No, it can not be zero. The "extreme" case is when you start with a diagonal matrix. In this case all circles have diameter zero, so contain exactly their midpoint = the entry on the associated diagonal. I hope that makes some sense!
Glad that you like the theorem (and maybe the video). Gershgorin's circle theorem is one of these theorems where I fall for the hindsight bias: "I could have done that myself!". That is of course not true - hindsight bias. But it illustrates why this theorem is so great: the statement and proof are surprisingly simple and one wonders why no one has done that earlier than the 1930s. In short, easy statement, beautiful mathematics.
Yes, it is very surprising. As an heuristic why this is supposed to be true: "Every" matrix is diagonalizable, and for a diagonal matrix the theorem is clear. Now continuously move from the diagonal form to any other incarnation of the matrix. The eigenvalues will stay the same, but the diagonal entries will change and at the same time the off-diagonal ones will change at about the same rate. The theorem describes the worst case how these two changes could differ.
@@VisualMath Yes, I was imagining an "almost" diagonal matrix with non-diagonal elements approaching zero as you said. The actual computation is still really surprisingly simple!
Sir can you give me an example to show that converse of pythagorean doesnot holds in case of complex space?
I do not understand the question, sorry. What does this have to do with Gershgorin circles?
@@VisualMath sorry this question is related to functional analysis
@@saffiaayub838 No worries! But I also can't help you - I do not know!
Can a greschgorin disc be empty??
No, it can not be zero. The "extreme" case is when you start with a diagonal matrix. In this case all circles have diameter zero, so contain exactly their midpoint = the entry on the associated diagonal. I hope that makes some sense!
Alright
@@saffiaayub838 Welcome!
Thanks for showing me
Glad that you like the theorem (and maybe the video).
Gershgorin's circle theorem is one of these theorems where I fall for the hindsight bias: "I could have done that myself!". That is of course not true - hindsight bias. But it illustrates why this theorem is so great: the statement and proof are surprisingly simple and one wonders why no one has done that earlier than the 1930s.
In short, easy statement, beautiful mathematics.
That. Is. Weird.
Yes, it is very surprising. As an heuristic why this is supposed to be true: "Every" matrix is diagonalizable, and for a diagonal matrix the theorem is clear. Now continuously move from the diagonal form to any other incarnation of the matrix. The eigenvalues will stay the same, but the diagonal entries will change and at the same time the off-diagonal ones will change at about the same rate. The theorem describes the worst case how these two changes could differ.
@@VisualMath Yes, I was imagining an "almost" diagonal matrix with non-diagonal elements approaching zero as you said.
The actual computation is still really surprisingly simple!
@@shafey Yes, I agree: That the theorem and its (not discussed) proof are surprisingly simple is spot on.