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I've been learning how to determine and find a basis (which I do understand). For a set to be a basis, it must span the space and be linearly independent. However, I'm not acknowledging why I'm doing this.

Why do we need a basis? And how do I know I need to find a basis? I'm missing the basic concepts in discrete mathematics because I can't seem to connect the dots between all the material I've learned.

Any help would be greatly appreciated!

Killua
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  • Maybe this link can help you https://math.stackexchange.com/q/1817242/798113. –  Mar 14 '21 at 22:26

1 Answers1

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The reason why we care about a basis is because many interesting functions between vector spaces are linear transformations, and linear transformations interact very nicely with bases. Let's say you have two vector spaces $V, W$ and you have a basis $(v_1, v_2, ..., v_n)$ for $V$. Let's also say that you have a linear transformation $T:V \to W$. Consider any element $v$ of the domain. $v$ can be written as $v = c_1 v_1 + \cdots + c_n v_n$. Now $$T(v) = T( c_1 v_1 + \cdots + c_n v_n ) = c_1 T(v_1) + \cdots + c_n T(v_n)$$ The punchline is that if you can evaluate easily $T(v_i)$ for all the basis vectors, then you can easily evaluate $T(v)$ for any vector $v$. Here's an example of where this is useful.

Imagine the vector $(3, 6)$ in $\mathbb{R}^2$. If we rotate this vector by $45^{\circ}$ counterclockwise, where will it land? You might be tempted to draw a complicated diagram and break out your trig functions, but there is a better way to do this.

The function $T: \mathbb{R}^2 \to \mathbb{R}^2$ that rotates vectors by $45^{\circ}$ is a linear transformation (should be somewhat intuitive if you picture it). Now $e_1 = (1, 0)$ and $e_2 = (0, 1)$ form a basis for $\mathbb{R}^2$. And we can easily compute $T(e_1) = (\sqrt{2}/2, \sqrt{2}/2)$ and $T(e_2) = (-\sqrt{2}/2, \sqrt{2}/2)$. Thus $$T(3, 6) = T(3 e_1 + 6e_2) = 3(\sqrt{2}/2, \sqrt{2}/2) + 6(-\sqrt{2}/2, \sqrt{2}/2)$$

$$= (- 3 \sqrt{2}/2, 9 \sqrt{2}/2).$$


Now, you may wonder "why do we need any other base than $e_1$ and $e_2$? Well, if $T$ is a rotation like above, it is easy to calculate $T(e_1)$ and $T(e_2)$. But for other transformations, this may not be so. For example: $T$ is projection on the line $y = \frac13 x$. Then it's much easier to calculate what $T$ does to the basis $\{(3, 1), (-1, 3) \}$ than to $e_1, e_2$.

Ovi
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