Exercise 2.1.4. Consider the set of all 3 by 3 symmetric matrices and the set of all 3 by 3 lower triangular matrices, each of which are subspaces of the space of all 3 by 3 matrices. What is the smallest subspace of the space of 3 by 3 matrices that contains both of those subspaces? What is the largest subspace contained in both of those subspaces?
Answer: Any subspace must be closed under vector addition and scalar multiplication. So if a subspace of the set of 3 by 3 matrices contains both the subspace of 3 by 3 symmetric matrices and the subspace of 3 by 3 lower triangular matrices, then it has to also contain all possible linear combinations of those two subspaces. (Otherwise it wouldn’t be a vector space.)
In particular, the subspace we’re looking for has to contain all matrices of the form where is a 3 by 3 symmetric matrix and is a 3 by 3 lower triangular matrix. Now, in general symmetric matrices can have nonzero entries above the diagonal, to match corresponding nonzero entries below the diagonal. Therefore in general the sum will produce a matrix that has entries both above and below the diagonal.
The question then becomes, are there any restrictions on what can look like? Or can end up being any possible 3 by 3 matrix? To find out, we take a random 3 by 3 matrix
and try to represent it as the sum of a 3 by 3 symmetric matrix and a 3 by 3 lower triangular matrix . Since must have zeros above the diagonal (by definition), the values for , , and (the entries of above the diagonal) have to come from . We can also take the diagonal entries , , and from as well, since we can just set the diagonal entries of to be zero. (It will still be lower triangular in this case.)
Since is a symmetric matrix the entries below the diagonal must be the same as the entries above the diagonal, so we have
What then should look like? We know the entries above the diagonal are zero (because is lower triangular) and also that the diagonal entries of are zero as well (from the previous paragraph). So we have
and we have to find suitable values for , , and .
For to equal the sum of and we must have
where the first term in each sum above comes from and the second term from . Solving for , , and we have
We then have
So, for any 3 by 3 matrix we can find a 3 by 3 symmetric matrix and a 3 by 3 lower triangular matrix for which . Put another way, adding 3 by 3 symmetric matrices and 3 by 3 lower triangular matrices together can produce any possible 3 by 3 matrix. So the smallest possible subspace containing both all 3 x 3 symmetric matrices and all 3 by 3 lower triangular matrices is the space of all 3 by 3 matrices.
What about the largest subspace contained in both the subspace of 3 by 3 symmetric matrices and 3 by 3 lower triangular matrices? If a matrix is in the subspace of lower triangular matrices then we must have for . If is also in the subspace of symmetric matrices then we must also have for all and , and thus for . So we have for and is a diagonal matrix.
The sum of two diagonal matrices is itself a diagonal matrix, so the set of all diagonal matrices is closed under vector addition. If is a diagonal matrix and is a scalar then is also a diagonal matrix (even if ), so the set of all diagonal matrices is closed under scalar multiplication. The set of all 3 by 3 diagonal matrices is therefore a subspace of the space of all 3 by 3 matrices, and it is the largest subspace that is contained in both the subspace of 3 by 3 symmetric matrices and the subspace of 3 by 3 lower triangular matrices.
UPDATE: I completely rewrote the answer to the first part of the problem to provide a more complete explanation, in response to a question from a reader. (I also corrected a typo in the definition of .)
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.