4
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Someone voted to close this as "lacking context". I don't think that is appropriate -- the OP explains enough of how they're stuck that it is possible to write an answer that addresses their specific lack of understanding. It is not just asking for a solution that can be handed in. $\endgroup$ – Henning Makholm Apr 15 '18 at 14:47
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ We really haven't learned yet about dimensions, just span for now. To make sure I understood correctly, the three vectors become (1,1,0) , (-1, 0, 1) and (0,1,2). And from these three I could create the 3x3 matrix and then solve normally? $\endgroup$ – Kode Ch Apr 15 '18 at 15:12
  • 1
    $\begingroup$ Yes exactly! from here you can use RREF $\endgroup$ – gimusi Apr 15 '18 at 15:15
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $\{(a,b,b,c)\mid a,b,c\in\mathbb R\}$.

$\endgroup$
  • $\begingroup$ Then that would mean I would be working with an nxm matrix to find the solution? $$\begin{pmatrix} 1 & -1 & 0\\ 1 & 0 & 1\\ 1 & 0 & 1\\ 0 & 1 & 2\\ \end{pmatrix}$$ I proceed normally after here? $\endgroup$ – Kode Ch Apr 15 '18 at 14:51
  • 1
    $\begingroup$ @KodeCh: Yes, if that is the way around you would normally compute the span of three length-4 vectors. My instinct would be to look at the tanspose of your matrix, and then do ordinary Gaussuan elimination where the row operations do not change the row space (aka the span), but the other way around would also work, and I believe there are textbooks that teach it that way. $\endgroup$ – Henning Makholm Apr 15 '18 at 14:54
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.