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?


2 Answers 2


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).

  • $\begingroup$ Yea, now I see. I somehow missed that completely. Thank you. $\endgroup$ Sep 6, 2019 at 12:15
  • $\begingroup$ Actually as a follow up, What other assignment does it produce a dual basis? They being relatively prime for a given $i$? $\endgroup$ Sep 6, 2019 at 12:19
  • $\begingroup$ @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$. $\endgroup$ Sep 6, 2019 at 12:33
  • $\begingroup$ I mean, basis for dual space. But nevermind. Thanks for the help :) $\endgroup$ 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 $(0)a_1+ (0)a_2+ (0)a_3+ (1)a_4= a_4$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .