Linear Algebra and Its Applications, Review Exercise 2.9

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 (a_{ij} > 0 for all i and j)?

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:

\begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \qquad \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}

If we add these two matrices

\begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} + \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}

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

P_1 = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \qquad P_2 = \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}

with P_1 equal to the identity matrix, i.e., leaving the order of rows unchanged.

Linear combinations of P_1 and P_2 are of the form:

c_1P_1 + c_2P_2

= c_1 \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} + c_2 \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}

= \begin{bmatrix} c_1&c_2 \\ c_2&c_1 \end{bmatrix}

Thus the space spanned by P_1 and P_2 is all 2 by 2 matrices for which a_{11} = a_{22} and a_{12} = a_{21}.

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

E_1 = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \qquad E_2 = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}

E_3 = \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix} \qquad E_4 = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}

Linear combinations of E_1 through E_4 are of the form:

c_1E_1 + c_2E_2 + c_3E_3 + c_4E_4

= c_1 \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} + c_2 \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}

+ c_3 \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix} + c_4 \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}

= \begin{bmatrix} c_1&c_2 \\ c_3&c_4 \end{bmatrix}

Thus the space spanned by E_1 through E_4 is the vector space of all 2 by 2 matrices.

Now clearly E_1 through E_4 are not positive matrices, since they contain zero entries. However each of E_1 through E_4 can be expressed as linear combinations of positive matrices as follows:

E_1 = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 2&1 \\ 1&1 \end{bmatrix} - \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix}

E_2 = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 1&2 \\ 1&1 \end{bmatrix} - \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix}

E_3 = \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix} = \begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix} - \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix}

E_4 = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 1&1 \\ 1&2 \end{bmatrix} - \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix}

Since each of E_1 through E_4 can be expressed as a linear combination of positive matrices, and any 2 by 2 matrix can be expressed as a linear combination of E_1 through E_4, 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:

A_1 = \begin{bmatrix} 2&1 \\ 1&1 \end{bmatrix} \qquad A_2 = \begin{bmatrix} 1&2 \\ 1&1 \end{bmatrix}

A_3 = \begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix} \qquad A_4 = \begin{bmatrix} 1&1 \\ 1&2 \end{bmatrix}

A_5 = \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix}

However since the space of 2 by 2 matrices is spanned by E_1 through E_4 the dimension of the space is only 4. We therefore conclude that the matrices A_1 through A_5 are linearly dependent and that one of them can be expressed as a linear combination of the others.

If we add A_1 through A_4 we see that

A_1+A_2+A_3+A_4

= \begin{bmatrix} 2&1 \\ 1&1 \end{bmatrix} + \begin{bmatrix} 1&2 \\ 1&1 \end{bmatrix} + \begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix} + \begin{bmatrix} 1&1 \\ 1&2 \end{bmatrix}

= \begin{bmatrix} 5&5 \\ 5&5 \end{bmatrix} = 5A_5

We therefore have A_5 = \frac{1}{5} \left(A_1+A_2+A_3+A_4\right). Note that this implies that A_1 through A_4 by themselves (i.e., without A_5) span the space of 2 by 2 matrices. We also know that A_1 through A_4 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 A_1 through A_4 therefore form a basis for the space of 2 by 2 matrices, an alternative basis to E_1 through E_4.

Second, note that any 2 by 2 matrix is equivalent to a vector in \mathbb{R}^4; thus, for example the matrix A_1 is equivalent to the vector v_1 = \left(2, 1, 1, 1\right) and similarly for A_2 through A_4. 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 \mathbb{R}^4.

(We can extend this argument to show that the set of positive vectors in \mathbb{R}^N spans \mathbb{R}^N, and to derive a basis for \mathbb{R}^N consisting of N positive vectors, analogous to A_1 through A_4 above. However we leave that as an exercise for the reader.)

d) Consider the following two invertible matrices:

\begin{bmatrix} 1&1 \\ 1&0 \end{bmatrix} \qquad \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}

If we subtract the second matrix from the first we have

\begin{bmatrix} 1&1 \\ 1&0 \end{bmatrix} - \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} = E_1

where E_1 is one of the matrices from part (c) above.

We can similarly express each of E_2 through E_4 as the difference between two invertible matrices:

\begin{bmatrix} 1&1 \\ 0&1 \end{bmatrix} - \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} = E_2

\begin{bmatrix} 1&0\\ 1&1 \end{bmatrix} - \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix} = E_3

\begin{bmatrix} 0&1\\ 1&1 \end{bmatrix} - \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix} = E_4

So we can express each of the matrices E_1 through E_4 as a linear combination of invertible matrices. But from (c) above we know that E_1 through E_4 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 E_1 through E_4 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.

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