# Vectors and invertible matrix

• Oct 25th 2009, 11:50 AM
noles2188
Vectors and invertible matrix
Consider two vectors v1 and v2 in R^n. They form the matrix G =

[(v1 dot v1) (v1 dot v2)]
[(v2 dot v1) (v2 dot v2)]. For which choices of v1 and v2 is the matrix G invertible?
• Oct 25th 2009, 03:31 PM
Jester
Quote:

Originally Posted by noles2188
Consider two vectors v1 and v2 in R^n. They form the matrix G =

[(v1 dot v1) (v1 dot v2)]
[(v2 dot v1) (v2 dot v2)]. For which choices of v1 and v2 is the matrix G invertible?

As long as $\displaystyle ||v_1 \times v_2|| \ne 0$.

The determinant of your matrix is

$\displaystyle (v_1 \cdot v_1)(v_2 \cdot v_2) - (v_1 \cdot v_2)^2 \ne 0$ for invertibility.

Thus,

$\displaystyle ||v_1|||^2 ||v_2||^2 - ||v_1||^2 ||v_2||^2 \cos^2 \theta \ne 0$

$\displaystyle = ||v_1||^2 ||v_2||^2 \sin^2 \theta \ne 0$

$\displaystyle = ||v_1 \times v_2||^2 \ne 0$
• Oct 25th 2009, 03:38 PM
tonio
Quote:

Originally Posted by noles2188
Consider two vectors v1 and v2 in R^n. They form the matrix G =

[(v1 dot v1) (v1 dot v2)]
[(v2 dot v1) (v2 dot v2)]. For which choices of v1 and v2 is the matrix G invertible?

The matrix G is NOT invertible iff its rows are linearly independent iff the second row is a multiple scalar of the first one iff $\displaystyle v_2\cdot v_1=k(v_1\cdot v_1)=kv_1\cdot v_1\,,\,\,v_2\cdot v_2=k(v_1\cdot v_2)=kv_1\cdot v_2$ $\displaystyle \Longrightarrow (v_2-kv_1)\cdot v_1=0\,,\,\,(v_2-kv_1)\cdot v_2=0$

So if both vectors are non-zero then the above means that it must be...

Tonio