Show that the product of two upper (lower) triangular matrices is again upper (lower) triangular.
I have problems in formulating proofs – although I am not 100% sure if this text requires one, as it uses the verb "show" instead of "prove". However, I have found on the internet the proof below but my problem is not just that I can't do one by myself, but also that it happens that I don't understand a proof which is already written – thing which let me think that I should quit these studies.
My first question is: how to choose the right strategy for proving something, in this case as in others?. I guess it is also matter of interest…Interest for numbers and their properties. I've never had such interest, frankly – although, perhaps, it is not nice to say this here. The reason for which I started studying math and some physics over an year and a half ago is because I dreamed about and still dreaming to do a kind of research for which I am interested and I need scientific background.
Sorry for the digression but I am so upset and I need some general advice too. Coming back to the topic of the present proof, my second question is: could you explain me in plain English the idea and logic behind the following proof, please? Is this proof general enough to cover also my case and what changes do I need to add for it? – I ask this because the text to which I refer speaks about upper and lower triagular matrices, while the text of the proof I report here speaks about upper triangular matrices only. I also notice that the following proof considers only square matrices but also rectangular matrices can be upper/lower triangular.
The only thing I know about all this is what upper/lower triangular matrices are and how to perform multiplication between matrices. This is the proof found on the internet:
"Suppose that U and V are two upper triangular n × n matrices. By the row-column rule for matrix multiplication we know that the (i,j)-th entry of the product UV is $ui1v1j + ui2v2j +···+ uinvnj$. We need to show that if i > j then this expression evaluates to 0. In fact, we will show that every term $uikvkj$ of this expression evaluates to 0. To prove this, we consider two cases: • If i > k then $uik = 0$ since U is upper triangular. Hence $uikvkj = 0$. • If k > j then $vkj = 0$ since V is upper triangular. Hence $uikvkj = 0$. Since i > j, for every k weeither have $i > k$ or $k > j$ (possibly both) so the set wo cases cover all possibilities for k."
Best Answer
I was interested on the same question, so allow me to exploit my logic, hopping of course to get comments for possible flaws. Suppose you have two lower triangular matrices $\mathbf{L}_1$ and $\mathbf{L}_2$ illustrated bellow
$$\mathbf{L}_1 = \begin{bmatrix} l_{11}^{(1)} & l_{12}^{(1)} & \dots & \dots & l_{1n}^{(1)} &\\ & l_{22}^{(1)} & l_{23}^{(1)} & \dots & \vdots &\\ & & l_{33}^{(1)} & & \vdots &\\ & & & \ddots & \vdots &\\ & & & & l_{nn}^{(1)} &\\ \end{bmatrix}~~~~~\mathbf{L}_2 = \begin{bmatrix} l_{11}^{(2)} & l_{12}^{(2)} & \dots & \dots & l_{1n}^{(2)} &\\ & l_{22}^{(2)} & l_{23}^{(2)} & \dots & \vdots &\\ & & l_{33}^{(2)} & & \vdots &\\ & & & \ddots & \vdots &\\ & & & & l_{nn}^{(2)} &\\ \end{bmatrix}$$
We want to prove that the following product is a lower triangular matrix,
$$\mathbf{L}_1 \mathbf{L}_2 = \mathbf{L}_1 \big[ \mathbf{l}_1, \mathbf{l}_2^{(2)}, \dots, \mathbf{l}_n^{(2)} \big] = \big[ \mathbf{L}_1 \mathbf{l}_1^{(2)}, \mathbf{L}_1 \mathbf{l}_2^{(2)}, \dots, \mathbf{L}_1 \mathbf{l}_n^{(2)} \big]$$
As we can see, the $k$-th column of product matrix $\mathbf{L}_1 \mathbf{L}_2$ is given by $\mathbf{L}_1 \mathbf{l}_k^{(2)}$ which is the linear combination of the $\mathbf{L_1}$ matrix columns with coefficients defined by the $k$-th column vecor $\mathbf{l}_k^{(2)}$. Each of the product matrix columns $\big(\mathbf{L}_1 \mathbf{l}_k^{(2)}\big)$ have possible non-zeros entries only above the $k$-th element.
This is because, the new columns are linear combinations of the first $k$ columns $\mathbf{l}_k^{(1)}$ which by their turn have possible non-zero values above their $k$-th entry. This property comes form the fact that columns $\mathbf{l}_k^{(2)}$ have zero entries after their $k$-th element.
$\mathcal{Thanks~for~reading}$.