Standard/Non-standard basis (vectors)?

Question

I have a problem understanding the classification of standard and non standard basis. The question is:

x = V*a = [ -1 1 ]  [ -1 ] = [ 5 ]
...............[ 1 1 ].  [ 4 ] ... [ 3 ]

Match the properties of x = V*a
End of question.

We are given 4 choices to classify each sub question:
a) standard basis
b) non-standard basis
c) coordinate vector in standard basis
d) coordinate vector in non standard basis

* Ignore all full stops only, they're there for easier viewing
The subquestions are as follows:

1)
[ -1 ] , [ 1 ]
[ 1 ] .. [ 1 ]

2)
[ 5 ]
[ 3 ]

3)
[ 1 ] , [ 0 ]
[ 0 ] . [ 1 ]

4)
5i + 4j

5)
- [ -1 ] + 4 [ 1 ]
. [ 1 ] ...... [ 1 ]

6)
[ -1 ]
[ 4 ]

Match a choice with a subquestion. Some explanation would be appreciated. Thanks.

husoski · Accepted Answer

The standard basis is specifically {(1, 0, 0,..., 0), (0, 1, 0, ..., 0), ..., (0, 0, 0, ..., 1)} by definition, so in R^2 that is {(1, 0), (0, 1)}.  Only (3) fits that description, where the 2x1 layout:

[ 1 ]
[ 0 ] ... is being used instead of (1, 0), and likewise for the (0, 1) vector.

(1) is not the standard basis, but any two independent vectors do make a basis for a 2D space, so that makes this a "non-standard basis".

(2) Not a standard basis vector, but any non-zero vector can be a part of a basis, so "non-standard basis vector"

(3) see above.

(4) Assuming i=(1,0) and j=(0,1) are the usual shorthand for the R^2 standard basis vectors, this is another name for (5,4).  As in (2), this is a "non-standard basis vector".

(5) and (6) are also nonzero vectors, (5, 0) and (-1, 4) respectively, and are both "non-standard basis vectors".

It was necessary to calculate the result in (5) to be sure it didn't turn out to be (1, 0) or (0, 1) in disguise.

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Standard/Non-standard basis (vectors)?

1 Answer