $c_0,c_1,..,c_n$ such that $g = c_0f + c_1f'+ ... + c_nf^{(n)}$

Published on 27 Jun 2014 - 8:39

score 84 · Answer 10 · 2014-06-27 09:11:06

84

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$T (x_1,x_2,x_3,...,x_n) = (-x_3,x_3,x_4,x_5,...) $ then $ W \ne ker T$

Published on 27 Jun 2014 - 9:11

score 1.5k · Answer 11 · 2026-03-25 14:21:29

1.5k

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What are Basis images?

Published on 25 Mar 2026 - 14:21

score 3.8k · Answer 12 · 2014-06-27 19:27:41

3.8k

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Isomorphism implies direct sum of Kernel and Image

Published on 27 Jun 2014 - 19:27

score 353 · Answer 13 · 2014-06-27 21:55:35

353

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what's a homogeneous transformation?

Published on 27 Jun 2014 - 21:55

score 85 · Answer 14 · 2014-06-28 17:26:11

85

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$\ker S$ is not contained in $\ker T$ implies $\dim \Im T \ge 1$

Published on 28 Jun 2014 - 17:26

score 3k · Answer 15 · 2014-06-29 02:29:00

3k

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$V = \operatorname{Im} T + \ker T $ then $ \operatorname{Im} T \cap \ker T = \{0\}$

Published on 29 Jun 2014 - 2:29

List Question

Linear transformation matrix representation with differentiation answer confirmation

Diagonalizable operator of a finite vector space

$V$ and $W$ finite vector spaces with dimension $n$ and $r$ with $\{ v_1,v_2,\ldots,v_n \} \subset \ker T$ and $\{ u_1,u_2,...,u_s \} \subset V$

$\{ v_1,v_2,...,v_n\}$ is basis of $V$ if and only if $\{ v_1,v_1 + v_2,...,v_1 + v_2+...+v_n,\}$ is a basis of $V$

Showing a set is a basis

How does the z transform work in practice?

Matrix transformations

Matrices as linear transformations

If $f\in V$ of degree $n$ then for every $g \in P_n(\Bbb R)$ there exist scalars $c_0,c_1,..,c_n$ such that $g = c_0f + c_1f'+ ... + c_nf^{(n)}$

$T (x_1,x_2,x_3,...,x_n) = (-x_3,x_3,x_4,x_5,...) $ then $ W \ne ker T$

What are Basis images?

Isomorphism implies direct sum of Kernel and Image

what's a homogeneous transformation?

$\ker S$ is not contained in $\ker T$ implies $\dim \Im T \ge 1$

$V = \operatorname{Im} T + \ker T $ then $ \operatorname{Im} T \cap \ker T = \{0\}$

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