Let $d1, d2, d3, d4, d5$ denote the dimensions of the subspaces of all real $29 × 29$ matrices which are diagonal, upper triangular, trace zero, symmetric, and skew symmetric, respectively. Write down the 5-tuple $(d1, d2, d3, d4, d5).$
I'm completely stuck, please help me. Hints are appreciated.
Thanks in advance
If I understand the question correctly, you work in the space of real $29\times 29$ matrices and you have to compute the respective dimensions of the subspaces which are given by matrices that are
1) diagonal
2) upper triangular
3) trace-zero
4) symmetric
5) skew-symmetric
In your notation, you denote by $d_i$ the dimension of the $i$-th subspace (as numbered above) and (for some reason) you should write the dimensions as a $5$-tuple $ (d_1, d_2, \dots, d_5)$. So let us compute.
1) Diagonal matrix is such that all non-diagonal entries of the matrix are zero. This means that the diagonal entries may be arbitrary. We have $29$ rows and $29$ columns, thus $29$ diagonal entries which are independent of each other. Hence the dimension is $$d_1 = 29 \ .$$
2) Upper triangular matrix is such that all the entries on the diagonal and below the diagonal are zero. Again, the entries above the diagonal are arbitrary real numbers. So we have to compute how many positions we have above the diagonal for a $29 \times 29$ matrix. The number of such positions is given by the following sum $$ 28 + 27 + 26 + \dots + 3 + 2 + 1$$ and corresponds to the number of free slots in the first row, second row, third row, and so on. Check for example arbitrary $4\times 4$ upper triangular matrix: $$ \begin{pmatrix} 0 & a & b & c \\ 0 & 0 & d & e \\ 0 & 0 & 0 & f \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ and the number of the above diagonal entries is $3+ 2 + 1 = 6$. For the case of $29 \times 29$ matrix, the corresponding sum may be computed by the formula $$ \frac{n}{2}(a_1 + a_n) \ ,$$ where in our case $n = 28, a_1 = 28, a_n = 1$. Thus we have $$ d_2 = 28 + 27 + 26 + \dots + 3 + 2 + 1 = \frac{24}{2}\cdot (28 + 1) = 406 \ .$$
3) Trace-zero matrix is the matrix having zero trace (does make sense right? :P) In other words, if we sum up the diagonal entries (this corresponds to computing the trace) we get zero. This implies two things: i) all the entries but one on the diagonal are arbitrary ; ii) the of the diagonal entries may be arbitrary. The later should be clear. The former is true because of the following: denote by $a_1, a_2, \dots, a_n$ the diagonal entries of a given matrix. The trace-zero condition means $$a_1 + a_2 + \dots + a_n = 0 \ .$$ Moving arbitrary $a_i$ (for simplicity the last one, $a_n$) on the right side of the equation gives $$a_1 + a_2 + \dots + a_{n-1} = -a_n \ .$$ This implies that the first $n-1$ diagonal entries may be chosen arbitrarily, forcing the last one to have the value $-(a_1 + a_2 + \dots + a_{n-1})$. Altogether we have that all but one entries of a trace-zero matrix are arbitrary, meaning the dimension of the coresponding subspace is $$d_3 = n^2 - 1 = 29^2 - 1 = 840\ ,$$ where the $n^2$ in the formula arises from the fact that each $n \times n$ matrix consists of $n^2$ entries.
4) Symmetric matrix is such that $A^t = A$ where $A^t$ is the transpose of $A$. In other words, $A$ is symmetric if the entries below and above the diagonal are the same on the corresponding places, i.e. the following condition holds (see the wiki article about the transpose of a matrix) $$a_{ij} = a_{ji} \ .$$ So the entries on the diagonal are arbitrary and the entries above the diagonal completely determine the entries below the diagonal. We have already computed the number of diagonal entries as well as the above diagonal entries (see cases 1) and 2) ). Thus the dimension of the subspace of $29 \times 29$ symmetric matrices is $$d_4 = 29 + 406 = 435 \ ,$$ where the first summand corresponds to the diagonal and the second summand to the above diagonal entries.
5) Skew-symmetric matrix (see wiki article on skew-symmetric matrices) is such that $A^t = -A$ which amounts to those matrices which has entries $a_{ij}$ satisfying the condition $$ a_{ij} = - a_{ji} \ .$$ The above condition implies
i) $a_{ii} = -a_{ii} $ which means that the diagonal entries must be zero;
ii) the elements above the diagonal completely determines the elements below the diagonal.
This means that the number of skew-symmetric matrices is the same as the number of upper triangular matrices (since only the elements above the diagonal may be arbitrary), that is $$d_5 = 406 \ . $$
Wrapping it all up, the $5$-tupple you are searching for is $$ (29,406,840 ,435 ,406) \ .$$