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Properties about vectors

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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. :)

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? 🤔

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 $v\cdot b_i$... 🤔

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 $a$. 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
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 $\displaystyle{v=\sum_{i=1}^n(v\cdot b_i)b_i}$.

🤓
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Great!! Thanks a lot!! 🥳