310 CHAPTER 13. MATRICES AND THE INNER PRODUCT

• • AFnAu

y

Ax

Theorem 13.3.3 Let y ∈ Fm and let A be an m×n matrix. Then there exists x ∈ Fn mini-mizing the function x 7→ |y−Ax|2 . Furthermore, x minimizes this function if and only if

((y−Ax) ,Au) = 0

for all u ∈ Fn.

Proof: First consider the characterization of the minimizer. Let u ∈ Fn. Let |θ |= 1,

θ̄ (y−Ax,Au) = |(y−Ax,Au)|

Now consider the function of t ∈ R

p(t)≡ |y−(Ax+tθAu)|2 = ((y−Ax)− tθAu,(y−Ax)− tθAu)

= |y−Ax|2 + t2 |Au|2−2t Re(y−Ax,θAu)≥ |y−Ax|2

= |y−Ax|2 + t2 |Au|2−2t Re θ̄ (y−Ax,Au)

= |y−Ax|2 + t2 |Au|2−2t |(y−Ax,Au)|

Then if |y−Ax| is as small as possible, this will occur when t = 0 and so p′ (0) = 0. Butthis says

|(y−Ax,Au)|= 0

You could also use Lemma 13.3.2 to see this is 0. Since u was arbitrary, this proves onedirection.

Conversely, if this quantity equals 0,

|y−(Ax+Au)|2 = |y−Ax|2 + |Ax−Au|2 +2Re(y−Ax,Au)= |y−Ax|2 + |Ax−Au|2

and so the minimum occurs at any point z such that Ax = Az.Does there exist an x which minimizes this function? From what was just shown,

it suffices to show that there exists x such that ((y−Ax) ,Au) for all u. By the GrammSchmidt process there exists an orthonormal basis {Axk} for A(Fn). Then for a given y,

(y−

r

∑k=1

(y,Axk)Axk,Ax j

)= (y,Ax j)−

r

∑k=1

(y,Axk)

δ k j︷ ︸︸ ︷(Axk,Ax j) = 0.

In particular, (y−A

(r

∑k=1

(y,Axk)xk

),w

)= 0

310 CHAPTER 13. MATRICES AND THE INNER PRODUCT\AXTheorem 13.3.3 Let y € F” and let A be anm Xx n matrix. Then there exists x € F" mini-mizing the function x> |y—Ax|”. Furthermore, x minimizes this function if and only if((y—Ax) ,Au) =0forallu€ F".Proof: First consider the characterization of the minimizer. Let u € F”. Let |6| = 1,6 (y — Ax, Au) = |(y — Ax, Au)|Now consider the function of f € Rp(t) =|y—(Ax+t@Au)|* = ((y —Ax) —t0Au, (y — Ax) —1@Au)= |y — Ax|* +27 |Au|? — 2tRe (y — Ax, @Au) > |y — Ax|?= |y —Ax|? +2? |Au|? — 2Re 6 (y — Ax, Au)= |y —Ax|? +27 |Au|? — 2 |(y — Ax, Au)|Then if |y —Ax| is as small as possible, this will occur when t = 0 and so p’ (0) = 0. Butthis says\(y —Ax,Au)| =0You could also use Lemma 13.3.2 to see this is 0. Since u was arbitrary, this proves onedirection.Conversely, if this quantity equals 0,ly—(Ax+Au)|? = |y—Ax|? + |Ax—Aul|” + 2Re(y —Ax,Au)= |y—Ax|?+|Ax—Aul’and so the minimum occurs at any point z such that Ax = Az.Does there exist an x which minimizes this function? From what was just shown,it suffices to show that there exists x such that ((y—Ax),Au) for all u. By the GrammSchmidt process there exists an orthonormal basis {Ax;} for A (F”). Then for a given y,S,;. . kjy- y? (y, Axx) Axx, AX; (y,Ax;) -y( y, Axx) (Axx, Ax;) = 0.k=l k=1[4 e sax) ) =0k=1In particular,