Question #204180

2 Let A [ 1 𝑖 −𝑖 2 ] And let g be the form (on the space of 2x1 complex matrices) defined by g(X,Y) =Y*AX.Is g an inner product ? 


Expert's answer

Let X=(x1,x2),Y=(y1,y2)X=(x_1,x_2), Y=(y_1,y_2). Then g(X,Y)=Y∗AX=x1yˉ1+2x2yˉ2+i(x2yˉ1−x1yˉ2)g(X,Y)=Y^*AX=x_1\bar{y}_1+2x_2\bar{y}_2+i(x_2\bar{y}_1-x_1\bar{y}_2).

1) Check that g(X,Y)g(X,Y) is a sesquilinear form:

g(X+X′,Y)=Y∗A(X+X′)=Y∗AX+Y∗AX′=g(X,Y)+g(X′,Y)g(X+X',Y)=Y^*A(X+X')=Y^*AX+Y^*AX'=g(X,Y)+g(X',Y)

g(X,Y+Y′)=(Y∗+Y′∗)AX=Y∗AX+Y′∗AX=g(X,Y)+g(X,Y′)g(X,Y+Y')=(Y^*+Y'^*)AX=Y^*AX+Y'^*AX=g(X,Y)+g(X,Y')

g(cX,Y)=Y∗A(cX)=cY∗AX=cg(X,Y)g(cX,Y)=Y^*A(cX)=cY^*AX=cg(X,Y),

g(X,cY)=(cY)∗AX=cˉY∗AX=cˉg(X,Y)g(X,cY)=(cY)^*AX=\bar{c}Y^*AX=\bar{c}g(X,Y).

2) Positivity: g(X,X)=∣x1∣2+2∣x2∣2≥0g(X,X)=|x_1|^2+2|x_2|^2\geq 0 and g(X,X)>0g(X,X)>0 , if X≠(0,0)X\ne(0,0).

3) Symmetry: g(Y,X)=y1xˉ1+2y2xˉ2+i(y2xˉ1−y1xˉ2)=g(X,Y)‾g(Y,X)=y_1\bar{x}_1+2y_2\bar{x}_2+i(y_2\bar{x}_1-y_1\bar{x}_2)=\overline{g(X,Y)}

Since all of the conditions to be a hermitean inner product are satisfied, we conclude that g(X,Y)g(X,Y) is a hermitean inner product on the space of 2x1 complex matrices.



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