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I am taking a class in Algebra but I am having a problem grasping exactly what it is I am being asked to do -- I think I am having a problem with the vocabulary being used. I have a couple of questions below and I would appreciate it, if you could take time to explain(in layman's terms or maybe programmer terms, exactly what is being asked. thanks in advance.

1) Let $[e_1; e_2; e_3]$ be the standard basis vectors in $\mathbb{R}^{3}$ and consider the ordered basis: $[e_2; e_1; e_3 + e_1]$ Verify that this is actually a basis and find the coordinates of the vector $(1; 1; 1)^T$ with respect to that basis.

2) Let $T$ be the linear map from $\mathbb{R}^{2}$ to $\mathbb{R}$ defined by: $T ((x, y)^T ) = x-y$ Find its matrix (with respect to the standard bases) and a basis for its kernel

3) Find a spanning vector set for the image of the linear map from $\mathbb{R}^{2}$ to $\mathbb{R}^{3}$ defined by: $T ((x, y)^T) = (2x-y, x + y, y)$

4) Consider the subspace U of $\mathbb{R}^3$ defined by: $ U = \{(x; y; z)^T: 2x- y + z = 0; x + y = 0 \} $. Express U as the kernel of an appropriately defined linear map and the matrix of that map with respect to the standard bases of the corresponding $\mathbb{R}^n$'s

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    @user10695: Neither the map in (2) nor in (3) is a linear map. Did you copy them correctly, or are you missing a $+$ plus sign somewhere?2011-06-22
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    @user10695: The problem isn't the use of commas versus semicolons. The problem is that mapping $(x,y)^T$ to $xy$ is not linear: the image of $(1,1)^T$ is $1$, but the image of $(1,1)^T+(1,1)^T = (2,2)^T$ is not the sum of the images.2011-06-22
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    ^^^ Now the above is part of what I would like to have broken down a bit more. :) . I made changes, does it work now? Could you also please explain yourself a bit more please?2011-06-22
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    @user10695: Yes, mapping $(x,y)$ to $x-y$ *is* linear. The reason the previous one was not linear is that, remember, if $T$ is linear, then $T(\mathbf{v}+\mathbf{w}) = T(\mathbf{v}) + T(\mathbf{w})$ and $T(\alpha\mathbf{v}) = \alpha T(\mathbf{v})$ for all vectors $\mathbf{v}$ and $\mathbf{w}$, and all scalars $\alpha$. To see $T(x,y) = xy$ is not linear, take $\mathbf{x}=(1,1)$, $\mathbf{y}=(1,1)$, and compare $T(\mathbf{x}+\mathbf{y})$ with $T(\mathbf{x}) + T(\mathbf{y})$. You have:$$T((1,1)+(1,1))=T(2,2)=4\neq 1+1 = T(1,1)+T(1,1).$$Not equal, not linear.2011-06-22
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    I think I understand now. Thanks2011-06-22

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