9. Let D = Diag(d 1 ,d 2 , . . . ,d n ) be a diagonal matrix. a. Prove by induct

Question

9. Let D = Diag(d1,d2, . . . ,dn ) be a diagonal matrix.
a. Prove by induction that D^k = Diag(d1k ,d2k , . . . ,dnk), for all positive integers
k.
b. Prove that D is invertible if and only if all of the di are non-zero. Hint:
Use the fact that any matrix A is invertible if and only if the reduced row
echelon form of A is the identity matrix.
c. Prove that if D = Diag(d1,d2, . . . ,dn) is invertible, then:
D-1= Diag(d1-1,d2-1, . . . ,dn-1)

18. Let T : R3 R3, with [T]= A =

3 5 4
0 2 1
0 0 7
.
a. Compute T(e1), T(e2), and T(e3).
b. Use your computations in (a) to find three vectors v1, v2, and v3, such
that T(v1) = e1, T(v2) = e2, and T(v3) = e3. Hint: fully exploit the
linearity properties of T.
c. Use your answers in (b) to write A-1.

Explanation / Answer

a. base case: D^2 = Diag ( d1^2 , d2^2 , d3^2 .... ) Now, for the inductive hypothesis assume D^(k-1) = Diag(d1^(k-1), d2^(k-1)....) need to show that D^k = Diag(d1^(k), d2^(k)....) D^k = D^(k-1) * D apply the inductive hypothesis to simply D^(k-1) = Diag(d1^(k-1), d2^(k-1)....) * Diag (d1, d2, d3, ....) = Diag ( d1^(k-1)*d1 , d2^(k-1)*d2, ....) = Diag (d1^k , d2^k, .... ) this completes the induction. b. Prove that D is invertible if and only if all of the di are non-zero. so iff needs to be proved both ways first lets prove : if all the di are non zero then D is invertible. we can prove this by construction. lets call a new matrix C = diag (1/d1 , 1/d2, 1/d3 ....) now C * D = diag (1/d1 , 1/d2, 1/d3 ....) * diag (d1, d2 , d3 ...) = diag ( 1 , 1 ,1 ,1 ,1 ) = I so aha! , C is acutally D^-1 , so D must be invertible. Now lets prove if D is invertible then all the di must be non-zero. D is n x n matrix D is invertible, so the rank of D must be n , it has n linearly independent vectors. Thus, all the di must be non-zero... or else wouldnt be enough linearly independent vectors. c . lets call a new matrix C = diag (1/d1 , 1/d2, 1/d3 ....) now C * D = diag (1/d1 , 1/d2, 1/d3 ....) * diag (d1, d2 , d3 ...) = diag ( 1 , 1 ,1 ,1 ,1 ) = I so C is infact D^-1 Your problem is too long... i gotta do my own hw now... maybe i'll work on it later =P good luck have fun

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