- It is in row echelon form.
- The leading entry in each nonzero row is a 1 (called a leading 1).
- Each column containing a leading 1 has zeros in all its other entries.

# NULLSPACE - RREF Command in Matlab bugs

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Hello everyone, can someone help me debug this code? Like I have a matrix *A*, and I need to find A^k by diagonalization method. Below is my original code, but there is a problem with this part:

m = (M - R(i,1)*eye(NumRowsR));

disp(rref(m));

t = null(rref(m));

disp(t);

This part can't give me the nullspace of matrix t after reduced for some reasons (I recently did some research on the Internet and see some people said about the bug of **rref** and **null**). The problem is that it keeps showing me elementary matrix

1 0 0

0 1 0

0 0 1

for the eigenvalues. Does anyone know how to fix it? I will be very much appreciated!

Below is my original code, for a better view:

syms x NumRowsM NumColsM NumRowsR NumColsR NumRowst NumColst k numeigen

M = input('Input a square matrix M: ');

k = input('Input the power of matrix M: ');

[NumRowsM, NumColsM]=size(M);

if (NumRowsM ~= NumColsM)

disp('Not valid input. Matrix must be a square matrix!')

else

%Find eigenvalues:

R = solve(det(M - x*eye(NumRowsM)), x);

[NumRowsR, NumColsR] = size(R);

if (or(NumRowsR == 0,NumColsR == 0))

disp('No eigenvalues. The matrix is not diagonalizable')

else

numeigen = 0;

F = zeros(NumRowsR, NumRowsR);

d = zeros(NumRowsR,1);

for i = 1:NumRowsR

m = (M - R(i,1)*eye(NumRowsR));

disp(rref(m));

t = null(rref(m));

disp(t);

[NumRowst, NumColst] = size(t);

if (NumColst == 0)

if (i == NumRowsR && numeigen > NumRowsR)

disp('Matrix not diagonalizable due to invalid eigenvalue');

return

else

continue

end

else

numeigen = numeigen + 1;

if (NumColst == 1)

for j = 1:NumRowsR

[n, d(j)] = numden(sym(t(j,1)));

end

for j = 1:NumRowsR

F(j,i) = sym(t(j,1)) * lcm(sym(d));

end

else

for k = 1:NumColst

for j = 1:NumRowsR

[n, d(j)] = numden(sym(t(j,k)));

end

for j = 1:NumRowsR

F(j,k) = sym(t(j,k)) * lcm(sym(d));

end

end

end

end

end

disp(F);

D = (F\M)*F;

disp('The power k of the matrix M is: ');

T = F*(D^k)/(F);

disp(T)

end

end

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### Answers (1)

Abhijeet
on 1 Jul 2022

I understand your problem that you want to find out A^k of a square matrix A using diagonalization method. In the code attached, you have used rref(), for calculating row reduced echelon form and it returns an identity matrix.

I went through the code and ran it on MALAB version R2022b but was not able generate the output you are mentioning. The code itself is doing its task correctly of calculating A^k for the square matrix A.

Although, the reason rref() returns the identify matrix is hidden in its definition itself. Conditions for a matrix to be in the row reduced echelon form are:

Since you are working with symbolic variables and MATLAB treats symbolic variables as non-zero element, it ends up removing them from the final result using row tranformations.

For more information, I kindly encourage you to visit below mentioed hyperlinks:

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