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?

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    $\begingroup$ 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
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    $\begingroup$ 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


Think to a possible standard basis for the two subspaces.

Case 2-by-2 - General case


Case 2-by-2 - Symmetric


  • $\begingroup$ 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
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    $\begingroup$ 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

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.

  • $\begingroup$ 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

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