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Find bases and dimensions of the following subspaces:

$\{(x, y, z) | x = y + 2z\}$ in $\mathbb{R}^3$ over $\mathbb{R}$

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    @TheGeekGreek Same context, different question about the context. In this, it is already assumed that the student knows that it is a subspace and is being asked about further details about it. In the linked question, that assumption is what is being asked to be proven. – 2017-01-26
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    How many linearly independent vectors are needed to form a basis for that subspace? (I might add that as a linear system, you should be able to find particularly simple answers. This is a basic question; you just need to understand the definition.) – 2017-01-26

3 Answers 3

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$$x = y + 2z\Leftrightarrow \begin{cases}x=\alpha+2\beta\\y=\alpha\\z=\beta \end{cases}\;(\alpha,\beta\in\mathbb{R})$$ $$\Leftrightarrow \begin{pmatrix}{x}\\{y}\\{z}\end{pmatrix}=\alpha\underbrace{\begin{pmatrix}{1}\\{1}\\{0}\end{pmatrix}}_{v_1}+\beta\underbrace{\begin{pmatrix}{2}\\{0}\\{1}\end{pmatrix}}_{v_2}\;(\alpha,\beta\in\mathbb{R}).$$ $B=\{v_1,v_2\}$ spans the given subspace and is clearly linearly independent so, $B$ is also a basis.

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Well it has dimension $2$ since it has two free variables. Do you know how to go about constructing a basis for a space? Where in that do you get confused?

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    {(x, y, z) | (y+2z, y, z) } and now i can see that that this vector can be done i terms of two variables ? – 2017-01-26
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Hint: The linear map \begin{align} f\colon \mathbf R^2&\longrightarrow \mathbf R^3 \\ (y,z)&\longmapsto (y+2z, y,z) \end{align} induces an isomorphism from $\mathbf R^2$ onto the mentioned subspace.

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    y[1,1,0] z[2,0,1] is the base ? – 2017-01-26
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    Almost. What do $y$ and $z$ do here? B. t. w., welcome to TeX SX! – 2017-01-26
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    respresents the vector z and y ? – 2017-01-26
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    $z$ and $y$ are scalars, not vectors. You're almost at it! – 2017-01-26
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    Ok guy i think it will be right that, You firstly have to make an basis which is constructed from y+2z,y,z so span {v1,v2} = {x,y,z | x=y+2z} q1v1+q2v2=0 is in V=(0,0,0) then (q1,q1,0)(2q2,0,q2) q1+2q2=0 q1=0 q2=0 Anybody else, what can i do more? Is it good explenation? – 2017-01-26
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    But what is the proposed basis? – 2017-01-26
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    [0 1 0] [0 0 1] ? – 2017-01-26
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    No. Rewrite $(y+2z,y,z)$ as $y(?,?,?)+z(?, ?,?)$. – 2017-01-26
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    y(1,1,0) + z(2,0,1) ? – 2017-01-26
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    Yes. So what's the basis? – 2017-01-26
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    So what exactly is a base. – 2017-01-26
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    A family of vectors such that every vector in the (sub)space is a unique linear combination of the vectors in the family. Note that an isomorphism maps a basis of the domain to a basis of the image. – 2017-01-26
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    How am i suposed to find it? Isn't [1.1.0] and [2.0.1] unique? – 2017-01-26
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    $\{(1,1,0),(2,0,1)\} $ *is* the seeked basis (it is the image of the basis $\{(1,0), (0,1)\}$ of $\mathbf R^2$). – 2017-01-26
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    {(1,0.0),(0,1,0),(0,0,1)} of R^3 ? – 2017-01-26
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    Did you read my previous comment? – 2017-01-26
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    span (1,1,0) and span(2,0,1) ? – 2017-01-26
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    A span is not a basis, it's a subspace. – 2017-01-26
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    v1 [1,1,0] v2 [2,0,1] ? – 2017-01-26
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    If you mean v1=…, v2=…, yes. – 2017-01-26