# Proof explanation: Theorem 2, Section 15, Dual Spaces, Halmos

This theorem claims, that there is a $$X'$$ which a dual basis of $$X$$ such that $$[x_i, y_i] = \delta_{ij}$$ and subsequently dual space is n-dimensional if the space is n-dimensional.

But I am confused at the proof step that, $$0 = \sum _j a_j[x_i, y_j] = \sum _j a_j \delta_j = a_i$$

How did that suddenly become $$a_i$$, If the proof setup claimed that we choose $$\delta_{ij} = 1$$ iff $$i == j$$ otherwise $$0$$ then I am convinced, but Halmos didn't assume anything about $$\delta_{ij}$$. I am confused, Can anyone help me here?

In standard notations $$\delta_{ij} =1$$ if $$i=j$$ and $$0$$ if $$i\neq j$$.

This notation, called Kronecker delta, has in fact been introduced by Halmos in Section 7 (Bases).

• Yea, now I see. I somehow missed that completely. Thank you. Sep 6, 2019 at 12:15
• Actually as a follow up, What other assignment does it produce a dual basis? They being relatively prime for a given $i$? Sep 6, 2019 at 12:19
• @VinothkumarRaman By definition $(y_i)$ is a dual basis for $(x_i)$ id $[x_i,y_j]=1$ for $i=j$ and $0$ for $i\neq j$. Sep 6, 2019 at 12:33
• I mean, basis for dual space. But nevermind. Thanks for the help :) Sep 6, 2019 at 12:52

Of course this will make no sense if you do not know what $$\delta_{ij}$$ means! (Surely it is defined in your text?)

$$\delta_{ij}$$ is defined to be $$0$$ if $$i\ne j$$ and 1 if $$i= j$$. So if our space is 4 dimensional, then $$\sum_{j= 1}^4 \delta_{ij}a_j= \delta_{i1}a_1+ \delta_{i2}a_2+ \delta_{i3}a_3+ \delta_{i4}a_4$$.

If i= 1, that is $$(1)a_1+ (0)a_2+ (0)a_3+ (0)a_4= a_1$$

If i= 2 that is $$(0)a_1+ (1)a_2+ (0)a_3+ (0)a_4= a_2$$

If i= 3, that is $$(0)a_1+ (0)a_2+ (1)a_3+ (0)a_4= a_3$$

If i= 4, that is $$(0)a_1+ (0)a_2+ (0)a_3+ (1)a_4= a_4$$