Identiy matrix with lambda

2018-03-12 20:20:57

I need some help with my homework in a subject called "Matrices in statistics".

The task is to show that

$$ P_{ij}(\lambda)^{-1} = P_{ij}\left(\frac{1}{\lambda}\right), $$ where $P_{ij} $ is an identity matrix, where the value of i-th row and j-th column is $\lambda.$

Here's my initial solution (which wasn't enough for the teacher):

--

We need to show, that $$ P_{ij}\left(\frac{1}{\lambda}\right) P_{ij}({\lambda}) =

P_{ij}({\lambda}) P_{ij}\left(\frac{1}{\lambda}\right) = I_n $$

So:

1) $$ P_{ij}\left(\frac{1}{\lambda}\right) P_{ij}({\lambda}) = P_{ij}\left(\frac{1}{\lambda} \cdot\lambda\right) = P_{ij}(1) = I_n$$

And

2) $$ P_{ij}({\lambda})P_{ij}\left(\frac{1}{\lambda}\right) = P_{ij}\left(\lambda\cdot\frac{1}{\lambda}\right) = P_{ij}(1) = I_n $$

Q.E.D

I submitted this solution, but my lecturer gave it back and wrote, that I need to supplement this solution. More precisely, I need to show, that

$$ \left(P_{ij}(\lambda)P_{ij}\left(\frac{1}{\lambda}

  • Our goal is to show that $P_{ij}(\lambda) P_{ij} (\dfrac{1}{\lambda}) = P_{ij} (\dfrac{1}{\lambda})P_{ij}(\lambda) = I_n$. To do this, we use the definition of matrix multiplication. That is, for $n \times n$ matrices $A$ and $B$, the product $AB$ of these matrices is the matrix $C$ with $ij$ element

    $$

    C_{ij} = \sum_{k = 1}^n A_{ik} B_{kj}

    $$

    If we let $A = P_{ij}(\lambda)$ and $B = P_{ij} (\dfrac{1}{\lambda})$ in this definition, we see that if $i \neq j$, then

    $$

    A_{ik} B_{kj} = 0

    $$

    for all $k$, since these matrices are zero everywhere except for on the diagonal. This tells us that the sum

    $$

    \sum_{k = 1}^n A_{ik} B_{kj}

    $$

    is equal to zero if $i \neq j$. That is, the product $AB = P_{ij}(\lambda) P_{ij} (\dfrac{1}{\lambda})$ is a matrix $C$ with $C_{ij} = 0$ for all $i \neq j$.

    Now, if we have $i = j$, then we see that

    $$

    \sum_{k = 1}^n A_{ik} B_{kj} = 1

    $$

    since $A_{ik} B_{kj}$ is zero everywhere except when $k = i = j$. When $k = i = j$, then we have that $A_{ik}B_

    2018-03-12 21:40:36