A question about the rank of the sum of linear transformations

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The discussion focuses on proving that the rank of the sum of two linear transformations, τ and σ, is less than or equal to the sum of their individual ranks, expressed as rk(τ + σ) ≤ rk(τ) + rk(σ). The reasoning provided highlights that each linear transformation can be represented by a matrix, with the ranks corresponding to the number of basis vectors in their column spaces. It is noted that the image of the sum of the transformations is a subspace of the sum of their images, supporting the rank inequality. The conclusion emphasizes that understanding the dimensions of the images is key to establishing the relationship between the ranks. Overall, the explanation is deemed simpler and effective for solving the question.
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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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