Chittesh14
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We have the standard inner/dot product for Euclidean space in R^n.

The standard inner/dot product of two vectors a = (a1, a2, …, an)T and b = (b1, b2, …, bn)T in R^n is defined as:

.

Here, I am defining these vectors as column vectors by saying a is the transpose of the row vector (a1, a2, …, an) and similarly for b.

But, I've come across things saying we can have an inner product e.g. on R^2 as: Let’s define, for x, yR^2: where x = (x1, x2)T and y = (y1, y2)T <x, y> = x_1.y_1 + 2x_2.y_2 as opposed to the standard inner/dot product: <x, y> = x_1.y_1 + x_2.y_2

So, I was wondering, wouldn't this cause confusion or is there something I am missing? For example, usually we say the length of a vector in R^2 e.g. (1,3)T is \sqrt{1^2 + 3^2} in agreement to the definition of the length of a vector using the inner product of R^2: ||(1,3)T|| = \sqrt {<(1,3)^T,(1,3)^T>} where ||x|| denotes the length of the vector x and in general: ||x|| = \sqrt {<x,x>}.

So, now if I use the other inner product <x, y> = x_1.y_1 + 2x_2.y_2, I get: ||(1,3)T|| = \sqrt {<(1,3)^T,(1,3)^T>} = \sqrt{1^2 + 2.(3^2)}, giving a different length.

So, is it always that the length of a vector in a vector space is ALWAYS dependent on how the inner product for the vector space is defined or given to you, and that if no SPECIFIC inner product is specified, you assume the length of the vector is calculated using the 'normal' method or standard inner product?
Last edited by Chittesh14; 10 months ago
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DFranklin
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Yes, the length will depend on the inner product you're using.

Btw: I can see significant improvement in the way you're writing things up. Good work.
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Chittesh14
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(Original post by DFranklin)
Yes, the length will depend on the inner product you're using.

Btw: I can see significant improvement in the way you're writing things up. Good work.
Thank you for your reply. I appreciate your positive comments.
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Chittesh14
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DFranklin

Hello, I was just wondering, a few days ago we had a discussion about a matrix being invertible.
What would be the chain of reasoning to show:
A matrix having a non-zero determinant implying that it has full row rank (for a square matrix, so full rank in general), so if |A| \neq 0, then A has full row rank.

Would this be good?

If a square matrix A has a non-zero determinant i.e. |A| \neq 0, this implies that A is invertible, and so the equation Ax = 0 has the trivial solution x = 0 implying the vector x has no free variables, or the general vector parametric solution of Ax = 0 doesn't contain any free parameters. As a result, we can say matrix A has full column rank and therefore as the matrix is square and the number of columns is equal to the number of rows, this implies that the matrix A has full row rank.

Or would you directly say: Since |A| ≠ 0, A is invertible and so it has full rank implying A has full row rank (and full column rank).
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DFranklin
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(Original post by Chittesh14)
DFranklin

Hello, I was just wondering, a few days ago we had a discussion about a matrix being invertible.
What would be the chain of reasoning to show:
A matrix having a non-zero determinant implying that it has full row rank (for a square matrix, so full rank in general), so if |A| \neq 0, then A has full row rank.

Would this be good?

If a square matrix A has a non-zero determinant i.e. |A| \neq 0, this implies that A is invertible, and so the equation Ax = 0 has the trivial solution x = 0 implying the vector x has no free variables, or the general vector parametric solution of Ax = 0 doesn't contain any free parameters. As a result, we can say matrix A has full column rank and therefore as the matrix is square and the number of columns is equal to the number of rows, this implies that the matrix A has full row rank.

Or would you directly say: Since |A| ≠ 0, A is invertible and so it has full rank implying A has full row rank (and full column rank).
It's very hard to answer this kind of question without knowing exactly what you've proved in your course and in what order.

But I think it would be very unlikely you proved det A \neq 0 implies A is invertible without proving A has full rank first (or during). In which case the 2nd method is better than the first.
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Chittesh14
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(Original post by DFranklin)
It's very hard to answer this kind of question without knowing exactly what you've proved in your course and in what order.

But I think it would be very unlikely you proved det A \neq 0 implies A is invertible without proving A has full rank first (or during). In which case the 2nd method is better than the first.
Thanks, I appreciate it
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