$\begingroup$

I am trying to prove that $a_1$, $a_2$, $a_3$ are linearly independent.

I am asked to use vector product and prove that if $c_{1}a_{1} + c_{2}a_{2} + c_{3}a_{3} = 0$ then $c_1 = c_2 = c_3 = 0$

I am completely stuck on where to go with this problem. I would think that linearly independent then the null space of the space can only be equal to $0$. Would this prove it?

$\endgroup$ 2

4 Answers

$\begingroup$

Dot through by a1. We get $$c_1(a1\cdot a1)=0$$ so $c_1=0$. The same holds for the other two constants. (I'm assuming that when you say orthogonal, you are not allowing any vectors to have zero magnitude.)

$\endgroup$ 2 $\begingroup$

Let $(v_1,v_2,v_3)$ be a set of orthogonal non zero vectors. Let $\alpha_1,\alpha_2,\alpha_3\in\Bbb R$ such that $$\sum_{k=1}^3 \alpha_k v_k=0$$ then $$\require{cancel}\alpha_j \cancelto{\ne0}{||v_j||^2}=\bigg\langle \sum_{k=1}^3 \alpha_k v_k,v_j\bigg\rangle=0,\;\;\forall j=1,2,3$$

Conclude.

Remark: You can generalize this result for a set of $n$ orthogonal vectors.

$\endgroup$ $\begingroup$

You are given

$$c_1a_1+c_2a_2+c_3a_3=0$$

As $a_1,a_2,a_3$ are orthogonal, we end up with the following three equalities, where $a\cdot b$ denotes the vector product between vectors $a$ and $b$.

$$(a_1\cdot c_1a_1+c_2a_2+c_3a_3)=c_1(a_1\cdot a_1)=0$$ $$(a_2\cdot c_1a_1+c_2a_2+c_3a_3)=c_2(a_2\cdot a_2)=0$$ $$(a_3\cdot c_1a_1+c_2a_2+c_3a_3)=c_3(a_3\cdot a_3)=0$$

But $(a_k\cdot a_k)>0$, for $k\in\{1,2,3\}$.

We can satisfy the above equalities only if $c_1=c_2=c_3=0$, thus proving that the set of orthogonal vectors are linearly independent.

$\endgroup$ $\begingroup$

For any orthonormal vector, we know the following 2 properties:

1. $a_i^Ta_j=0$ when $i\neq j$, because they are perpendicular to each other.
2. $a_i^Ta_j=1$ when $i=j$, because the dot product of a vector with itself is the $length^2$ of that vector, and by definition, an orthonormal vector has a length of 1.

If a matrix Q containing only orthonormal column vectors $a_1 , ..., a_n$, then from the 2 properties we have:$Q^TQ=I$. Just try a few calculations and you will know it's true!

Proofing $c_1=c_2=c_3=0$ is the same as proofing $Qx=0$ only when $x = zero\space vector$.

$Q^TQx=Q^T\times0$

$Ix=0$

$x=zero\space vector$

Thus $c_1=c_2=c_3=0$

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy