Prove that diagonally dominated matrices are always invertible. Now, what is a diagonally dominated matrix? It is a square matrix that has each of it’s diagonal values larger in magnitude than all the other values in the rows combined.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Prove that diagonally dominated matrices are always invertible. Now, what is a diagonally dominated matrix? It is a square matrix that has each of it’s diagonal values larger in magnitude than all the other values in the rows combined.

 

To be mathematically precise, for every diagonal value a;,i,
lai,i| > Eitj lai,j| ...(1)
|ai,il > Jai,i[+ |ai,2| + ...
For example,
[3
+ |ai,i-1|+ |ai,i+1|+....
+ lai,n|
1
1
5 2
1
1
3
1
2
6.
-3
1
1
3
1
1
4
Hints:
• Remember that, a matrix A is invertible if the only solution to Ax = 0 is the 0 vector, i.e.
x = 0.
Suppose, for a DDM (diagonally dominated matrix), there exists a non-zero solution to
|x1
X2
Ax
= 0. Since x =
Let's say, x; has the greatest magnitude here which means |x;|> |xj| for all j #i . .(2).
• Since, Ax = 0, we must have,
Ai,1X1+ ai,2X2+ • · ·
In other words,
+ ai,nxn = 0 for the i’th row.
ai,1x1+ a¿.2x2+ • · ·
+ ai,i–1xi–1+ai,i+1Xi+1 +·…+ ai,n Xn = -ai,ix;
or,
|ai,1x1 + ai,2X2 + · · ·+ ai,i-1Xi-1+ai,i+1Xi+1+•• •
+ ai,nºn| = |ai,iXi|
TEitj di,j¤j| = |a;,i||¤||
But we will prove that, this is not possible when x is non-zero.
itj
So, we have proved that, |Eit; ai,ja;| < |ai,iX;|
Explanation:
IEit; ai,jxj| < Eit; lai,j®;| because |æ + y| < |x| + ly| for all real numbers r, y, z.
Eitj lai,jaj| = Eit¡lai,j||x;| because |xy| = |x||y| for all real r, y.
Eits laij||a;| < (it; lai,j)|#:| using (2).
(Eitj lai,j))|æ:| < |a;,i||¤:| using (1)
Transcribed Image Text:To be mathematically precise, for every diagonal value a;,i, lai,i| > Eitj lai,j| ...(1) |ai,il > Jai,i[+ |ai,2| + ... For example, [3 + |ai,i-1|+ |ai,i+1|+.... + lai,n| 1 1 5 2 1 1 3 1 2 6. -3 1 1 3 1 1 4 Hints: • Remember that, a matrix A is invertible if the only solution to Ax = 0 is the 0 vector, i.e. x = 0. Suppose, for a DDM (diagonally dominated matrix), there exists a non-zero solution to |x1 X2 Ax = 0. Since x = Let's say, x; has the greatest magnitude here which means |x;|> |xj| for all j #i . .(2). • Since, Ax = 0, we must have, Ai,1X1+ ai,2X2+ • · · In other words, + ai,nxn = 0 for the i’th row. ai,1x1+ a¿.2x2+ • · · + ai,i–1xi–1+ai,i+1Xi+1 +·…+ ai,n Xn = -ai,ix; or, |ai,1x1 + ai,2X2 + · · ·+ ai,i-1Xi-1+ai,i+1Xi+1+•• • + ai,nºn| = |ai,iXi| TEitj di,j¤j| = |a;,i||¤|| But we will prove that, this is not possible when x is non-zero. itj So, we have proved that, |Eit; ai,ja;| < |ai,iX;| Explanation: IEit; ai,jxj| < Eit; lai,j®;| because |æ + y| < |x| + ly| for all real numbers r, y, z. Eitj lai,jaj| = Eit¡lai,j||x;| because |xy| = |x||y| for all real r, y. Eits laij||a;| < (it; lai,j)|#:| using (2). (Eitj lai,j))|æ:| < |a;,i||¤:| using (1)
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