# Finding orthonormal basis of subspace spanned by two functions

From S.L Linear Algebra:

Let $$V$$ be the subspace of functions generated by the two functions $$f$$, $$g$$ such that $$f(t)=t$$ and $$g(t)=t^2$$. Find an orthonormal basis for $$V$$.

In this case, $$V$$ is considered as space of continuous real-valued functions on the interval $$[0, 1]$$. Scalar product of two functions $$f$$ and $$g$$ is defined as:

$$⟨f, g⟩=\int_{0}^{1} f(t)g(t) \, \, dt$$

where $$t \in [0, 1]$$.

I have showed that the integral above is a valid scalar product by displaying it as Riemann sum approximation and verifying properties according to it (but this is not important here).

# Problem:

I know that I could use Gram-Schmidt orthogonalization process to ensure that two elements $$f(t)=t$$ and $$g(t)=t^2$$ mutually orthogonal as such:

$$t - \frac{\int_{0}^{1} t^3 \, \, dt}{\int_{0}^{1} t^2 \, \, dt}t = t - \frac{\frac{1^4}{4} - \frac{0^4}{4}}{\frac{1^3}{3}-\frac{0^3}{3}}t=t-\frac{3}{4}t=\frac{t}{4}$$

And then divide them by norm to ensure orthonormality:

$$s=(\frac{t}{\sqrt{\int_{0}^{1} t^2 \, \, dt}}, \frac{\frac{t}{4}}{\sqrt{\int_{0}^{1} \frac{t}{4}^2 \, \, dt}})$$

(I apologize if I made any mistakes in calculations).

But now the problem is to decide whether every element in $$s$$ is linearly independent, if we consider homogeneous equation:

$$c_1\frac{t}{\sqrt{\int_{0}^{1} t^2 \, \, dt}}+c_2\frac{\frac{t}{4}}{\sqrt{\int_{0}^{1} \frac{t}{4}^2 \, \, dt}}=0$$

And $$s$$ is a valid orthonormal basis for $$V$$, then $$c_1=c_2=0$$. But I don't know how to ensure linear independence between two vectors.

How can I finish the proof by showing linear independence?

• Note, that if you have an orthogonal set of vectors, then they are also linear independent. – blub Apr 5 at 21:30
• @blub Wow, didn't think it would be this easy. Thank you! – ShellRox Apr 5 at 21:41
• I've added an answer to elaborate on why this is true. – blub Apr 5 at 22:01

To give a little explanation on my comment: I'm doing this in a toy setting and you can easily extend this further. Let $$(V,\langle\cdot,\cdot\rangle)$$ be a real inner product space with $$\mathrm{dim}(V)\geq 2$$.

Take $$v,w\in V\setminus\{\mathbf 0\}$$ (as otherwise they are clearly not linear independent) and suppose that they are orthogonal, i.e. that $$\langle v,w\rangle=0$$. Now, suppose $$\alpha v+\beta w=\mathbf 0$$. We want to show that then $$\alpha,\beta=0$$.

To see this, first consider that $$\langle \alpha v+\beta w,v\rangle=0$$ as $$\alpha v+\beta w$$ is the null vector. By linearity in the first argument, symmetry and orthogonality, we have

$$0=\alpha\langle v,v\rangle+\beta\langle w,v\rangle=\alpha\langle v,v\rangle+\beta\langle v,w\rangle=\alpha\langle v,v\rangle$$

As $$v$$ is not the null vector, $$\langle v,v\rangle>0$$ by positive definiteness and thus $$\alpha=0$$.

The same argument also leads us to $$\beta=0$$ by considering $$\langle\alpha v+\beta w,w\rangle$$.

Thus, the vectors $$v,w$$ are linearly independent.