Linear algebra

What's this thing that the spanning vectors must be independent? What's required is just that any vector in the vector space can be expressed as a linear combination of these. To require linear independence in addition is to require they are a basis for the vector space.

So what? Read the question.

But this is false. We have [math]\mathbb{R}^3 = \langle \mathbb{R}^3 \rangle[math] as a counter example. Since OP is clearly pretty lost, I don't think he should be misinformed like this.

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But this is false. We have [math]\mathbb{R}^3 = \langle \mathbb{R}^3 \rangle[/math] as a counter example. Since OP is clearly pretty lost, I don't think he should be misinformed like this.

This user may be a retard but the question mentions only three vectors that also span R^3, because its dimension is 3, it does follow that the vectors are linearly independent.

Also kys weeb faggot

That's probably what he means. Yet, he didn't say "any n vectors spanning [math]\mathbb{R^n}[/math]", but "any vectors", and the latter is false. That's the problem in his post.

203 at GMU?

caught you

I was only thinking of bases when I was posting this pls no bully

If A is a 3x3 matrix, and the columns of A span R3, then they are linearly independent.
This means that the determinant of the matrix is nonzero.
This means you can invert the matrix.
This means you can solve for X: X = inverse(A)*B

Every statistic is actually 50% because everything either happens or it doesn't.