How can I derive inverse of the matrix with infinite determinant?

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Junho Kweon 2022-11-17

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评论： Walter Roberson 2022-11-18

myMatrix.mat

在 MATLAB Online 中打开

Hi guys, there is a matrix A which has finite dimension and finite-valued elements.

However, because of its large size and values, the determinant of A becomes infinite on MATLAB.

load("myMatrix.mat","A")
size(A)
ans = 1×2
   320   320
max(max(abs(A)))
ans = 2.2913e+04
det(A)
ans = Inf

Therefore, it cannot compute the inverse of the matrix precisely.

B = inv(A);
Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND =  8.336932e-19.
isdiag(A*B)
ans = logical
   0

Is there way to compute the inverse of the matrix A?

Cheers,

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Answer 1

Bruno Luong 2022-11-17

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编辑：Bruno Luong 2022-11-18

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Welcome to the world of numerical calculation. Every conclusion you made is wrong.

"det(A) = Inf Therefore, it cannot compute the inverse of the matrix precisely. ".

Wrong

A=eye(200)*1024;
det(A)
ans = Inf
B=inv(A);
all(A*B == eye(size(A)), 'all') % A*B is exatly I, despite det(A) is Inf due to overflow
ans = logical
   1

Check B is inverse of A using isdiag(A*B).

This is a bad numerically criteria as showed in the false negative answer example

A=[1 2; 3 4]
A = 2×2
     1     2
     3     4
B = inv(A)
B = 2×2
   -2.0000    1.0000
    1.5000   -0.5000
AB = A*B;
isdiag(AB) % Suppose to be TRUE
ans = logical
   0
AB(2,1) % Suppose to be 0, but not exactly
ans = 8.8818e-16

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Bruno Luong 2022-11-17

编辑：Bruno Luong 2022-11-17

The funny thing is the inconsistency of linear algebra - mostly det - in floating point world.

The matrix is numerical singular, so the determinant should be 0, yet it is Inf.

Junho Kweon 2022-11-18

@Bruno Luong Yeah, it is quite interesting that detmerninant is not zero though it is mathematically 0. May be in the mathematical computation of determinant, (very large number)- (very large number), and it is over the computational limitation of MATLAB. Therefore, it might be Inf comes out, and MATLAB just conclude that the determinant is infinite.

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Answer 2

Torsten 2022-11-17

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编辑：Torsten 2022-11-17

在 MATLAB Online 中打开

What do you get if you type

rank(A)

?

If it's smaller than 320, no inverse exists (or is almost impossible to calculate).

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Junho Kweon 2022-11-18

I was silly not to check that one. Thanks

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Answer 3

Bruno Luong 2022-11-17

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移动：Bruno Luong 2022-11-17

"Is there way to compute the inverse of the matrix A?"

No.

The sum of all columns of your matrix A is numerically 0, therefore your matrix is not invertible.

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Steven Lord 2022-11-17

The first of Cleve's Golden Rules of computation applies here. "The hardest things to compute are things that do not exist."

Bruno Luong 2022-11-17

Hmm read these two rules from Cleve I wonder this: things that do not exist are they unique?

It sounds like a non-sense question, but why it is non-sense?

If it does make sense I have no preference for the answer.

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Answer 4

Walter Roberson 2022-11-17

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编辑：Walter Roberson 2022-11-18

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In theory if you have the symbolic toolbox, you could

digits(16);
sA = sym(A, 'd');
digits(32);
Ainv = inv(sA);

... But test it with a much smaller matrix first, as it might take very very long.

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Bruno Luong 2022-11-18

编辑：Bruno Luong 2022-11-18

在 MATLAB Online 中打开

I don't know symbolic at all but is there any approximation going on as well (digits=32)? Why it does give out Ainv for the matrix that has numerical singular?

How to exploit the bunch of fractions in the result and be sure it is correct with still a finite truncation (digits=32)?

It doesn't look like it is any better than pure numerical method.

