## 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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