## 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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