# Math Help - 3*2 Matrix with complex elements

1. ## Re: 3*2 Matrix with complex elements

Originally Posted by bugatti79
Sorry, I was editing my post while you replied to me. See update. Thanks
What is the question:

Are all matrices obtained from the given matrix by addition and multiplaction a subset? First case? Yes

Are all matrices with 1,1 in the first row closed under addition? No

2. ## Vector Space over Field F

Vector Space over Field F:

An association of scalars a, b, c.... from a field F with Vectors A, B, C... (undefined) from an Abelian group G.

The association has certain properties, for ex, aA is defined and belongs to G.

EDIT: A typical question might be, is multiplication (rule of association) of a Matrix (Vector) by a scalar from a field F (real or complex) a Vector Space?
1) Give the scalar and define the matrix.
2) Are the matrices an Abelian Group?
3) Are the rules of association satisfied? Especially does aA belong to G?

3. ## Re: Vector Space over Field F

ok, if you can explain why the first of the following is a subset and the second is not then I
will mark thread as solved. Thanks :-)

Let $W=([a_{ij}]:a_{ij} \in \mathbb{R})$. Let $A=\begin{bmatrix}1&1\\ 1&0\\0 & 1\end{bmatrix} \in W$

Take take scalar $\alpha \in \mathbb{R}$

Therefore $\alpha*A=\begin{bmatrix}\alpha &\alpha \\ \alpha&0 \\ 0 & \alpha\end{bmatrix} \in W$. Hence is a subset.

Is W a subspace for the set of matrices with first row (1,1)

$W=\begin {bmatrix} 1&1\\a_{21} & a_{22}\\a_{31}&a_{32}\end {bmatrix}:a_{ij} \in \mathbb{C}$

$2A \not \in W since 2A=\begin{bmatrix} 2&2\\A*a_{21}&A*a_{22}\\A*a_{31}&A*a_{32} \end{bmatrix}$. Why is this NOT a subset because '2' is scalar in $\matthbb{R}$ just like $\alpha$ was above....ie the '1' is 'in' the set when you just take out the factor 2

4. ## Re: Vector Space over Field F

Originally Posted by bugatti79
ok, if you can explain why the first of the following is a subset and the second is not then I
will mark thread as solved. Thanks :-)

Let $W=([a_{ij}]:a_{ij} \in \mathbb{R})$. Let $A=\begin{bmatrix}1&1\\ 1&0\\0 & 1\end{bmatrix} \in W$

Take take scalar $\alpha \in \mathbb{R}$

Therefore $\alpha*A=\begin{bmatrix}\alpha &\alpha \\ \alpha&0 \\ 0 & \alpha\end{bmatrix} \in W$. Hence is a subset.

Is W a subspace for the set of matrices with first row (1,1)

$W=\begin {bmatrix} 1&1\\a_{21} & a_{22}\\a_{31}&a_{32}\end {bmatrix}:a_{ij} \in \mathbb{C}$

$2A \not \in W since 2A=\begin{bmatrix} 2&2\\A*a_{21}&A*a_{22}\\A*a_{31}&A*a_{32} \end{bmatrix}$. Why is this NOT a subset because '2' is scalar in $\matthbb{R}$ just like $\alpha$ was above....ie the '1' is 'in' the set when you just take out the factor 2
Vector Space over Field F:
An association of scalars a, b, c.... from a field F with Vectors A, B, C... (undefined) from an Abelian group G, which satisfy certain properties, including aA is defined and belongs to G.

First case:
1) What are the vectors? All matrices aA where a is real.
Are they a group? Yes. So they are a sub-group Gs of W.
Define multiplication of members B of Gs by a real scalar. aB is defined and closed (belongs to Gs). Therefore it is a subspace of W.

Second Case:
1) What are the vectors? All matrices with 1 in the first row. This is not a group so it is not a subgroup..
or
2) All real matrices with equal real numbers in the first row. This is a subgroup of Gs. Define multiplication of members B of Gs by a real scalar a. aB is defined and closed (first row equal, belongs to Gs). Therefore it is a subspace of W.

If W has real components, nothing with complex components is a sub-group of W.
The difference between first and second case is what you define as a vector.

Complex numbers are all ordered pairs of real numbers (a,b). (1,0) is a complex number, 1 isn't unless you explicitly state it is a complex number, in which case it is really (1,0)

EDIT: As a trivia lbut interesting example:
Define the vector A as all 3X2 matrices with 1 in the first row.
Define vector addition as usual for second 2 rows but always 1 for first row.
Define multiplication by a scalar as usual for second two rows but always 1 for the first two rows.

Is it a vector space? yes. For all practical purposes the first two rows are just decoration.
Is it a subspace of W? No. Because it is not a subspace for the same opeations of addition and multiplication.

EDIT: Factoring out to give ones in the top row. You can define a vector as a factored matrix with one's in the first row. But then you are not using the same rule of scalar multiplication as W, which says that a scalar times a matrix is identical to the matrix with all elements multiplied by the scalar. So you could define a vector as all matrices factored to give ones in the first row, it would be a space, but not a subspace because you are changing the rules of scalar multiplication by saying that the factored matrix is not identical to the unfactored matrix.

5. ## Re: 3*2 Matrix with complex elements

Been meaning to clean up my posts:

1) Both elements of an mXn matrix and matrix scalar multipliers are from a field F, real or complex: Matrix has dimension mxn.

eg, basis of a 1X2 matrix (a11,a12) is (1,0), (0,1) and any 1X2 matrix can be expressed as z1x(1,0) + z2x(0,1), z1 and z2 real or complex, but not mixed.

2) Elements of an mXn matrix are (u,v), u and v and scalar multipliers are real or complex. Matrix has dimension 2xmxn.

eg, basis of a 1X2 matrix (a11,a12) is ((1,0),(0,0)), ((0,1),(0,0)), ((0,0),(1,0)), ((0,0),(0,1)), and any 1X2 matrix (a11,a12) can be expressed as z1x((1,0),(0,0))+ z2x((0,1),(0,0)) + z3x((0,0),(1,0)) + z4x((0,0),(0,1)).

The source of confusion in my previous posts was failure to distinguish between (u,v) as shortcut notation for the complex number u+iv with (u1,v1)x(u2,v2) = (u1+iv1)x(u2+iv2), and (u,v) as a 2D vector with real or complex components and (u1,v1)x(u2,v2) not defined but z1x(u1,v1) = (z1xu1,z1xv1).

EDIT

With regard to the questions in #33

If (x,y,z) is a 3d vector space:

a(1,0,1) is a 1d vector subspace. (analogy of first part)

(1,1,0) + a(0,0,1) = (1,1,a) is a linear manifold of dimension 1 (no 0 vector).
a(1,1,1) = (1,1,a) is not a vector unless defined as above, = (1,1,0) + (0,0,a). So the answer to second part is that it is not a subspace as defined.

[a(1,1,x) is not the same as the vector whose first two components are always 1,1]

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