Determine if the three matrices span the vector space of $2\times 2$ matrices

Question

Thus far I've seen vectors and polynomials but this the first and only exercise I find that introduces matrices.

The question is as follows:

Determine whether the three matrices

$\begin{pmatrix} 1 & 1 \\ 1 & 0\\ \end{pmatrix}$, $\begin{pmatrix} -1 & 0 \\ 0 & 1\\ \end{pmatrix}$, $\begin{pmatrix} 0 & 1 \\ 1 & 2\\ \end{pmatrix}$

span the vector space of all 2x2 symmetric matrices.

I am stuck at this stage because previously I would find the matrix of the vectors or polynomials and work on that, but this time it's 3 matrices, what are the steps that I should follow to always get it right?

Answer 1

HINT

Since we are dealing with symmetric matrices $\begin{pmatrix} a & b \\ b & c\\ \end{pmatrix}$ the dimension of the space is 3 and we can consider the equivalent vectors $(a,b,c)$. Therefore to find the dimension of the subspace spanned by the three matrices let arrange each matrix as a vector row in a 3-by-3 matrix and perform the RREF.

Answer 2

When you're viewing a set of matrices as a vector space, you're ignoring the matrix multiplication, so the rectangular arrangement of the elements of each matrix is irrelevant. Eseentiall you can treat them as belonging in $\mathbb R^{nm}$. So the question is equivalent to asking whether $$\{(1,1,1,0),(-1,0,0,1),(0,1,1,2)\}$$ spans the subspace of $\mathbb R^4$ that corresponds to the symmetric matrices, which is $\{(a,b,b,c)\mid a,b,c\in\mathbb R\}$.

Answer 3

Hint:

The vector space of all s $2\times 2$ symmetric matrices has dimension $3$. To show three symmetric matrices span this vector space, you can make do with proving they're linearly independent.

Answer 4

Take an arbitrary symmetric matrix $$\begin{pmatrix} x & y \\ y & z \end{pmatrix}$$ and compute $a, b, c$ such that \begin{align*}\begin{pmatrix} x & y \\ y & z \end{pmatrix} &= a\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} + b\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} + c\begin{pmatrix} 0 & 1 \\ 1 & 2 \end{pmatrix} \\ &= \begin{pmatrix} a - b & a + c \\ a + c & b + 2c \end{pmatrix}. \end{align*} Equating entries, this produces the system of linear equations, \begin{align*} a - b &= x \\ a + c &= y \\ b + 2c &= z, \end{align*} which can be solved using whichever methods you like. Either way, we get \begin{align*} a &= -x + 2y - z \\ b &= -2x + 2y - z \\ c &= x - y + z. \end{align*} In particular, at least one solution exists regardless of the value of $x, y, z$. That is, the arbitrary symmetric matrix is spanned by the the three matrices.