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Diagonally dominant matrix

From Wikipedia, the free encyclopedia

In mathematics, a square matrix is said to be diagonally dominant if, for every row of the matrix, the magnitude of the diagonal entry in a row is larger than or equal to the sum of the magnitudes of all the other (non-diagonal) entries in that row. More precisely, the matrix A is diagonally dominant if

where aij denotes the entry in the ith row and jth column.

This definition uses a weak inequality, and is therefore sometimes called weak diagonal dominance. If a strict inequality (>) is used, this is called strict diagonal dominance. The unqualified term diagonal dominance can mean both strict and weak diagonal dominance, depending on the context.[1]

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Transcription

So in this case we will look at what is a diagonally dominant matrix. So N by N matrix A, so it is a square matrix, is diagonally dominant. If, what happens is that each of the elements which are on the diagonal, the absolute value of that will be greater than the sum of the absolute elements of the rest of the elements which are in that row. So we will explain it through example and also we will go through this formula again. So we have summation here: J is equal to 1 to N. I is not equal to J. Absolute value of AIJ, I is equal to 1 through N. And this is greater than or equal to. And then we say AII is strictly greater than the summation which we just wrote. J is equal to 1 through N. I not equal to J. AIJ for at least 1 I. So let's concentrate on this part here. So what this means is that you have a diagonal element, you take the absolute value of that. Then you look at the row, I, and you take the sum of the absolute elements, the absolute value of all the elements in that row except for AII. That is why I is not equal to J. And then you add all of them up and then you find of out if the diagonal element is greater than or equal to that sum. Now here we see greater than or equal to and here we see greater than, so at least for one of the values of I, it can be 1, 2, 3, N, any of the values of I, for at least one row, that's what we mean for at least one I. So for at least one row this particular inequality is strictly greater than. That's when we will consider A matrix to be diagonally dominate. So let's look at an example and see whether a particular matrix is diagonally dominated or not. So what we are going to do is, we are going to take several examples and to make sure that is the case. So let's look at this particular matrix here and see if we can classify it as a diagonally dominated matrix. 15, 6, 7, 2, -4, -2, 3, 2, 6. So here what you are finding out is that, let's look at row 1. So an I has a 1. Row number 1. The absolute value of the diagonal element is 15. You are finding out right here. So absolute value of A 1 1 is equal to 15. Let's see what the sum of the rest of the elements are, which is A 1 2 plus A 1 3. The absolute value of the rest of the elements in that row will be the absolute value of A 1 2 and absolute of A 1 3. You are going to add those up. And what do we get? We get the absolute value of 6 plus the absolute value of 7. That is 13. So it is greater than or equal to. So let's go for I is 2. I is equal to 2. The second row, second column absolute value is absolute value -4. Which is 4. Let's see if it's greater than or equal to the rest of the elements being added together with their absolute values. So we have A 2 1, which is right here. Absolute value of that plus the absolute value of A 2 3, which is right there. A 2 3. Those are the rest of the elements in the A matrix right there. So this one will be A 1 2 is 2. Absolute value of 2. A 2 3 is -2. So that gives me 4. And 4 is greater than or equal to 4. So that inequality is satisfied. Let's go for I is equal to 3. Let's go to the 3rd row, 3rd column. That number is 6. Absolute value of 6 will be 6. Let's see whether it is greater than or equal to the sum of the absolute value of the elements which are left over. So is it greater than or equal to A 3rd row, 1st column -absolute value- which is this one. Plus A 3rd row 2nd column, which is this one here. Absolute value of 3 plus absolute value of 2 which gives 5. And 6 is greater than or equal to 5. So we have satisfied the inequality, which we had for the 1st condition and for each row the diagonal element absolute value has to be greater than the sum of the rest of the elements in that particular row. But are we meeting the strictly greater than condition? Yes we are. We are meeting the condition for the 1st row. Not for the 2nd row but for the 3rd row. But we need this strictly greater than inequality to be satisfied just for one row. So this particular matrix is diagonally dominant. Let's look at this matrix. Somebody gives us a matrix like this. 25, 5, 1, 64, 8, 1, 144, 12, 1. Somebody is saying: hey can you describe if this matrix is diagonally dominant or not? So again what I will do is I will take the absolute value of the first row and first column, which is C 1 1, which is 25 here. And now let me look at the rest of the elements of that particular row, which is this element and this element. So that is the absolute value of the first row 2nd column plus the absolute value of the 1st row 3rd column, which is 5, absolute value, plus absolute value 1 which is 6. And surely is greater than or equal to. Let's look at the 2nd row. This is I equal to 1. That's what we talked about, I equal to 1. Let's look at I equal to 2. I equal to 2 will have 2nd row and 2nd column, which is this one right here. The absolute value of 8 is 8. Let's see if this 8 is greater than the sum of the absolute elements in the rest of the row here. So that will be C 2 1, 2nd row 1st column. And then left is 2nd row 3rd column. Which is absolute value of 64 plus the absolute value of 1, which is 65. IS that greater than or equal to 65? Is 8 greater than or equal to 65? No. So C is automatically not diagonally dominant. I don't need to look at the 3rd row because it has already violated the inequality for the 2nd row. So it is not diagonally dominant. So that is the end of this segment.

