# Basic linear algebra

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This module will give a very brief tutorial on some of the basic terms and ideas of linear algebra. These will include linear independence, span, and basis.

This brief tutorial on some key terms in linear algebra is not meant to replace or be very helpful to those of you trying togain a deep insight into linear algebra. Rather, this brief introduction to some of the terms and ideas of linear algebra ismeant to provide a little background to those trying to get a better understanding or learn about eigenvectors andeigenfunctions, which play a big role in deriving a few important ideas on Signals and Systems. The goal of theseconcepts will be to provide a background for signal decomposition and to lead up to the derivation of the Fourier Series .

## Linear independence

A set of vectors $\forall x, {x}_{i}\in \mathbb{C}^{n}\colon \{{x}_{1}, {x}_{2}, \dots , {x}_{k}\}()$ are linearly independent if none of them can be written as a linear combination of the others.

Linearly Independent
For a given set of vectors, $\{{x}_{1}, {x}_{2}, \dots , {x}_{n}\}()$ , they are linearly independent if ${c}_{1}{x}_{1}+{c}_{2}{x}_{2}+\dots +{c}_{n}{x}_{n}=0$ only when ${c}_{1}={c}_{2}=\dots ={c}_{n}=0$

We are given the following two vectors: ${x}_{1}=\left(\begin{array}{c}3\\ 2\end{array}\right)()$ ${x}_{2}=\left(\begin{array}{c}1\\ 2\end{array}\right)()$ These are linearly independent since ${c}_{1}{x}_{1}=-({c}_{2}{x}_{2})$ only if ${c}_{1}={c}_{2}=0$ . Based on the definition, this proof shows that these vectors are indeed linearly independent. Again, wecould also graph these two vectors (see [link] ) to check for linear independence.

Are $\{{x}_{1}, {x}_{2}, {x}_{3}\}()$ linearly independent? ${x}_{1}=\left(\begin{array}{c}3\\ 2\end{array}\right)()$ ${x}_{2}=\left(\begin{array}{c}1\\ 2\end{array}\right)()$ ${x}_{3}=\left(\begin{array}{c}-1\\ 0\end{array}\right)()$

By playing around with the vectors and doing a little trial and error, we will discover the followingrelationship: ${x}_{1}-{x}_{2}+2{x}_{3}=0$ Thus we have found a linear combination of these threevectors that equals zero without setting the coefficients equal to zero. Therefore, these vectors are not linearly independent !

As we have seen in the two above examples, often times the independence of vectors can be easily seen through a graph.However this may not be as easy when we are given three or more vectors. Can you easily tell whether or not thesevectors are independent from [link] . Probably not, which is why the method used in the above solution becomesimportant.

A set of $m$ vectors in $\mathbb{C}^{n}$ cannot be linearly independent if $m> n$ .

## Span

Span
The span of a set of vectors $\{{x}_{1}, {x}_{2}, \dots , {x}_{k}\}()$ is the set of vectors that can be written as a linear combination of $\{{x}_{1}, {x}_{2}, \dots , {x}_{k}\}()$ $\mathrm{span}(\{{x}_{1}, \dots , {x}_{k}\}())=\{\forall \alpha , {\alpha }_{i}\in \mathbb{C}^{n}\colon {\alpha }_{1}{x}_{1}+{\alpha }_{2}{x}_{2}+\dots +{\alpha }_{k}{x}_{k}\}()$

## Basis

Basis
A basis for $\mathbb{C}^{n}$ is a set of vectors that: (1) spans $\mathbb{C}^{n}$ and (2) is linearly independent.
Clearly, any set of $n$ linearly independent vectors is a basis for $\mathbb{C}^{n}$ .

We are given the following vector ${e}_{i}=\left(\begin{array}{c}0\\ ⋮\\ 0\\ 1\\ 0\\ ⋮\\ 0\end{array}\right)()$ where the $1$ is always in the $i$ th place and the remaining values are zero. Then the basis for $\mathbb{C}^{n}$ is $\{\forall i, i=()\}()$

1 2 n
e i
$\{\forall i, i=()\}()$
1 2 n
e i is called the standard basis .

${h}_{1}=\left(\begin{array}{c}1\\ 1\end{array}\right)()$ ${h}_{2}=\left(\begin{array}{c}1\\ -1\end{array}\right)()$ $\{{h}_{1}, {h}_{2}\}()$ is a basis for $\mathbb{C}^{2}$ .

If $\{{b}_{1}, \dots , {b}_{2}\}()$ is a basis for $\mathbb{C}^{n}$ , then we can express any $x\in \mathbb{C}^{n}$ as a linear combination of the ${b}_{i}$ 's: $\forall \alpha , {\alpha }_{i}\in \mathbb{C}\colon x={\alpha }_{1}{b}_{1}+{\alpha }_{2}{b}_{2}+\dots +{\alpha }_{n}{b}_{n}$

Given the following vector, $x=\left(\begin{array}{c}1\\ 2\end{array}\right)()$ writing $x$ in terms of $\{{e}_{1}, {e}_{2}\}()$ gives us $x={e}_{1}+2{e}_{2}$

Try and write $x$ in terms of $\{{h}_{1}, {h}_{2}\}()$ (defined in the previous example).

$x=\frac{3}{2}{h}_{1}()+\frac{-1}{2}{h}_{2}$

In the two basis examples above, $x$ is the same vector in both cases, but we can express it in many different ways (we give only two out of many, manypossibilities). You can take this even further by extending this idea of a basis to function spaces .

As mentioned in the introduction, these concepts of linear algebra will help prepare you to understand the Fourier Series , which tells us that we can express periodic functions, $f(t)$ , in terms of their basis functions, $e^{i{\omega }_{0}nt}$ .

#### Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
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Abhi
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salma
Commplementary angles
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Uday
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salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
I'm interested in nanotube
Uday
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
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Azam
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Prasenjit
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Azam
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Uday
I'm interested in Nanotube
Uday
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Prasenjit
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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