Question

<Problem 2>

Answer the following questions about the square matrix A of order 3:

A= $\bigl(\begin{smallmatrix} 3 \0 \,\,1 \\ -1 \2 \ -1 \\ -2\ -2\ 1 \\ \end{smallmatrix}\bigr)$

III. The square matrix B of order 3 is diagonalizable and meets AB=BA.

prove that any eigenvector p of A is also an eigenvector of B.

IV. Find the square matrix B of order 3 that meets B² = A, where B is

diagonalizable and all eigenvalues of B are positive.

V. The square matrix X of order 3 is diagonalizable and meets AX = XA.

When tr(AX) = d, find the maximum of det(AX) as a function of d.

Here, d is positive real and all eigenvalues of X are positive. In addition,

tr(M) is the trace (the sum of the main diagonal elements) of the square matrix M,

and det(M) is the determinant of the matrix M.

Answer 1

Given, 1 2 0 1 A Here de tA) =6 and eigen values of A are: 1,2,3. So A is diagonalizable and each eigen space corresponding to each eigen value has dimension=1 rrrrrrrrrrrrrrr Bis diagonalizable. BA =AL To prove: Any eigen vector p of A is also an eigen vector of B. Proof: Ap=18 ( p is an eigen vector of A) =) B(AP) = B(AP) - ОАР АВР Y ABP = (BP) 7 A (BP) = A (BP) .. By is also an eigen vector of a corresponding to eigen valve d. But the eigen space corresponding to eigen value i has dimension I. so Bp is a scadar moltiple of p. = BP = 1P for some scalar to - p is also an eigen vector of B. So any eigen vector p is also an eigen vector of B. (poored, of A

(1) To find B sit ² A First we have to find eigen vectors of A. For Ar! A-1769) - 7 "23)-(0) /201 -1.1- 1-2- 2012 2017030 ni-N2 + N₂ = 0 anitar2 = 0 n=-X2 =-X3 Solving an eigen vector for dial For A2=2 & (4-21) (59 ) = » - O 1-2-2- 1 AJ3) - 3 solving 4, +13=0 an, +282 +13=o n=-23 "2 = 2 I is an eigen vector ford by=2 for $3=35 Similarly we geti G) is an eigen vector for A223

let Then A= PD pt where p= diag (1,2,3) het B = PVD p . where ro= on il O OV3 Then p² = (PVD pt 72 - pppy A so required B= Ploo IPT loors Where DFH - 21 Ans) I 2 O

A = 120 Given -21 . .) - P diag (1,2,3) p! Ax=XA , To Qx)=0 = x must be of the form P Do PT for some diagonal matrin Di= diag (2, 4, 2) C. Any matrimx that commuta witha diagonal matria PD pt with distinct eigen vadre must be of the form PD, pl ) so Ax=PoDipl, where b-diag(1,2,3) - Di=diag (219, 2) so too (4x) = 2x+2y+37=d Det AX) = xx 24x9. By AM-GM inequality, x+2y+32 (n. 24.32) 3 > (det (4x) ; 13 2 Tr (4x) - detax) L 40 x ² 3 27

> det(x) = 2 So manimum value of de+(Axis d3 when na 24=32 = 0/3 ie when (ny, z) = (as de ida).