Let $B=\{v_1,v_2,\dots,v_n\}$ is an orthonormal basis. I'm given that $v$ can be expressed as $v=\sum_{j=1}^n\lambda_jv_j $.
My task is to then show that $v=\sum_{i=1}^n \langle v_i,v\rangle v_i$.
So far I have said the following:
Taking the inner product with $v_i$ gives $$\langle v_i,v \rangle=\left\langle v_i, \sum_{j=1}^n\lambda_j v_j\right\rangle = \sum_{j=1}^n\lambda_j \langle v_i, v_j \rangle = \sum_{j=1}^n\lambda_j \delta_{ij}$$
The solution then says $$\sum_{j=1}^n\lambda_j \delta_{ij}=\lambda_i$$
Could someone explain why this is the case?