Find a matrix P such that P'AP orthogonally diagonalizes A. Verify that P'AP gives the proper diagonal form. (Enter each matrix in the form [[row 1], [row 2], ...], where each row is a comma-separated list.)2 30 03 200A =0 0 2 30 0 3 2(P, PTAP) =

We have A=2300320000230032 Now, we find the eigenvalues and corresponding eigenvectors. Consider: A-λI=0⇒2300320000230032-λ1000010000100001=0⇒2-λ30032-λ00002-λ30032-λ=0⇒2-λ2-λ0002-λ3032-λ-330002-λ3032-λ=0⇒2-λ2-λ2-λ332-λ-3×32-λ332-λ=0⇒2-λ22-λ2-9-92-λ2-9=0⇒2-λ2-92-λ2-9=0⇒2-λ2-92=0⇒4+λ2-4λ-92=0⇒λ2-4λ-52=0⇒λ2-5λ+λ-52=0⇒λ(λ-5)+1λ-52=0⇒λ+1(λ-5)2=0⇒λ+12=0, (λ-5)2=0⇒λ=-1, -1, 5, 5 So, we get two eigenvalues -1, 5 with algebraic multiplicity 2.Eigenvector for λ=-1: Let the eigenvector be X=xyzw Then: A-λIX=0000 Hence, consider: A-λIX=0000⇒2300320000230032-λ1000010000100001xyzw=0000⇒2300320000230032--11000010000100001xyzw=0000⇒2300320000230032+1000010000100001xyzw=0000⇒3300330000330033xyzw=0000⇒3x+3y3x+3y3z+3w3z+3w=0000⇒3x+yx+yz+wz+w=0000⇒x+yx+yz+wz+w=0000 This gives: x+y=0z+w=0 Which further gives: x=-yz=-w Hence, X=xyzw=-yy-ww=y-1100+w00-11 When y=1, w=0 we get one eigenvector is -1100 When w=1, y=0 we get another eigenvector is 00-11 Hence, the eigenvectors corresponding to λ=-1 are -1100 and 00-11Eigenvector for λ=5: Let the eigenvector be Y=x'y'z'w' Then: A-λIY=0000 Hence, consider: A-λIY=0000⇒2300320000230032-λ1000010000100001x'y'z'w'=0000⇒2300320000230032-51000010000100001x'y'z'w'=0000⇒-33003-30000-33003-3x'y'z'w'=0000⇒-3x'+3y'3x'-3y'-3z'+3w'3z'-3w'=0000⇒3-x'+y'x'-y'-z'+w'z'-w'=0000⇒-x'+y'x'-y'-z'+w'z'-w'=0000 This gives: -x'+y'=0x'-y'=0-z'+w'=0z'-w'=0 This further gives: x'-y'=0z'-w'=0 And this gives: x'=y'z'=w' Hence, X=x'y'z'w'=y'y'w'w'=y'1100+w'0011 When y'=1, w'=0 we get one eigenvector is 1100 When w'=1, y'=0 we get another eigenvector is 0011 Hence, the eigenvectors corresponding to λ=5 are 1100 and 0011Now, we got: The eigenvectors corresponding to λ=-1 are -1100 and 00-11 The eigenvectors corresponding to λ=5 are 1100 and 0011 The matrix P is made of columns which are the eigenvectors of matrix A Therefore, matrix P is: P=-101010100-1010101Now, we find the value of PTAP which will be a diagonal matrix. Now P=-101010100-1010101 Therefore, PT=-110000-1111000011 Hence, PTAP=-110000-11110000112300320000230032-101010100-1010101=-2+3-3+20000-2+3-3+22+33+200002+33+2-101010100-1010101=1-100001-155000055-101010100-1010101=-1-101-11-10-1-101-1-5+505+500-5+505+5=-20000-2000010000010 Hence, PTAP is a diagonal matrix. Answer: We get P=-101010100-1010101 and PTAP=-20000-2000010000010 Hence, P,PTAP=-101010100-1010101, -20000-2000010000010