# What subspace of 3 x 3 matrices is spanned by the invertible matrices? Rank 1 matrices? [GStrang P181, 3.5.29(a)(b)]

What subspace of $$3$$ by $$3$$ matrices is spanned (take all combinations) by
(a) the invertible matrices? (b) the rank one matrices?

Answer: (a) The invertible matrices span the space of all $$3$$ by $$3$$ matrices.
(b) The rank one matrices also span the space of all $$3$$ by $$3$$ matrices. $$\quad \square$$

P144: The rank of a matrix is its number of pivots.
P171: A set of vectors spans a space if their linear combinations fill the space.

How'd you divine that these matrices fulfill the questions? The answers don't explain.

For (a), I recalled that invertible matrices have $$n$$ pivots (1 in each row) and so $$n$$ linearly-independent columns.

(b) Rank one matrices must've only 1 pivot. Thus, its $$n - 1$$ columns are linearly dependent. Then what?

This question precedes dimensions/theorems of the 4 subspaces, Orthogonality, Determinants, eigenvalues and eigenvectors, and linear transformations. Please pretermit them.

The standard basis for $$\mathbb{M}_{n \times n}$$ is $$\{E(i, j)\}_{1 \le i, j\le n}$$ with $$1$$ in the $$i$$th row and $$j$$th column and $$0$$ elsewhere. Call this $$S$$.

1. Is the next step rewriting, in terms of $$S$$, all (a) the invertible matrices and (b) rank one matrices.

2. Then, how'd I describe the space of all the invertible matrices (neither a subspace nor a vector space)? Since these two matrices in Deven Ware's last equation have 3 and 2 pivots respectively, they're invertible, but don't span the set of all invertible matrices? $$\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right), \left( \begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$$

3. I know that each of the $$n\cdot n$$ matrices of size $$n \times n$$ in $$S$$ are rank one, but how'd I describe the set of all rank $$1$$ matrices?

• If you know eigenvalues, for any matrix $A$ there exists a $\lambda \neq 0$ such that $A- \lambda I$ is invertible. Since $\lambda I$ is also invertible.... Nov 26, 2013 at 23:44

The standard basis is composed of matrices with a $1$ in one position and $0$'s elsewhere. It is easy to see that matrices of this form are independent and span all $M_3$. So to show both (a) and (b) it suffices to show that all of these matrices are in the space spanned by either invertible matrices or rank one matrices.

For example, if you wanted to show that
$$\left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$$ is in the space spanned by invertible matrices you could write $$\left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) = \left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right) + \left( \begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$$

You have understood slightly backwards. What you want to do is write all of $S$ in terms of (a) invertible matrices and (b) rank one matrices.

The reason is this: If you can write all of S in terms of (a) invertible matrices, then you know that you can write anything in terms of invertible matrices. Thus invertible matrices span the entire space. Similarly for rank one matrices.

• Thank you. Will you please to respond to my supplementary posted in my OP, in your answer and not as a comment? Nov 26, 2013 at 6:41
• @LePressentiment Done. Nov 26, 2013 at 22:30

Let $E_{i,j}$ be the matrix with a $1$ on the $i^{th}$ line of the $j^{th}$ column and $0$s everywhere else.

It is easy to show that the set of all $E_{i,j}$ for a basis for the space of all matrices (and we just need to prove they generate the space for this particular exercise).

All the $E_{i,j}$ have rank one so the set of all rank $1$ matrices generate the whole space.

Look at $I+E_{i,j}$. Now matter what $i$ and $j$ are, the matrix will be upper or lower triangular with non-zero diagonal elements so it will have a non-zero determinant and therefore be invertible. So both $I+E_{i,j}$ and $I$ are invertible so the subspace generated by the invertible matrices contains $E_{i,j}=(I+E_{i,j})-I$ so it's the whole space.

• Thank you. Will you please to respond to my supplementary posted in my OP, in your answer and not as a comment? Also, could you please answer without determinants? Nov 26, 2013 at 6:41
• One does not need determinants to show that a triangular matrix is invertible if and only if all of its diagonal entries are nonzero. Many introductory texts contain this as an example or an exercise. The idea is that if there is a zero entry on the diagonal you can row reduce to get a zero row; if not, you can row reduce to get the identity. Nov 27, 2013 at 11:51

There's an easier way to show that the invertible matrices span the whole space.

Let $A=[a_{ij}]\in\mathbb{M}_{n\times n}$; then define the matrices $B=[b_{ij}]$ and $C=[c_{ij}]$ by the following procedure: $$b_{ij}=\begin{cases} a_{ij} & \text{if i>j}\\ 0 & \text{if i<j}\\ a_{ii}/2 & \text{if i=j and a_{ii}\ne0}\\ 1 & \text{if i=j and a_{ii}=0} \end{cases} \qquad c_{ij}=\begin{cases} 0 & \text{if i>j}\\ a_{ij} & \text{if i<j}\\ a_{ii}/2 & \text{if i=j and a_{ii}\ne0}\\ -1 & \text{if i=j and a_{ii}=0} \end{cases}$$ Then $B$ is lower triangular with nonzero diagonal entries, while $C$ is upper triangular with nonzero triangular entries. Therefore $B$ and $C$ are invertible and, since $A=B+C$, we have that $A$ belongs to the span of the invertible matrices.

For the rank one matrices, just consider $A=\begin{bmatrix}a_1 & a_2 & \dots & a_n\end{bmatrix}$, where $a_i$ are column vectors.

For $i=1,2,\dots,n$ define $\hat{a}_i$ as the matrix which has all zero columns except for the $i$-th column, equal to $a_i$. Then $\hat{a}_i$ is either the zero matrix or has rank one. But, evidently, $$A=\hat{a}_1+\hat{a}_2+\dots+\hat{a}_n$$

In particular the set of invertible matrices is not a subspace, because there are noninvertible matrices. Nor it is the set of rank one matrices, because there is a rank zero matrix.

A rank one matrix can be described as one having at least a nonzero column (or row) and all other columns (or rows) are scalar multiples of this one. So, apart from a permutation of the columns it is of the form $$\begin{bmatrix} v & \alpha_2 v & \alpha_3 v & \dots &\alpha_n v \end{bmatrix}$$ where $v$ is a nonzero column vector.

Here is a more parsimonious approach:

Let $$A$$ be any square matrix. Note that $$A_n = A-{1 \over n}I$$ is invertible for $$n$$ sufficiently large (since $$A$$ has only a finite number of eigenvalues. Then, since $$A = A_n+{1 \over n}I$$ we see that $$A$$ lies in the span of invertible matrices.

Note that $$I = \sum_k e_k e_k^T$$, where $$e_k$$ is the $$k$$th unit vector. Each $$e_k e_k^T$$ is a rank one matrix. Then $$A= A I = \sum_k A e_k e_k^T$$ which is also the sum of rank one matrices. Hence $$A$$ lies in the span of rank one matrices.