math111_logo Composition and Matrix Multiplication


Exercise What are the matrices A satisfying AX = XA for any X?

Answer If A is an m by n matrix, then in order for both AX and XA to make sense, X has to be an n by m matrix. Then AX is an m by m matrix, and XA is an n by n matrix. Now for the two matrix to be equal, we must have m = n. In other words, A has to be a square matrix.

Now assume A is an n by n matrix. Let Eij be the n by n matrix with 1 as the ij-entry and 0 as all the other entries. For example, in case n = 3 we have

E11 = [ 1 0 0 ], E23 = [ 0 0 0 ], E31 = [ 0 0 0 ]
0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 1 0 0

Then AEij is obtained by moving the i-th column of A to j-th column, and leaving all the other columns zero. For example,

AE23 = [ a11 a12 a13 ] [ 0 0 0 ] = [ 0 0 a12 ]
a21 a22 a23 0 0 1 0 0 a22
a31 a32 a33 0 0 0 0 0 a32

And EijA is obtained by moving the j-th row of A to i-th row, and leaving all the other rows zero. For example,

E23A = [ 0 0 0 ] [ a11 a12 a13 ] = [ 0 0 0 ]
0 0 1 a21 a22 a23 a31 a32 a33
0 0 0 a31 a32 a33 0 0 0

Thus from AEij = EijA, we have

Since the above should hold for any i, j, we conclude that aij = 0 for ij, and aii = ajj for any i, j. This is the same as A = aI, which is a constant multiple of the identity.