Problem 6
If an n n matrix A is diagonalizable (over R), then there must be a basis of Rn consisting of eigenvectors of A.
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MATH 2450 WEEK 7 Strategy On a close domain, look for all critical points inside the domain. Look for the boundary points then compare. Look for all critical points inside the domain: F = 0 x Fy= 0 Look for all critical points on the boundary g’(t) = 0 Look at the boundary points of the boundary EX. Find the absolute max/min f(x,y) = e(2) - over the disk x + y <= 1 1) x + y < 1 2) x + y = 1 x^(2) - y^(2) Fx= 0 2xe = 0 x = 0 x^(2) - y^(2) Fy= 0 -2ye = 0 y = 0 Po= (0,0) Forms of absolute minimum and maximum 1) Y = sqrt(1-x ) -1<= x <= 1 Y = -sqrt
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Chapter 7, Problem 6 is Solved
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