Understanding Vectors: Properties and Applications

In summary, we discussed the properties of vectors $b_i$ in a $n$-dimensional vector space $V$ and determined their linear independence and the maximum number of vectors they can have. We also showed that if $k=n$, then the set of vectors $B=(b_1, \ldots , b_n)$ forms a basis for $V$ and every element of $V$ can be written as a linear combination of these vectors. Lastly, we demonstrated that the matrix $a$ formed by these vectors is orthogonal.
  • #1
mathmari
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Hey!

Let $1\leq n\in \mathbb{N}$, $V=\mathbb{R}^n$ and $\cdot$ the standard scalar multiplication. Let $b_1, \ldots , b_k\in V$ such that $$b_i\cdot b_j=\delta_{ij}$$
  1. Let $\lambda_1, \ldots , \lambda_k\in \mathbb{R}$. Determine $\displaystyle{\left (\sum_{i=1}^k\lambda_i b_i\right )\cdot b_j}$ for$1\leq j\leq k$.
  2. Show that $b_1, \ldots , b_k$ are linear independent and that $k\leq n$.
  3. Let $k=n$. Show that $B=(b_1, \ldots , b_n)$ is a basis of $V$ and it holds that $\displaystyle{v=\sum_{i=1}^n(v\cdot b_i)b_i}$ for all $v\in V$.
  4. Let $k=n$. Show that $a=(b_1\mid \ldots \mid b_n)\in O_n$.

I have done the following:

For 1:
We have that $$\left (\sum_{i=1}^k\lambda_i b_i\right )\cdot b_j=\sum_{i=1}^k\lambda_i \left (b_i\cdot b_j\right )=\lambda_j$$ or not? :unsure: For 2:
We have that $$\sum_{i=1}^k\lambda_i b_i=0 \ \overset{\cdot b_j}{\longrightarrow} \ \left (\sum_{i=1}^k\lambda_i b_i\right )\cdot b_j=0\cdot b_j \ \overset{\text{ Question } 1.}{\longrightarrow} \ \lambda_j=0$$ for all $1\leq j\leq k$, and so $b_1, \ldots , b_k$ are linear independent.
Is this correct?

How can we show that $k\leq n$? :unsure: For 3:
We have that the vectors of $B$ are linear independent, according to question 2, and the number of vectors equals the dimension of $V$. This imply that $B$ is a basis of $V$, right?
Since $B$ is a basis of $V$, every element of $V$ can be written as a linear combination of the elements of $B$. But why is this linear combination $\displaystyle{v=\sum_{i=1}^n(v\cdot b_i)b_i}$ ? Is this because of the definition of $b_i$, i.e. that $b_i\cdot b_i=1$ ? :unsure:For 4:
To show that the matrix $a$ is orthogonal, we have to show that $a^Ta=I=aa^T$ using the definition os the vectors $b_i$, i.e. that $b_i\cdot b_i=1$ and $b_i\cdot b_j=0$ fr $i\neq j$, right? :unsure:
 
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  • #2
mathmari said:
For 1:
We have that $$\left (\sum_{i=1}^k\lambda_i b_i\right )\cdot b_j=\sum_{i=1}^k\lambda_i \left (b_i\cdot b_j\right )=\lambda_j$$ or not?

Hey mathmari!

Yep. :)

mathmari said:
For 2:
Is this correct?

How can we show that $k\leq n$?

Correct yes.

Hmm... I don't see an easy way to show that $k\le n$. :unsure:
Then again, perhaps we can use the property that a set of independent vectors in an $n$-dimensional vector space can have at most $n$ independent vectors? 🤔

mathmari said:
For 3:
We have that the vectors of $B$ are linear independent, according to question 2, and the number of vectors equals the dimension of $V$. This imply that $B$ is a basis of $V$, right?
Since $B$ is a basis of $V$, every element of $V$ can be written as a linear combination of the elements of $B$. But why is this linear combination $\displaystyle{v=\sum_{i=1}^n(v\cdot b_i)b_i}$ ? Is this because of the definition of $b_i$, i.e. that $b_i\cdot b_i=1$ ?

Yep.
Since $B$ is a basis we can write $v=\sum \lambda_j b_j$, can't we?
Suppose we use that to calculate $v\cdot b_i$... 🤔

mathmari said:
For 4:
To show that the matrix $a$ is orthogonal, we have to show that $a^Ta=I=aa^T$ using the definition os the vectors $b_i$, i.e. that $b_i\cdot b_i=1$ and $b_i\cdot b_j=0$ fr $i\neq j$, right?

Yep.
$a^T$ has each vector $b_i$ as a row doesn't it?
And $a$ has each $b_i$ as a column.
So if we calculate $a^Ta$ we multiply indeed each $b_i$ row in $a^T$ with each $b_j$ column in $a$. 🤔
 
  • #3
As for 3:
SInce $B$ is a basis of $V$ then we can write $\displaystyle{v=\sum_{i=1}^n\lambda_ib_i}$.
Then we get $$v\cdot b_j=\left (\sum_{i=1}^n\lambda_ib_i\right )\cdot b_j=\lambda_j$$
That means that the linear combination can be written as $\displaystyle{v=\sum_{i=1}^n(v\cdot b_i)b_i}$.

🤓
 
  • #4
Yep. :cool:
 
  • #5
Great! Thanks a lot! 🥳
 

Related to Understanding Vectors: Properties and Applications

What are vectors?

Vectors are mathematical objects that have both magnitude (size) and direction. They are commonly represented as arrows in a coordinate system.

What are the properties of vectors?

The properties of vectors include magnitude, direction, addition, subtraction, scalar multiplication, and dot product. Vectors can also be represented in different coordinate systems such as Cartesian, polar, or cylindrical.

How are vectors added?

Vectors are added by adding their corresponding components. For example, if vector A has components (3, 2) and vector B has components (1, 4), the resulting vector C would have components (3+1, 2+4) = (4, 6).

What is scalar multiplication of vectors?

Scalar multiplication of vectors involves multiplying a vector by a scalar (a number). This operation results in a new vector with the same direction as the original, but with a different magnitude. For example, multiplying a vector by 2 would result in a vector with twice the magnitude of the original.

What is the dot product of vectors?

The dot product of two vectors is a scalar value that represents the projection of one vector onto the other. It is calculated by multiplying the corresponding components of the two vectors and then adding them together. The dot product can be used to find the angle between two vectors and to determine if they are perpendicular.

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