A question about the rank of the sum of linear transformations

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SUMMARY

The discussion focuses on proving the inequality rk(τ + σ) ≤ rk(τ) + rk(σ) for linear transformations τ and σ in the vector spaces V and W. The user correctly identifies that the rank of a linear transformation corresponds to the dimension of its image, represented by matrices. The conclusion drawn is that the image of the sum of two linear transformations is a subspace of the sum of their respective images, reinforcing the established relationship between ranks and images in linear algebra.

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  • Understanding of linear transformations and vector spaces
  • Familiarity with the concept of rank in linear algebra
  • Knowledge of matrix representation of linear transformations
  • Basic understanding of subspaces and their properties
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  • Learn about the rank-nullity theorem in linear algebra
  • Explore the concept of image and kernel of linear transformations
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Students and professionals in mathematics, particularly those studying linear algebra, as well as educators looking to clarify concepts related to linear transformations and their ranks.

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Notations:
L(V,W) stands for a linear transformation vector space form vector space V to W.
rk(?) stands for the rank of "?".

Question:
Let τ,σ ∈L(V,W) , show that rk(τ + σ) ≤ rk(τ) + rk(σ).
I want to know wether the way I'm thinking is right or not, or there's a better explanation.
My thought is:
Since every linear tansformation is reprensted by a matrix, also is τ + σ. rk(τ) and rk(σ) roughly means the numbers of basis vectors of their own matrix column spaces, so the combination τ + σ only preserves the distinct basis vectors from the column spaces of the matirces of τ and σ, namely, rk(τ + σ) ≤ rk(τ) + rk(σ) holds.

Thanks for any help.
 
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Show that the im(τ + σ) is a subspace of im(τ) + im(σ). Remember that in general, rk(τ) = dim(im(τ)).

(By im, I mean image.)
 
adriank said:
Show that the im(τ + σ) is a subspace of im(τ) + im(σ). Remember that in general, rk(τ) = dim(im(τ)).

(By im, I mean image.)

Indeed, it's a simpler way to solve the question.

Thanks!
 

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