I have a non-singular matrix $K$ where $k_{ij} = k(x_i, x_j)$, and two vectors $\boldsymbol{\alpha} = \{\alpha_1,...,\alpha_m\}$ and $\boldsymbol{\beta} = \{\beta_1,...,\beta_m\}$.

Since $K$ is non-singular, there exists $K^{-1}$ such that:

$$ K^{-1} \times K = I $$

If I have an equation:

$$ K \boldsymbol{\alpha} = K \boldsymbol{\beta} $$

The fact that $K$ is non-singular implies that $\boldsymbol{\alpha} = \boldsymbol{\beta}$ and $\alpha_i = \beta_i$ (since I can multiply both sides of the equation by $K^{-1}$).

I understand everything so far, but I am having problems trying to figure out how I would write this element-wise. If I have an equation:

$$ \sum_i \alpha_i k(x_i, x_j) = \sum_i \beta_i k(x_i, x_j) $$

how can I show that $\alpha_i = \beta_i$ without converting it into matrix form?

Edited to make the question more generic.


The equation you have given,

$$K\alpha = K\beta$$

for an invertible $K$ and $\alpha, \beta \in \mathbb{R}^n$ can be expressed as a set of $n$ scalar equations. Letting $i,j$ denote a specific row-column index of $K$, these equations are,

$$\sum_j K_{i,j} \alpha_j = \sum_j K_{i,j} \beta_j,\ \ \ \text{for}\ \ i = 1, 2, \ldots, n$$

Or in less compact notation,

$$\sum_j K_{1,j} \alpha_j = \sum_j K_{1,j} \beta_j$$

$$\sum_j K_{2,j} \alpha_j = \sum_j K_{2,j} \beta_j$$


$$\sum_j K_{n,j} \alpha_j = \sum_j K_{n,j} \beta_j$$

If you wrote out all of these equations, you would notice that right and left hand sides of every single one of them match except that $\alpha_j$'s are on the left while $\beta_j$'s are on the right, so by simple inspection $\alpha = \beta$ is a solution, and it is the only solution because none of the equations are linearly dependent.

Alternatively, if you really wanted to, you could solve the system by substitution, and you'd quickly find that as you group like-terms, the coefficients will start canceling on both sides of each equation and leave you explicitly with $\alpha_j = \beta_j$ in all $n$ equations.

If you just had one equation with $n$ unknowns,

$$\sum_j K_{1,j} \alpha_j = \sum_j K_{1,j} \beta_j$$

there would be an infinite number of solutions. But this is only one of the $n$ equations that $K\alpha = K\beta$ compactly represents.

Hopefully that clears things up. Linear algebra is super cool. Good luck!


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