Consider \(\mathbf{a}' = \begin{pmatrix}a_1 & a_2 & \dots a_n \end{pmatrix}\)
Thus, \[\mathbf{a}'\mathbf{1}_n = \begin{pmatrix}a_1 & a_2 & \dots a_n \end{pmatrix} \times \begin{pmatrix} 1\\ 1\\ \vdots\\ 1\end{pmatrix} = a_1 + a_2 + a_3 + \dots a_n = \mathbf{a'} \mathbf{1_n} \]
Row sum of \(j^{th}\) column of \(A = \sum_{i=1}^na_{ij}\)
Column sum of \(i^{th}\) row of \(A= \sum_{j=1}^n a_{ij}\)
And hence \(AB = a_1b_1' + A_2B_2 = \begin{pmatrix} 6 & 7 & 3 & 6\\ 4 & 2 & 2 & 4\\ 2 & 3 & 1 & 2\\ \end{pmatrix}\) # Problem 3
\(A\) is \(n \times p\)
\(AA'\) is symmetric if \(AA' = A'A\)
Proof: \((AA')' = (A')'A' = AA'\) and hence \(AA'_{n \times n}\) is symmetric
Similarly, for \(A'A\) consider the following:
\((A'A)' = A'A'' = A'A\) and hence \(A'A_{p \times p}\) is symmetric too.
\((A'A)_{ij} = \sum_{k=1}^n a_{ki}a_{kj}\) and hence any diagonal element of \(A'A\) is a sum of perfect squares(the case when \(i=j\) \(\implies\) \((A'A)_{ii} = \sum_{k=1}^n a_{ki}^2\))
If \(\sum_{k=1}^n a_{ki}^2 =0\) \(\implies\) \(a_{ki} =0 \forall 1 \leq k \leq n \forall i\) Hence \(A\) is a zero matrix.
\(A_2 = \begin{pmatrix} 1 & 1 \\ 1 & 1\\ 1 & 0\\ 0 & 0 \end{pmatrix}\)
\(A_3 = \begin{pmatrix} 1 & 1 & 1\\ 1 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix}\)
$A_4 =$