Review exercise 2.9. Answer the following questions for the vector space of 2 by 2 matrices:

a) Does the set of 2 by 2 matrices with rank 1 form a subspace?

b) What is the subspace spanned by the 2 by 2 permutation matrices?

c) What is the subspace spanned by the 2 by 2 matrices with all positive entries ( for all and )?

d) What is the subspace spanned by the 2 x 2 matrices that are invertible?

Answer: a) Consider the following two matrices of rank 1:

If we add these two matrices

we obtain a matrix with rank 2. So the set of rank 1 matrices is not closed under addition and is therefore not a subspace.

b) The two 2 by 2 permutation matrices are

with equal to the identity matrix, i.e., leaving the order of rows unchanged.

Linear combinations of and are of the form:

Thus the space spanned by and is all 2 by 2 matrices for which and .

c) To start answering this question we begin with the following matrices:

Linear combinations of through are of the form:

Thus the space spanned by through is the vector space of all 2 by 2 matrices.

Now clearly through are not positive matrices, since they contain zero entries. However each of through can be expressed as linear combinations of positive matrices as follows:

Since each of through can be expressed as a linear combination of positive matrices, and any 2 by 2 matrix can be expressed as a linear combination of through , we conclude that any 2 by 2 matrix can be expressed as a linear combination of positive matrices. The set of positive matrices therefore spans the space of 2 by 2 matrices.

Two final notes: First, from above it is clear that the following five positive matrices span the space of 2 by 2 matrices:

However since the space of 2 by 2 matrices is spanned by through the dimension of the space is only 4. We therefore conclude that the matrices through are linearly dependent and that one of them can be expressed as a linear combination of the others.

If we add through we see that

We therefore have . Note that this implies that through by themselves (i.e., without ) span the space of 2 by 2 matrices. We also know that through are linearly independent: if they were linearly dependent then there would be at most three linearly independent matrices in the set, and three linearly independent matrices would not be sufficient to span the 4-dimensional space of 2 by 2 matrices. The matrices through therefore form a basis for the space of 2 by 2 matrices, an alternative basis to through .

Second, note that any 2 by 2 matrix is equivalent to a vector in ; thus, for example the matrix is equivalent to the vector and similarly for through . Since the set of positive matrices spans the space of 2 by 2 matrices, we can also conclude that the set of positive vectors spans .

(We can extend this argument to show that the set of positive vectors in spans , and to derive a basis for consisting of positive vectors, analogous to through above. However we leave that as an exercise for the reader.)

d) Consider the following two invertible matrices:

If we subtract the second matrix from the first we have

where is one of the matrices from part (c) above.

We can similarly express each of through as the difference between two invertible matrices:

So we can express each of the matrices through as a linear combination of invertible matrices. But from (c) above we know that through span the entire space of 2 by 2 matrices. Therefore the set of invertible 2 by 2 matrices also spans the entire space of 2 by 2 matrices.

UPDATE: I corrected the answer to (c); in my original answer I mistakenly identified through as positive matrices. (They are instead merely non-negative.)

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.