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# 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):

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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