Show that the unit matrix commutes with any matrix.?
Seems simple enough….
The unit matrix here is the one with 1’s across the main diagonal and 0’s everywhere else.
If you are referring to the identity matrix, I, then isn’t it in its definition that it commutes with any matrix of the same order?
I is the identity matrix of order nxn iff IxM=M=MxI, where M is any nxn matrix.
Or, may be we don’t know yet it is the identity matrix.
Then, there is some work to be done… 
Let M be a nxn matrix where aij is the element at row i column j, where i,j = 1..n, and n is an integer.
Let I be the nxn unit matrix, where bij is the element at row i column j. Thus bij = 0 for i=/=j, and aij=1 for i=j.
Element at row i column j of the matrix IxM will be
bi1xa1j + bi2xa2j + … + bi(i-1)xa(i-1)j + biixaij + bi(i+1)xa(i+1)j + … + binxanj = 0 + 1xaij + 0 = aij
Element at row i column j of the matrix MxI will be
ai1xb1j + ai2xb2j + … + ai(j-1)xb(j-1)j + aijxbjj + ai(j+1)xb(j+1)j + … + ainxbnj = 0 + aijx1 + 0 = aij
Thus, IxM = M = MxI.
Review of the Dahon Matrix for Dahon by David Griffiths
