Column vectors as entries of a column vector
The representation in terms of vectors it is just a more compact way to express a linear system as the one you are given. $S$ in particular is the matrix of coefficients, which contains in its row and columns all the coefficients of the linear system, namely $s_{11},s_{12},...$ :
\begin{bmatrix} s_{11}\ s_{12}\ s_{13}\ ... \ s_{1n}\\ ........\\ ........\\ s_{n1}\ s_{n2} \ s_{n3} \ \dots s_{nn} \end{bmatrix}
in such a way that if you multiply the matrix $S$ times the vector $\bf{B}$ you obtain the RHS of your linear system.
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Updated on February 29, 2020Comments
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Admin over 3 years
I'm having a hard time understanding notation. Because of notation, I've lost hours and hours trying to understand a simple concept. I'm going to post a picture of the pdf that I've so that you can see exactly what I'm seeing and therefore have eventually the same doubts that I've (I hope not).
As far as I understand from the explanation $b_1, b_2, \cdots, b_n$ (in bold) of the basis $B$ are column vectors.
I understand everything until the "or" (which comes after the system of equations). I don't understand this
I've a few questions that might help you to help me.
What's the meaning of putting column vectors in a column vector. This quite blows my stupid mind.
How is the linear system of equations
(before the "or") equivalent to this? Maybe a step my step explanation would clarify.
Note that I understood well the linear system of equations and everything else, except, again, the equivalence between the linear system of equations and the equation that comes immediately after, which apparently should be equivalent.
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la flaca over 7 yearsI see, you're right, is it possible that text is mistaken? It seems that they wrote $b_1, \ldots ,b_n$ instead of $v_1, \ldots ,v_n$ when they defined the basis $B$. Coordinates are usually represented with the letters $a,b,c$ and vectors with $v,w,u$, so i think they just put it the other way round at the begining of the text
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Silvia over 7 yearsAgain they tried to write it in a compact form. Just try to look at the first line of your linear system $ \bf{\tilde{b}}=s_{11} \bf{b_1} +\dots s_{1n} \bf{b_n}$ : this means that each of the entry of $ \bf{\tilde{b}}$ is a linear combination of the entries of the vectors of the other basis $\bf{b}$. If you want to write it explicitly each line of the system should be expanded to include the $\tilde{n}$ entries of each $ \bf{\tilde{b_1}}, \bf{\tilde{b_2}}, \bf{\tilde{b_n}}$. But you agree with me that it is quite tedious to write, that is why they work with vectors..
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la flaca over 7 yearsActually i don't think it has any meaning to put elements of $R^n$ as entries of a matrix (in this case of a column vector), entries of a matrix should be elements of a field
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Silvia over 7 yearsSo basically you don't have to multiply the S matrix times a' vector of vector' just consider the vector $\bf {b}$ as normal entrie of the a vector.
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Silvia over 7 yearsSo the formalism is the same, in a sense that whenever you a have a linear system e.g. $\tilde{x}= ax +by+cz, \tilde{y}= a'x +b'y+c'z, \tilde{z}= a''x +b''y+c''z$ you ca always write it down as $ \bf {\tilde{v}}= \bf A \bf v$ where ${\bf{\tilde{v}}}=(\tilde{x},\tilde{y}, \tilde{z})$ and ${\bf{v}}=(x,y,z)$ and A contains the coefficients a,b,c... In your case the only difference is that $x,y,z$, and $\tilde{x},\tilde{y}, \tilde{z}$ are vectors (it's a linear system of vectors instead of simple variables and the solution would be a vector in stead of a single value..). I hope this helps!
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Silvia over 7 yearsYes, it is correct. I didn't know which was the formula you were referring to in the other resource before. Anyway as a general advice try to multiply always the matrices and vector to see if you can recover the original linear system.