alVerify that $X^TX=\sum^3_{i=1}X_iX^T_i$ letting $X=\begin{bmatrix} 1&2 \\ 2&5 \\ 0&3\end{bmatrix}$ and $X^T_i$ be the ith row of X.

Question

Verify that $X^TX=\sum^3_{i=1}X_iX^T_i$ letting $X=\begin{bmatrix} 1&2 \\ 2&5 \\ 0&3\end{bmatrix}$ and $X^T_i$ be the ith row of X.

We were actually asked to verify this using an application called gretl which i tried but i'm not sure this is even feasible and like i said thought the machine didn't understand.

So $X^TX$ is easy enough $$X^TX=\begin{bmatrix} 1&2&0 \\ 2&5&3 \end{bmatrix} \begin{bmatrix} 1&2 \\ 2&5 \\ 0&3\end{bmatrix} = \begin{bmatrix} 5&12 \\ 12&38 \end{bmatrix} $$

So we just gotta prove that the sum $X_1X^T_1+X_2X^T_2+X_3X^T_3$ is the same

But you can't add things that have different dimensions!!! $$\begin{bmatrix} 1&2 \end{bmatrix} \begin{bmatrix} 1\\ 2 \end{bmatrix}+\begin{bmatrix} 1&2 \\ 2&5 \end{bmatrix} \begin{bmatrix} 1&2 \\ 2&5 \end{bmatrix} +\begin{bmatrix} 1&2 \\ 2&5 \\ 0&3\end{bmatrix} \begin{bmatrix} 1&2&0 \\ 2&5&3 \end{bmatrix}$$

$$5+\begin{bmatrix} 5&12 \\ 12&29 \end{bmatrix}+\begin{bmatrix} 5&12 \\ 12&38 \end{bmatrix}=n/a$$

Answer 1

So catalina you mean $$XX^T = \begin{bmatrix} 1&2 \end{bmatrix} \begin{bmatrix} 1\\ 2 \end{bmatrix}+ \begin{bmatrix} 2&5 \end{bmatrix} \begin{bmatrix} 2\\ 5 \end{bmatrix}+ \begin{bmatrix} 0&3 \end{bmatrix} \begin{bmatrix} 0\\ 3 \end{bmatrix} = 5+29+9=43$$

Which is not \begin{bmatrix} 5&12 \\ 12&38 \end{bmatrix}

What did i do wrong?