Matrix subtraction within Matlab

I am generating a 3X1 matrix and name it 'xT'. Simultaneously I am generating another 3X1 matrix and name it 'x'. When I am computing the difference between these two i.e. Er = xT - x within Matlab, the answer is different than individual subtraction. Therefore when I am subtracting xT(2,1) - x(2,1), I am getting a different answer when Matlab is computing the Er. Please help.

[madhan ravi]

Illustrate.

[KALYAN ACHARJYA]

Is this way, are you trying?

xT=randi(10,3,1)  % Random integer 3x1 mat
x=randi(20,3,1)   % % Random integer 3x1 another mat
Er=xT-x

Command Window:

xT =
     7
     8
     3
x =
     1
     3
    16
Er =
     6
     5
   -13

[Monish Sengupta]

The xT is generated as follows:

%Update the true state

xt = phi*xt;

xt11(T) = xt(1,1);

xt21(T) = xt(2,1);

xt31(T) = xt(3,1);

and x is generated by

x = xA*pKAp + xB*pKBp + xC*pKCp;

x11(T) = x(1,1);

x21(T) = x(2,1);

x31(T) = x(3,1);

This cycle runs at every instance T. At the end of every cycle,

I derive the Er as follows:

Er = xt - x;

Er11(T) = Er(1,1);

Er21(T) = Er(2,1);

Er31(T) = Er(3,1);

The problem is when I carry a physical difference for example xt(2,1) - x(2,1), they do not match with Er which is xt-x.

When I run my code, at instance T = 230, the xt is as follows:

xt =

1.0e+04 *

1.0869

-0.0000

-0.0001

and

x =

1.0e+04 *

1.0869

0.0000

-0.0001

If I exract the xt(2,1) from this, I get

xt(2,1)

ans =

-0.3000

When I extract x(2,1), I get

x(2,1)

ans =

0.3174

Therefore a manual subtraction gives:

xt(2,1)-x(2,1)

ans =

-0.6174

But the Matlab calculation of Er, which is xt - x, gives

Er(2,1)

ans =

-0.0674

How is that possible?

[James Tursa]

It isn't possible, of course. You are either not doing the coding the way you think you are, or you are misinterpreting the displayed results. Instead of "describing" to us what you think you are doing, just copy & paste the actual code and actual output. If you need to attach a mat file for the values, do that as well. You don't provide enough digits in your post for us to guess what is going on.

[Monish Sengupta]

   % Generate the state estimate
    x = xA*pKAp + xB*pKBp + xC*pKCp;
    x11(T) = x(1,1);
    x21(T) = x(2,1);
    x31(T) = x(3,1);
    
    % Generate the error
    Er = xt - x;
    Er11(T) = Er(1,1);
    Er21(T) = Er(2,1);
    Er31(T) = Er(3,1);
    
    %Update the true state
    xt = phi*xt;
    xt11(T) = xt(1,1);
    xt21(T) = xt(2,1);
    xt31(T) = xt(3,1);

This is exactly what I am doing. While the Er(3,1) matches with manual subtraction xt(3,1)-x(3,1), the Er(2,1) doesn't match with xt(2,1)-x(2,1).

Please see the attached graph (file) which shows the difference between the two calculations.

[Stephen23]

@Monish Sengupta : please also upload your data in a .mat file.

Not in a text file, an excel file, or anything else. In a .mat file.

We cannot check your results without your data.

[Monish Sengupta]

I have no uploaded the .mat file.

Answer 1

[John D'Errico]

whos
  Name       Size             Bytes  Class     Attributes

  E          2x1                 16  double              
  E11        1x1                  8  double              
  E21        1x1                  8  double              
  Er         3x1                 24  double              
  Er11       1x435             3480  double              
  Er21       1x435             3480  double              
  Er31       1x435             3480  double              
  ErA        3x1                 24  double              
  ErA11      1x434             3472  double              
  ErA21      1x434             3472  double              
  ErA31      1x434             3472  double              
  ErB        3x1                 24  double              
  ErB11      1x434             3472  double              
  ErB21      1x434             3472  double              
  ErB31      1x434             3472  double              
  ErC        3x1                 24  double              
  ErC11      1x434             3472  double              
  ErC21      1x434             3472  double              
  ErC31      1x434             3472  double              
  H          2x3                 48  double              
  I          3x3                 72  double              
  InA        2x1                 16  double              
  InA11      1x434             3472  double              
  InA21      1x434             3472  double             
  ...

This, by the way, is why you DON'T ever want to create such reams of numbered variables. That is incredibly bad programming style, as it will cuse you to make mistake after mistake, as you type the wrong variable name. LEARN TO USE VECTORS. Better, learn to use arrays.

Regardless, you only think it did something different, probably because of the many different variable names, and possibly because you are using a short display format.

format long g
>> x
x =
          10869.0885676116
         0.228888860503846
         -1.08890062860864
xt
xt =
                  10869.05
        -0.299999999999384
                      -1.1
                      
xt - x
ans =
       -0.0385676116566174
        -0.528888860503229
       -0.0110993713913601

xt(2) - x(2)
ans =
        -0.528888860503229

The results will be identical. The only problem is that sometimes when you display a vector as a result, it might get displayed so it looks differently, especially if you are only showing 4 digits.

A frequent and similar mistake made by new users at MATLAB, is they will display the numbers using format short, as:

format short
>> x
x =
   1.0e+04 *
    1.0869
    0.0000
   -0.0001
>> xt
xt =
   1.0e+04 *
    1.0869
   -0.0000
   -0.0001

and then either miss the 1.0e+4 at the top, or they will think those numbers are exactly what they see, then do a subtraction by hand.

Sorry, but these are NOT MATLAB mistakes. It is a user error, with 100% probability. I won't even call it probable user error, as that would imply some vague chance that it was not a user error. As you develop better skills at MATLAb, you will learn to make better use of your variables, but you will also learn to go more slowly, to check your work more carefully.

[Monish Sengupta]

Thank you for your helpful comments. I have managed to find the problem.