As usual start Maple and load all the linear algebra programs using > with(LinearAlgebra); Tell Maple about a new Matrix A: > A:=<<9, 6, -4>|<-9, -6, 6>|<-3, -3, 5 >>; Use Maple to check that the matrix is non-singular: > Determinant(A); and then form the matrix from which we will get the eigenvalues: > A1:=A-lambda*IdentityMatrix(3); ColumnOperation works in just the same way as RowOperation does. Use a pair of determinant row and column operations to create two zeros such that you can use SubMatrix to extract a 2x2 matrix S and evaluate > Determinant(S); Repeat with a different row/column pair to identify a different submatrix determinant and check that the product is the same as previously. We can find the polynomial whose roots are the eigenvalues using: > pA:=Determinant(A1); which is the so-called characteristic polynomial of A. Note that the coefficient of pA without any lambda in is the determinant of A. To manipulate polynomials you can use factor, expand, sort, simplify, etc. > factor(pA); Verify that all of the different linear polynomials in lambda are included in the factorisation. You can also plot pA to see the locations of the eigenvalues using > plot(pA,lambda=1.5..4); We can also get the system of equations to solve using this matrix: > Aj:=>; Pivot this matrix to create a row of zeros and get a solution. Compare what you just got with this vector solution: > v1:=LinearSolve(subs(lambda=2,A1),<0,0,0>,free=s); What value of s gives your solution? Verify that v1 when multiplied by A on the left gives the expected result (for any s). > Multiply(A,v1); > 2*v1; Find two different eigenvectors of the eigenvalue lambda=3 similarly and check them. 3 is an eigenvalue of multiplicity 2 since it has two different vector solutions. What are the homogeneous solutions if you try to substitute a non eigenvalue into A1 instead of 2 or 3? Try to repeat the row and column operations procedure for > B:=<<32, -36, -18>|<27, -31, -18>|<-6, 6, 2 >>; > B1:=B-lambda*IdentityMatrix(3); Some row/column operations give two easy zeros but some don't. Find one which works. Now try with this matrix > C:= <<-6, 2, -3>|<-13, 15, -15>|<-2, 8, -7>>; > C1:=C-lambda*IdentityMatrix(3); but note that each way you try to do operations on C1 no easy cancellation occurs. Use > Eigenvalues(C); to note what kind of factors you should get and try to create one. If after 5 minutes if you have failed to create two zeros in a row, just evaluate > fC:=factor(Determinant(C1)); to reassure yourself the answers Maple gave are indeed correct. Find one of the three eigenvectors in C using RowOperation and/or Pivot and check for it or a multiple of it in the columns of P in > (v,P):=Eigenvectors(C); Extract another eigenvector from P and check it has the expected eigenvalue. Make a diagonal matrix of eigenvalues using > D1:=DiagonalMatrix(v); Check that P and D1 multiplied is the same as C and P multiplied in that order. Create a matrix which has random eigenvalues from w as follows: > randomise():w:=RandomVector(3,generator=rand(1..9)); > Dw:=DiagonalMatrix(w); > Q:=RandomMatrix(3,3,generator=rand(-3..4)); We choose Q as a random matrix for the eigenvectors and (if Q is non-singular) use diagonalisation to make J from Dw as follows: > J:=Multiply(Multiply(Q,Dw),MatrixInverse(Q)); Check that the eigenvalues and eigenvectors of J are as you expect.