1
$\begingroup$

i) What is the dimension of the linear space $\mathbb{R}^{n\times n}$ of all $n\times n$ matrices?

ii) What is the dimension of the subspace of symmetric matrices in $\mathbb{R}^{n\times n}$?

Def: The dimension of the linear space $V$ is defined as the maximum number of linearly independent vectors in $V$.

I really don't understand att all how I can use that definition to answer i) and ii) above. How should I think through this?

$\endgroup$
  • 1
    $\begingroup$ The definition is only about vectors. Apply it to matrices by stacking the columns of the matrix into a single vector. $\endgroup$ – LinAlg Mar 20 '18 at 19:19
  • 1
    $\begingroup$ Imagine $A$ an $nxn$ matrix filled with $0$'s. Replace the $0$ in first line, first column by a $1$: matrix $A_{11}$. Then replace the zero in in$A$ line 1, column 2 by a 1: matrix $A_{12}$ and so on and so forth. How many matrices will you have? The proof of independance of this set of matrices should be rather straight-forward as should be the proof that any matrix in $\mathbb{R}^{n\times n}$ is a linear combination of the matrices in this set. $\endgroup$ – Bernard Massé Mar 20 '18 at 19:31
  • $\begingroup$ @BernardMassé I'd go all the way to $A_{1n}$, så $n-1$ different matrices? $\endgroup$ – Parseval Mar 20 '18 at 19:44
2
$\begingroup$

HINT

Think to a possible standard basis for the two subspaces.

Case 2-by-2 - General case

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}=av_1+bv_2+cv_3+dv_4=a\begin{bmatrix}1&0\\0&0\end{bmatrix}+b\begin{bmatrix}0&1\\0&0\end{bmatrix}+c\begin{bmatrix}0&0\\1&0\end{bmatrix}+d\begin{bmatrix}0&0\\0&1\end{bmatrix}$$

Case 2-by-2 - Symmetric

$$\begin{bmatrix}a&b\\b&c\end{bmatrix}=aw_1+bw_2+dw_3=a\begin{bmatrix}1&0\\0&0\end{bmatrix}+b\begin{bmatrix}0&1\\1&0\end{bmatrix}+c\begin{bmatrix}0&0\\0&1\end{bmatrix}$$

$\endgroup$
  • $\begingroup$ You lost me gimusi. Please elaborate. $\endgroup$ – Parseval Mar 20 '18 at 19:41
  • $\begingroup$ @Parseval I've added some detail. As you noted the dimension is given by the numbers of vectors of a basis. Now it easy to construct a basis by elementary matrices, once you have them you have dimension also. $\endgroup$ – gimusi Mar 20 '18 at 20:00
  • $\begingroup$ So, for $2\times 2$ matrx (general case), the dimension is $2^2$ and for an $n\times n$ it's $n^2$? The answer in the book for the i) is $\frac{n(n+1)}{2}.$ $\endgroup$ – Parseval Mar 20 '18 at 20:48
  • $\begingroup$ @Parseval For the general case yes we han $n\times n$ entries and thus we need $n\times n=n^2$ elementary matrix (imagine ordinary vectors in $\mathbb{R^{n^2}}$. How many entries there are for a symmetric matrix of order n? $\endgroup$ – gimusi Mar 20 '18 at 20:51
  • 1
    $\begingroup$ Sorry, I'm a bit tired, I was jumping back and fourth between i) and ii) without realizing it myself. Everything is clear now thank you! $\endgroup$ – Parseval Mar 20 '18 at 21:25
2
$\begingroup$

There are very basic $n\times n$ matrices consisting of a $1$ in position $ij$ and zeroes on the rest of the entries. Those allow you to write a convenient linear combinations for any generic $n\times n$ matrix.

$\endgroup$
  • $\begingroup$ Yes, but I can't write all of them and count them. What do you have in mind? $\endgroup$ – Parseval Mar 20 '18 at 19:42
  • $\begingroup$ @Parseval, there are only $n^2$ of these very simple basic matrices $\endgroup$ – janmarqz Mar 20 '18 at 19:56
  • $\begingroup$ Okay, I understand that. But the answer is $\frac{n(n+1)}{2}$. So there have to be other than those simple basic matrices to add to $n^2$? $\endgroup$ – Parseval Mar 20 '18 at 20:50
  • $\begingroup$ to count symmetric matrices observe that you only have to count diagonals from the principal diagonal up to the entry $1,n$: this sum is $n+(n-1)+(n-2)+...+2+1$ $\endgroup$ – janmarqz Mar 20 '18 at 20:57

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.