Minimising a linear function with a regularizer

Question

I have the following optimization problem.

$$\underset{w}{\text{minimize}} \ w^\intercal x + \lambda \|w\|^2$$

I am told to solution is:

$$w = -2 \lambda x$$

Why?

Also scribbled on the paper is

$$x + \lambda w = 0$$

Answer 1

Just differentiate in each component of $w$. You get:

$$x_i+2w_i=0.$$

From which it follows that $w_i=-x_i/2$. Now verify this is a minimum

Answer 2

If you are looking for the value of $w$ which minimizes this expression, there is a general result that says that the quadratic expression (with $A$ definite positive) :

$X^TAX+X^TB+c$ is minimized when its gradient is 0, i.e.,

$2AX+B=0$ which is equivalent to say that $X$ is the unique solution to $AX=-\frac{1}{2}B$.

Here $A=I$, $B=w$, thus the answer is $X=-\frac{1}{2}w$.

Answer 3

A huge thank you to @JeanMarie and @Alex R for getting me 99% of the way there.

The solution is:

$$w= \frac{x}{- 2 \lambda} $$

The $ \lambda $, at least for my problem,is very important.