[tex]\ Let T: V \rightarrow W [/tex] be a linear transformation, let [tex]b \in W [/tex]be a fixed vector, and let [tex]x_0 \in V [/tex] be a fixed solution of(adsbygoogle = window.adsbygoogle || []).push({});

[tex]T(x)=b.[/tex] Prove that a vector [tex]x_1 \in V [/tex]is a solution of [tex] T(x)=b,[/tex] if and only if [tex] x_1 [/tex]is of the form [tex]x_1=x_h +x_0 [/tex]where [tex]x_h \in kerT[/tex]

I started out by saying that

[tex] x_i \in X_i[/tex]

[tex](x_1... x_n) \in \prod [/tex] (where i=1 and h is at the top) [tex]X_0[/tex]

[tex](x_1... x_n) \in \prod X_i[/tex]

[tex] x_i \in X_i \rightarrow x_1 [/tex] is not equal to the empty set for all i.

I am not sure if I am doing this right. I'd appreciate any feedback.

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# Linear Transformation Proof

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