Variations

The definition in the first paragraph sums entries across each row. It is therefore sometimes called row diagonal dominance. If one changes the definition to sum down each column, this is called column diagonal dominance.

Any strictly diagonally dominant matrix is trivially a weakly chained diagonally dominant matrix. Weakly chained diagonally dominant matrices are nonsingular and include the family of irreducibly diagonally dominant matrices. These are irreducible matrices that are weakly diagonally dominant, but strictly diagonally dominant in at least one row.

Examples

The matrix

is diagonally dominant because

  since  
  since  
  since   .

The matrix

is not diagonally dominant because

  since  
  since  
  since   .

That is, the first and third rows fail to satisfy the diagonal dominance condition.

The matrix

is strictly diagonally dominant because

  since  
  since  
  since   .

Applications and properties

The following results can be proved trivially from Gershgorin's circle theorem. Gershgorin's circle theorem itself has a very short proof.

A strictly diagonally dominant matrix (or an irreducibly diagonally dominant matrix[2]) is non-singular.

A Hermitian diagonally dominant matrix with real non-negative diagonal entries is positive semidefinite. This follows from the eigenvalues being real, and Gershgorin's circle theorem. If the symmetry requirement is eliminated, such a matrix is not necessarily positive semidefinite. For example, consider

However, the real parts of its eigenvalues remain non-negative by Gershgorin's circle theorem.

Similarly, a Hermitian strictly diagonally dominant matrix with real positive diagonal entries is positive definite.

No (partial) pivoting is necessary for a strictly column diagonally dominant matrix when performing Gaussian elimination (LU factorization).

The Jacobi and Gauss–Seidel methods for solving a linear system converge if the matrix is strictly (or irreducibly) diagonally dominant.

Many matrices that arise in finite element methods are diagonally dominant.

A slight variation on the idea of diagonal dominance is used to prove that the pairing on diagrams without loops in the Temperley–Lieb algebra is nondegenerate.[3] For a matrix with polynomial entries, one sensible definition of diagonal dominance is if the highest power of appearing in each row appears only on the diagonal. (The evaluations of such a matrix at large values of are diagonally dominant in the above sense.)

Notes

  1. ^ For instance, Horn and Johnson (1985, p. 349) use it to mean weak diagonal dominance.
  2. ^ Horn and Johnson, Thm 6.2.27.
  3. ^ K.H. Ko and L. Smolinski (1991). "A combinatorial matrix in 3-manifold theory". Pacific J. Math. 149: 319–336.

References

External links

This page was last edited on 21 August 2023, at 08:09
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