How to manually determine if a matrix is symmetric?

Question

I tried to test if the matrix is symmetric, but this doesn't work.

> View(new.function)
> View(new.function)
> new.function <- function(a){
+     for(i in 1:nrow(a)){
+         for(j in 1:ncol(a)){
+             if(a==t(a)){
+                 print("true")
+             }else{print("false")}
+         }
+     }
+ }
> a <- matrix(1:3,3,3)
> a
     [,1] [,2] [,3]
[1,]    1    1    1
[2,]    2    2    2
[3,]    3    3    3
> new.function(a)
[1] "true"
[1] "true"
[1] "true"
[1] "true"
[1] "true"
[1] "true"
[1] "true"
[1] "true"
[1] "true"

https://stat.ethz.ch/R-manual/R-patched/library/base/html/isSymmetric.html — markus, Apr 20 '20 at 20:23

M-- · Accepted Answer · 2021-09-14T14:47:23.710

There is a function in base (as mentioned in the comments) that can get you what you need.

isSymmetric.matrix(a)
 # [1] FALSE

However, if you want to write up your own function, then you should know that you don't need loops. You can check if a matrix and it's transpose are the same (all of the entries are equal) simply by using all.

new.function <- function(a) {all(a==t(a))}

or using all.equal (less efficient).

new.function <- function(a) {all.equal(b, t(b))==1}

If you insist on using loops, then you have to use the indices within your comparison. Here, I changed the way I am comparing the entries. I assume the matrix is symmetric; if one of the indices and its relevant entry from the transposed matrix (a[i,j] will be compared with t(a)[i,j] or a[j,i]) were unequal, then I exit the loops and return FALSE.

new.function <- function(a){
                 ans = TRUE
                  for(i in 1:nrow(a)){
                  for(j in 1:ncol(a)){
                      if(a[i,j]!=a[j,i]){
                         ans = FALSE;break}
                      }
                  }
                return(ans)}

But this is not efficient at all.

How to manually determine if a matrix is symmetric?

1 Answers1