3*2 Matrix with complex elements

**Hartlw** · October 21st 2011, 02:47 PM

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

**Hartlw** · October 22nd 2011, 08:58 AM

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?

**bugatti79** · October 22nd 2011, 11:39 AM

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

**Hartlw** · October 22nd 2011, 03:26 PM

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.

Comments.
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.

**Hartlw** · November 25th 2011, 10:10 AM

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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