Let V be a finite-dimensional inner product vector space over the reals with dim
ID: 1888266 • Letter: L
Question
Let V be a finite-dimensional inner product vector space over the reals with dim V = n. Let W V be a subspace. We know from class that V = W W Define the operator (orthogonal projection) T : V rightarrow V by T(y + z) = y for y W and z W Show that T is self-adjoint. Find the characteristic polynomial of T. Does U have an ONB of eigenvectors? Prove your answer.Explanation / Answer
sol: suppose that as like this, (a) 1. draw the line x=y 2. take any point P=(x0,y0) of the plane 3. trace the line perpendicular to y=x passing through P: so The slope of y=x is 1, then the slope of a perpendicular is -1/1 = -1, then any perpendicular has an equation like y = -x + b. If (x0, y0) is a point of this perpendicular, y0 = - x0 + b=> b = x0+y0, then y = x0 + y0 - x. 4. find the intersection of the two lines: y = x y = x0 + y0 - x => 2x = x0+y0 => x = (x0+y0)/2 = y 5. finally, T(x0,y0) = (x,y): T(x0,y0) = ((x0+y0)/2, (x0+y0)/2) let the, Before this point we have used (x0,y0) for allowing to distinguish between the point P and its image coordinates. Now we can drop the subindices: T(x,y) = ((x+y)/2, (x+y)/2) (b) T^2(x,y) = T(T(x,y)) = T((x+y)/2, (x+y)/2) = = (((x+y)/2+(x+y)/2)/2, ((x+y)/2+(x+y)/2)/2) = = ((x+y)/2, (x+y)/2) = T(x,y) => T^2 = T (c) Eigenvalues: T(x,y) = c (x,y) => => ((x+y)/2, (x+y)/2) = c (x,y) => [1/2 1/2] [x] [1/2 1/2] [y]= = c[x] [y] For this system has a nontrivial solution, det [1/2-c 1/2] = 0 [1/2 1/2-c] => (1/2-c)^2 - 1/4 = 0 => (1/2 - c) = +/- 1/2 The eigenvalues are c1 = 0 c2 = 1 Eigenvectors (non-null solutions of the system) (1,1) (eigenvalue 1) (1,-1) (eigenvalue 0) Let's check: T(1,1) = ((1+1)/2, (1+1)/2) = (1,1) = 1 (1,1) T(1,-1) = ((1-1)/2, (1-1)/2) = (0,0) = 0 (1,-1) answer
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