MHB How Do You Prove Two Spans Are Equal?

[Dethrone]

I just want to make sure I'm understand this. If I want to show to spans are equal, say $\text{span}(X)=\text{span}(Y)$, then I think I read that if we can represent all the elements of $X$ as a linear combination of the elements of $Y$, then $\text{span}(X) \subseteq \text{span}(Y)$ and if we can show all the elements of $Y$ can be represented as a linear combination of the elements of $X$, then $\text{span}(Y) \subseteq \text{span}(X)$. Therefore, the spans are equal.

Now, this is my understanding on why that is true, can you let me know if it is right? If we can show that all the elements of $X$ can be represented as a linear combination of the elements of $Y$, that would mean that $X_{i}\in \text{span}(Y)$, where $X_i$ is an element of $X$. Furthermore, all linear combinations of $X_i$ are also in $\text{span}(Y)$, since the elements themselves are linear combinations of the elements in $Y$. Thus, $\text{span}(X) \subseteq \text{span}(Y)$. The same reasoning holds for the other way around...
Is this correct, and is there any better way to think about this? It hurts my brain :(

 

[Nono713]

Maybe an easier way to imagine it is that if you can express each vector in $X$ as a linear combination of vectors in $Y$, then any vector that can be expressed as a linear combination of $X$ can be expressed as a linear combination of $Y$, by simply expanding out the $x_i$'s as their representation as a linear combination of elements of $Y$, and collecting terms. So any vector that is in the span of $X$ is automatically in the span of $Y$.

For instance suppose that you have $X = \{ x_1, x_2, x_3 \}$ and $Y = \{ y_1, y_2 \}$. Furthermore suppose that:
$$x_1 = 2y_1 + y_2$$
$$x_2 = y_1 + (-3)y_2$$
$$x_3 = 2y_1 + 2y_2$$
Now let $u$ be any vector in the span of $X$, so that:
$$u = c_1 x_1 + c_2 x_2 + c_3 x_3$$
Then we expand out to get:
$$u = c_1 (2y_1 + y_2) + c_2(y_1 + (-3)y_2) + c_3 (2y_1 + 2y_2)$$
$$u = 2 c_1 y_1 + c_1 y_2 + c_2 y_1 + (-3) c_2 y_2 + 2 c_3 y_1 + 2 c_3 y_2$$
$$u = (2 c_1 + c_2 + 2 c_3) y_1 + ( c_1 - 3c_2 + 2 c_3) y_2$$
And so $u$ can be expressed as a linear combination of $y_1$ and $y_2$ and so is in the span of $Y$.

From there your reasoning on expressing $X$ in terms of $Y$ and $Y$ in terms of $X$ to show that their spans are equal is perfectly valid. Of course, if you think on it a bit you will find that $X$ and $Y$ can be linearly independent without loss of generality, and so for their spans to be equal there must be the same number of (linearly independent) vectors in $X$ and $Y$ anyway, which leads you straight to the idea of a linear transformation between $X$ and $Y$ representable as a square invertible matrix $A$, from which you can establish a bijection between the coefficients in the linear combinations of vectors in either span... or something like that. I need to go now, I might add more details later if someone else hasn't.​

 

[Dethrone]

Thanks Bacterius for the well-written answer! Your reasoning makes complete sense. I was worried that I might have had the wrong idea, but it seems that my ideas conform with yours, which is a great feeling. :D