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September 22

How can we briefly characterize a given set of vectors, as "linearly dependent - every proper sub set being linearly independent", while we only refer to the vectors rather than to their set?

For example: S is a set of the following vectors:

A=(1,1,0),

B=(1,0,0),

C=(0,1,0).

Note: A=B+C, and B=A-C, and C=A-B, so the set S is linearly dependent.

Using A,B,C only, i.e without using S, what's the shortest description, claiming that the set S is linearly dependent but every proper sub set of S is linearly independent? HOTmag (talk) 13:30, 22 September 2024 (UTC)[reply]

@HOTmag: I'd just say 'S has k linearly independent elements' (in the give example k=2). --CiaPan (talk) 14:46, 22 September 2024 (UTC)[reply]

Is your response a suggestion of rephrasing my question?

If it's intended to be an answer, then please note: My condition requires to be "using A,B,C only, i.e without using S". Additionally, where does your description claim, that S is linearly dependent? HOTmag (talk) 14:56, 22 September 2024 (UTC)[reply]

One way to characterise the set is "a set of vectors, any one of which can be written in terms of the others in a unique way". The set is just A, B and C, i.e. any property of them is a property of the set of them. --2A04:4A43:900F:F4C3:49F4:4EFB:C442:608F (talk) 15:15, 22 September 2024 (UTC)[reply]

Since my condition requires to be "using A,B,C only, i.e without using S", so I guess you mean the following: "Each vector, can be written as a unique linear combination of the other vectors". Thanks. HOTmag (talk) 15:52, 22 September 2024 (UTC)[reply]

Assuming the vectors are

v_{1},v_{2},\dots ,v_{k}\in R^{n}

, form the matrix

V

whose columns are

v_{1},\dots ,v_{k}

. The stated condition is then:

the matrix of $k\times k$ minors of $V$ is zero, and
each of the columns of the matrix of $(k-1)\times (k-1)$ minors of V has a non-zero entry.

- Tito Omburo (talk) 17:36, 22 September 2024 (UTC)[reply]

Your description, both uses sets, and also becomes longer than the original one indicated in the title. HOTmag (talk) 08:41, 23 September 2024 (UTC)[reply]

You can say, "each of the sets {A,B}, {A,C} and {B,C} is linearly independent". --Lambiam 21:05, 22 September 2024 (UTC)[reply]

You also have to add that the set {A,B,C} is linearly dependent, but then the description - both uses sets, and also becomes longer than the original one indicated in the title. HOTmag (talk) 08:40, 23 September 2024 (UTC)[reply]

There is a unique linear combination generating 0, and in this linear combination all coefficients are nonzero.2404:2000:2000:8:FDE8:8311:95E3:654D (talk) 00:00, 23 September 2024 (UTC)[reply]

Yes, and by symbolic notation the description even becomes shorter. Thanks. HOTmag (talk) 08:42, 23 September 2024 (UTC)[reply]