Subspace U = LH {a,b} subset of C^2,... what is dimension of U?

[Jenkins99]

"Given is the subspace

. . . . .\(\displaystyle U\, =\, LH\, \{a,\, b\}\, \subseteq \, \mathbb{C}^2,\, \mbox{ with }\, a\, =\, \left(\begin{array}{c}1\\i\end{array}\right)\, \mbox{ and }\, b\, =\, \left(\begin{array}{c}-i\\1\end{array}\right)\)

How big is the dimension of U?"

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I determined the number of independent vectors:

a (1, i) + b(-i, 1) = (0, 0)

a - ib = 0
ia + b = 0

a = ib
i(ib) + b = 0
i^2(b) ≠ -b.

Two independent vectors; therefore, a 2-dimensional subspace.

However, the answer given is dim(U) = 1.

 

[Ishuda]

"Given is the subspace

[see picture]

How big is the dimension of U?"

I determined the number of independent vectors:

a (1, i) + b(-i, 1) = (0, 0)

a - ib = 0
ia + b = 0

a = ib
i(ib) + b = 0
i^2(b) ≠ -b. <=========== wrong

Two independent vectors; therefore, a 2-dimensional subspace.

However, the answer given is dim(U) = 1.

See above. Assuming the i is the normal i² = -1, the second equation becomes
i² b + b = -b + b = 0.
for any b. Thus, using the original a and b in the problem
\(\displaystyle i\, a\, +\, b\, =\, i\, \begin{pmatrix} 1\\ i \end{pmatrix}\, +\, \begin{pmatrix} -i\\ 1 \end{pmatrix}\)
\(\displaystyle =\, \begin{pmatrix} i\\ i^2 \end{pmatrix}\, +\, \begin{pmatrix} -i\\ 1 \end{pmatrix}\)
\(\displaystyle =\, \begin{pmatrix} i\\ -1 \end{pmatrix}\, +\, \begin{pmatrix} -i\\ 1 \end{pmatrix}\)
\(\displaystyle =\, \begin{pmatrix} 0\\ 0 \end{pmatrix}\)
So the dimension of the space is one.

 

[Jenkins99]

Must one assume that i is an imaginary number? I treated i like any other
variable.

 

[Ishuda]

Must one assume that i is an imaginary number? I treated i like any other
variable.

Correct. As I said, assuming i² = -1, ...