Help my Linear Algebra Question

Question

f(x,y,z)=s(x² +y²+z²)-4(xy+xz+yz)

Express this function to

Also, find the condition which A is positive definite.

Answer 1

Assuming that s is a constant real number, we have the following:

Given f(x,y,z) = s (x² + y² + z²) - 4 (xy + xz + yz) = sx² + sy² + sz² - 4xy - 4xz - 4yz, we can let

[ a b c ] [ a b c ] [ x ] [ ax + by + cz ]

A = [ d e f ], so we can write: f(x,y,z) = [x y z] [ d e f ] [ y ] = [x y z] [ dx + ey + fz ], so:

[ g h i ] [ g h i ] [ z ] [ gx + hy + iz ]

f(x,y,z) = x (ax + by + cz) + y (dx + ey + fz) + z (gx + hy + iz) = ax²+ bxy + cxz + dxy + ey² + fyz + gxz +

hyz + iz² = ax²+ ey² + iz² + (b+d) xy + (c+g) xz + (f + h) yz = sx² + sy² + sz² - 4xy - 4xz - 4yz, so

a = e = i = s, and b+d = c+g = f+h = -4. Next, A is positive definite if and only if both A is symmetric

and f(x,y,z) is positive (also known as strictly > 0) for all vectors [x, y, z] =/= [0, 0, 0].

A being symmetric forces b to = d, c to = g, and f to = h, so they all must = -2 for each of those

sums earlier to be = -4, so now we just need to find s such that f(x,y,z) > 0 for [x,y,z] =/= [0,0,0].

If we first set s = t + 4, then now we have:

f(x,y,z) = t(x²+y²+z²) + 4(x²+y²+z²) - 4(xy+xz+yz) = t(x²+y²+z²) + 4x² + 4y² + 4z² - 4xy - 4xz - 4yz

= t(x²+y²+z²) + 4x² - 2xy - 2xz - 2xy + y² + yz - 2xz + yz + z² + 3y² - 6yz + 3z²

= t(x²+y²+z²) + 2x(2x-y-z) - y(2x-y-z) - z(2x-y-z) + 3(y²-2yz+z²)

= **************, so for all [x y z] =/= [0 0 0], the 1st term is always > 0

for all t > 0, the 2nd term is always >= 0, and the 3rd term is always >= 0, so f(x,y,z) is

always > 0 for all [x y z] =/= [0 0 0] and f(0,0,0) = 0 when t > 0, or s > 4, so, finally, we have the following:

[ s -2 -2 ]

A = [ -2 s -2 ], with s > 4.

[ -2 -2 s ]