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Let q = - 2xy - y² + 2xz+2yz + z² be a quadratic form on R³ viewed as a polynomial in 3 variables. Find a linear change of variables to u, v, w that puts q into the canonical form in `Sylvester's law of inertia'! What are the values of the associated indices s, t? Select one: O We let u = √3(x − y + z), v=x+y, w = (x - y)/2 to find that q = u²+ v². Hence s = 2, t = 0 in Sylvester's law of intertia. w². Hence s = 1, t = 2 in Sylvester's law of intertia. O We let u = x - 2y, v = z+y, w = √3y to find that q =u² - v² ○ we let u = x, v = √√√2(x + y), w =x+y+z to find that q = u² O None of the others apply v² +w². Hence s = 2, t = 1 in Sylvester's law of intertia. O The quadratic form does not obey the condition to be diagonalisable over R. This is because the minimal polynomial of the corresponding matrix is not a product of distinct linear factors. Hence Sylvester's law of inertia does not apply. By convention, we set s = t = ∞ when this happens.

Let q = - 2xy - y² + 2xz+2yz + z² be a quadratic form on R³ viewed as a polynomial in 3 variables. Find a linear change of variables to u, v, w that puts q into the canonical form in `Sylvester's law of inertia'! What are the values of the associated indices s, t? Select one: O We let u = √3(x − y + z), v=x+y, w = (x - y)/2 to find that q = u²+ v². Hence s = 2, t = 0 in Sylvester's law of intertia. w². Hence s = 1, t = 2 in Sylvester's law of intertia. O We let u = x - 2y, v = z+y, w = √3y to find that q =u² - v² ○ we let u = x, v = √√√2(x + y), w =x+y+z to find that q = u² O None of the others apply v² +w². Hence s = 2, t = 1 in Sylvester's law of intertia. O The quadratic form does not obey the condition to be diagonalisable over R. This is because the minimal polynomial of the corresponding matrix is not a product of distinct linear factors. Hence Sylvester's law of inertia does not apply. By convention, we set s = t = ∞ when this happens.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Let q = - 2xy - y² + 2xz+2yz + z² be a quadratic form on R³ viewed as a polynomial in 3 variables. Find a linear change of variables tou, v, w that puts q into the canonical form in `Sylvester's law of inertia'!. What are the values of the associated indices s, t? Select one: O We let u = √3(x − y + 2), v = x+y, w = (x - y)/2 to find that q = u²+². Hence s = 2, t = 0 in Sylvester's law of intertia. O We let u = x - 2y, v = z+y, w = √3y to find that q=u²v² w². Hence s = 1,t = 2 in Sylvester's law of intertia. O we let u = x, v= √2(x+y), w = x+y+z to find that q = u²v² + w²2. Hence s = 2, t = 1 in Sylvester's law of intertia. O None of the others apply O The quadratic form does not obey the condition to be diagonalisable over R. This is because the minimal polynomial of the corresponding matrix is not a product of distinct linear factors. Hence Sylvester's law of inertia does not apply. By convention, we set s = t = ∞ when this happens.
Transcribed Image Text:Let q = - 2xy - y² + 2xz+2yz + z² be a quadratic form on R³ viewed as a polynomial in 3 variables. Find a linear change of variables tou, v, w that puts q into the canonical form in `Sylvester's law of inertia'!. What are the values of the associated indices s, t? Select one: O We let u = √3(x − y + 2), v = x+y, w = (x - y)/2 to find that q = u²+². Hence s = 2, t = 0 in Sylvester's law of intertia. O We let u = x - 2y, v = z+y, w = √3y to find that q=u²v² w². Hence s = 1,t = 2 in Sylvester's law of intertia. O we let u = x, v= √2(x+y), w = x+y+z to find that q = u²v² + w²2. Hence s = 2, t = 1 in Sylvester's law of intertia. O None of the others apply O The quadratic form does not obey the condition to be diagonalisable over R. This is because the minimal polynomial of the corresponding matrix is not a product of distinct linear factors. Hence Sylvester's law of inertia does not apply. By convention, we set s = t = ∞ when this happens.
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