Answer: Image Direct Sum & Linear Operator: Is Union Equal?

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SUMMARY

The discussion clarifies that the image of the direct sum of two subspaces under a linear operator is not equal to the union of the images of those subspaces. A counterexample is provided using the vector space $V=\mathbb R^2$ with subspaces $X=\{(x,0):x\in\mathbb R\}$ and $Y=\{(0,y):y\in\mathbb R\}$, demonstrating that $I(X\oplus Y) \neq I(X) \cup I(Y)$ when using the identity operator $I$. The correct statement is that the image of the direct sum is equal to the sum of the images under the operator.

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Fermat1
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Given 2 subspaces and a linear operator, is the image of the direct sum of the subspaces equal to the union of the images under the operator?

Thanks
 
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Fermat said:
Given 2 subspaces and a linear operator, is the image of the direct sum of the subspaces equal to the union of the images under the operator?

Thanks
This is not true.

For a counterexample:
Let $V=\mathbb R^2$ and $X=\{(x,0):x\in\mathbb R\}$ and $Y=\{(0,y):y\in\mathbb R\}$.

Then $X$ and $Y$ are subspaces of $V$.

Let $I$ be the identity operator on $V$.

You can see that $I(X\oplus Y)\neq I(X)\cup I(Y)$.To make your statement true you can have:
Given 2 subspaces of a vector space $V$ and a linear operator on $V$, the image of the direct sum of the subspaces is equal to the sum of the images under the operator.
 

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