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.
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
Section: Chapter Questions
Problem 1RQ
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Question
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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd7266660-79aa-4d28-9ec6-518064a4e7b2%2Fcc81ed55-b7f5-4616-b672-7bdab8cb666f%2Fcwuht7h_processed.png&w=3840&q=75)
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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