Linear Algebra and Its Applications, Exercise 1.6.16

Exercise 1.6.16. (i) If A is an n by n symmetric matrix, how many entries of A can be chosen independently of each other?

(ii) If If K is an n by n skew-symmetric matrix, how many entries of K can be chosen independently?

Answer: (i) Since A is an n by n matrix, it has $n^2$ entries. We can select n independent entries to go on the diagonal, leaving $n^2 - n$ off-diagonal entries to be chosen. However since A is a symmetric matrix only half of those entries (e.g., those above the diagonal) can be chosen independently, since the corresponding entries in the other part of the matrix (in this case, below the diagonal) will be the same.

The total number of independent entries is then

$n + (n^2 - n)/2 = (2n + n^2 - n)/2 = (n^2 + n)/2 = n(n+1)/2$

(ii) Even though K is skew-symmetric rather than symmetric, the entries below the diagonal are still dependent on the corresponding entries above the diagonal. (They’re simply negatives of those entries rather than being the same.) The number of off-diagonal entries that can be chosen independently is thus the same as for A.

However for a skew-symmetric matrix $K = -K^T$ we must have $k_{ii} = -k_{ii}$ for all $i$ , so that all the entries on the diagonal must be zero. Since the entries on the diagonal cannot be chosen independently, the number of independent entries is just half the number of off-diagonal entries, or

$(n^2 - n)/2 = n(n-1)/2$

UPDATE: Corrected the answer to part (ii) to account for the diagonal entries being zero.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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2 Responses to Linear Algebra and Its Applications, Exercise 1.6.16

Daniel says:

May 22, 2012 at 10:38 pm

In the second part (ii) of the exercise 1.6.16 the number of entries that can be chosen independently is n*(n-1)/2 because the entries of the diagonal are cero. Thanks a lot for your post they are really handy.

- hecker says:
  
  May 22, 2012 at 11:05 pm
  
  You are absolutely right, I forgot about the diagonal entries being constrained. Thanks for the correction; I’ve updated the post to reflect the correct answer.