A = rand(8,7)*rand(7,8)

A = 8×8

1.1932 1.4076 1.2416 0.7862 1.4254 1.3012 1.7597 1.5144 1.6909 2.2753 2.1156 1.4951 2.5238 2.3885 2.7331 2.1692 1.1620 1.6231 1.5313 1.2976 1.9202 1.8231 2.1256 1.5481 1.1399 1.2803 1.2986 0.9485 1.6928 1.6402 2.0115 1.3554 1.7376 2.4798 2.3184 1.5994 2.6161 2.4092 2.8091 2.1660 1.3393 1.5057 1.3630 0.9368 1.4926 1.5373 2.0491 2.0210 1.3573 2.2611 1.8506 1.2858 2.4179 2.2090 2.5930 2.0655 1.8750 1.6381 2.0915 1.2237 2.0587 2.2006 2.3607 1.8280

B = A\eye(size(A)) % for those who think \ is better than B=inv(A)

Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 8.262494e-18.

B = 8×8

1.0e+15 * 0.3034 -0.4816 0.3936 -0.2782 -0.1810 -0.1330 0.3088 0.2056 0.7213 -1.1449 0.9358 -0.6614 -0.4303 -0.3162 0.7342 0.4889 -0.5342 0.8479 -0.6930 0.4898 0.3187 0.2342 -0.5437 -0.3620 0.0684 -0.1086 0.0888 -0.0628 -0.0408 -0.0300 0.0697 0.0464 -1.0208 1.6202 -1.3242 0.9360 0.6090 0.4475 -1.0390 -0.6918 0.8036 -1.2754 1.0425 -0.7368 -0.4794 -0.3523 0.8179 0.5446 0.2064 -0.3277 0.2678 -0.1893 -0.1232 -0.0905 0.2101 0.1399 -0.4766 0.7565 -0.6183 0.4370 0.2843 0.2090 -0.4852 -0.3230

Inum = A*B % Garbage

Inum = 8×8

0.8339 0.1022 -0.4563 -0.0586 0.1096 0.0810 -0.1324 -0.0255 -0.2712 1.1957 -0.2699 0.0979 0.1207 0.1741 0.0739 -0.0912 0.1829 -0.1298 0.8032 -0.1076 0.0034 0.0307 -0.1420 0.1355 -0.2903 0.1039 -0.6101 0.9326 0.0593 0.0702 -0.0003 -0.0164 -0.0857 -0.2977 0.0205 -0.0615 1.0466 0.1174 0.2904 0.2597 -0.0443 0.1953 -0.3567 0.0948 0.0285 1.1345 0.1275 0.0570 -0.0902 -0.1424 0.0200 -0.3266 -0.1365 0.1368 1.1092 0.1673 0.0828 -0.2810 -0.1661 -0.1480 0.0341 -0.0215 0.1781 1.1653

norm(Inum-eye(size(A)))

ans = 1.0418

digits(16);

sA = sym(A, 'd');

digits(32);

Ainv = inv(sA)

Ainv =

Isym = A*double(Ainv) % Still garbage, is it a correct test?

Isym = 8×8

1.1474 -0.4078 0.4030 -0.2153 -0.2193 -0.1139 0.1195 0.1647 -0.2932 1.1517 0.0988 -0.0115 0.0175 0.0423 -0.1806 -0.0345 -0.7159 1.2818 0.1346 0.6660 0.4150 0.2834 -0.8288 -0.4359 -0.0244 -0.1726 0.1465 0.9150 -0.0716 -0.0160 0.0375 -0.0374 1.0789 -1.5676 1.3428 -0.9192 0.3049 -0.4418 0.9138 0.6292 0.4085 -0.6508 0.6104 -0.2542 -0.2783 0.8002 0.5759 0.3289 -0.3197 0.3399 -0.5910 0.4714 0.3066 0.2587 0.4518 -0.2483 -0.6429 0.9187 -0.6401 0.6024 0.2725 0.2275 -0.5178 0.5798

norm(Isym-eye(size(A)))

ans = 4.2935

Bruno Luong 2022-11-18

编辑：Bruno Luong 2022-11-18

OK I make change accordingly to 'd' flag in my test code, and as you can see the symbolic is no bettrer than numerical method with "\", at least in this toy example.

Even if it use minor formula, there are a lot of cancellation going on in determinant so the summation order and tree (parenthesis) matter. The symbolic probably does the sumation in the implicit order of the implementation as it progres and no care about cancellation.

I'm not sure if the symbolic solution is any better than numerical method for precision point of view, let alone the computation time that goes as factorial of the dimesion.

Walter Roberson 2022-11-18

The point about this approach was that it should avoid the intermediate inf

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How can I derive inverse of the matrix with infinite determinant?

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