3
$\begingroup$

I was given this question in one of my interviews. I have tried a lot even after that but I am not able to get a hold of it. Can anybody help me with this.

Function $f$ has the following properties

  1. $f:$ Square Matrix -> Real
  2. $f(I_n) = 1$ // Identity Matrix of size n maps to one : independent of n
  3. If $A$ and $B$ both are square matrices of same size then $f(AB) = f(BA)$
  4. If $A$ and $B$ both are square matrices of same size then $f(mA+nB) = mf(A) + nf(B)$

Describe the family of functions $f$ that satisfies the above criteria.

The only property I was able to obtain was that $f(-A) = -f(A)$

Thanks

2 Answers 2

5

Consider the matrix units $\{E_{kj}\}$ where $E_{kj}$ is the matrix with $1$ in the $k,j$ entry and zeroes everywhere else. Note that $f$ is linear, so in particular $f(0)=0$. Then, if $k\ne j$, $f(E_{kj})= f(E_{kk}E_{kj})=f(E_{kj}E_{kk})=f(0)=0.$ When $k=j$, $ f(E_{kk})=f(E_{k1}E_{1k})=f(E_{1k}E_{k1})=f(E_{11}). $ Using 4, $ 1=f(I)=f(\sum_kE_{kk})=\sum_kf(E_{kk})=\sum_kf(E_{11})=nf(E_{11}), $ so $f(E_{kk})=1/n$ for all $k$.

Now, given any matrix $A$, $A=\sum_{kj}a_{kj}E_{kj}$. So $ f(A)=f(\sum_{kj}a_{kj}E_{kj})=\sum_{kj}a_{kj}f(E_{kj})=\frac1n\sum_{k}a_{kk}=\frac1n\,\text{Tr}(A). $

Comment: any function satisfying 1 and 4 is of the form $A\mapsto \text{Tr}(AH)$ for some matrix $H$. Properties 2 and 3 force $H=\frac1n\,I$.

  • 0
    $E_{kj}E_{kk}=0$ when $j\ne k$ (direct verification from the definition). And $f(0)=0$ because f is linear.2012-11-16
2

Denote by $A_{ij}$ the matrix with $ [A_{ij}]_{kl} = \begin{cases} 1 & k = l, i\ne k, j \ne k\\ 1 & k = j, l = i\\ 1 & l = j, k = i\\ 0 & \text{otherwise} \end{cases} $ And by $E_i$ the matrix $ [E_{ij}]_{kl} = \begin{cases} 1 & k = i, l = j\\ 0 & \text{otherwise}\end{cases} $ Then we have $A_{ij}E_{ii}A_{ij} = E_{jj}$ and $A_{ij}^2 = I_n$. As $\sum_{i=1}^n E_{ii} = I_n$, $ f(E_{ii}) = f(A_{ij}E_{ij}A_{ij}) = f(A_{ij}^2E_{jj}) = f(E_{jj}) $ and hence, by linearity $f(E_{ii}) = \frac 1n$ for $1 \le i \le n$. For $i \ne j$, we have $E_{ij}E_{ii} = 0$ and $E_{ii}E_{ij} = E_{ij}$, hence $ 0 = f(E_{ij}E_{ii}) = f(E_{ii}E_{ij}) = f(E_{ij}) $ Now let $A \in \mathrm{Mat}_n(\mathbb R)$, then we have \begin{align*} f(A) &= f\left(\sum_{i,j} a_{ij}E_{ij}\right)\\ &= \sum_{i,j} a_{ij}f(E_{ij})\\ &= \sum_i a_{ii} \frac 1n\\ &= \frac 1n \operatorname{tr}A \end{align*} That is $f = \frac 1n\operatorname{tr}$ on $n \times n$-matrices.