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Linear Transformation Question 
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#1
Sep1911, 08:39 PM

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Question about linear transformations if you have a matrix such as
 5 6 9   5 0 3   9 3 7  Can it be a matrix transformation? Or does it have to follow the identity matrix? Can be a transformation and the "y" transformation being just makes the it flat on the y axis? or does it have to be a form of the identity matrix? Or am I totally misunderstanding this? 


#2
Sep1911, 08:57 PM

P: 3

A matrix is a linear transformation expressed with respect to a basis for the source space and the target space.
Given a linear transformation [itex]T:\mathbb{F}^n \to \mathbb{F}^m[/itex], the corresponding matrix written with respect to a basis [itex]\alpha[/itex] for the source space and a basis [itex]\beta[/itex] for the target space is as follows: [itex] \left[ \begin{array}{cccc} [T(\alpha_1)]_\beta & [T(\alpha_2)]_\beta & ... & [T(\alpha_n)]_\beta \end{array} \right] [/itex] The theory behind this is as follows. Since any vector in a given vector space can be expressed as a linear combination of a set of basis vectors for that vector space, we need only transform an arbitrary basis to capture the transformation. Given some vector [itex]v \in \mathbb{F}^n[/itex] and a basis [itex]\alpha[/itex] we can write [itex]v = a_1\alpha_1 + a_2\alpha_2 + ... + a_n\alpha_n[/itex]. Then [itex]v[/itex] transformed is as follows [itex] \begin{eqnarray*} T(v) &=& T(a_1\alpha_1 + a_2\alpha_2 + ... + a_n\alpha_n) \\ &=& T(a_1\alpha_1) + T(a_2\alpha_2) + ... + T(a_n\alpha_n) \\ &=& a_1T(\alpha_1) + a_2T(\alpha_2) + ... + a_nT(\alpha_n) \end{eqnarray*} [/itex] 


#3
Sep1911, 10:36 PM

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#4
Sep2011, 10:26 AM

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Linear Transformation Question
Any m by n matrix is a linear transformation from [itex]R^m[/itex] to [itex]R^n[/itex].
What you have given is a perfectly good linear transformation although the way you have written it, with the "straight" vertical sides, makes it look more like a determinant than a matrix! The matrix you give represents the linear transformation that maps a vector, [itex]a\vec{i}+ b\vec{j}+ c\vec{k}[itex] into [itex]a(5\vec{i}+ 5\vec{j}+ 9\vec{k})+ b(6\vec{i} 3\vec{k})+ c(9\vec{i}+ 3\vec{j} 7\vec{kl})[/itex][itex]= (5a+ 6b+ 9c)\vec{i}+ (5a+ 3b)\vec{i}+ (9a 3b 7c)\vec{k}[/itex]. I wonder if you aren't confusing "matrix", in general, with "invertible matrix". 


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