Weights in the linear combination:

If $\{u_{1}, u_{2}, ... , u_{p}\}$ be an orthogonal basis for a subspace $W$ of $R^{n}$ , then for each $y$ in $W$, the weights for the linear combination $y = c_{1}.u_{1} + ⋯ + c_{p}.u_{p}$ can be defined as $$c_{j} = \frac{y.u_{j}}{u_{j}.u_{j}} \qquad j=1,2, ... ,p$$

I understood the geometrical interpretation of the above formula, but I want to know how to derive the above equation.

  • 3
    $\begingroup$ You can derive that formula using “Fourier’s trick” : starting from the formula for $y$, take the dot product of both sides with $u_j$, and observe the wonderful simplification that occurs. $\endgroup$
    – littleO
    Jun 3, 2022 at 4:39
  • 1
    $\begingroup$ That was quite simple, @littleO . Thank you! $\endgroup$ Jun 3, 2022 at 4:47

1 Answer 1


Since $y\in W$ and $\{u_{1}, u_{2}, ... , u_{p}\}$ form a basis for $W$, it is possible to write $y$ as a linear combination of $u_{1}, u_{2}, ... , u_{p}$: $$ y = c_{1}u_{1} + ⋯ + c_{p}u_{p} = \sum_{i=1}^{p}c_{i}u_{p} $$ Take dot product on both sides with respect to $u_j$: $$ \begin{align} y\cdot u_j &= \left[ \sum_{i=1}^{p}c_{i}u_{p}\right] \cdot u_{j}\\ &= \sum_{i=1}^{p}c_{i}u_{i}\cdot u_{j} \end{align} $$ Since $\{u_{1}, u_{2}, ... , u_{p}\}$ are also orthogonal, $u_{i}\cdot u_{j} = 0$ when $i\neq j$ and $$ y\cdot u_j = c_{j}u_{j}\cdot u_{j}\\ \Rightarrow c_{j} = \dfrac{y\cdot u_j}{u_{j}\cdot u_{j}} $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .