# Thread: 3*2 Matrix with complex elements

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

Originally Posted by Hartlw
No

If you denote the matrices in my previous post by E1, E2, E3 , E4, ...
Then an arbitrary complex matrix is expressed as a linear combination of E1, E2, E3,...,E12

(1,3) (5,3)
(3,5) (5,7) = 1xE1 + 3xE2 + 5xE3 + ..........
(3,3) (4,3)

Above is a matrix. The ordered pair (1,3) is just a std shortcut way of expressing 1+i3.

A complex matrix has to be expressed as a sum of complex matrices. Not as a sum of complex numbers. The addition of two complex matrices is a complex matrix by term by term addition, added just like complex numbers.
if the underlying field of a vector space is $\displaystyle \mathbb{C}$, which is what one normally means by a "complex vector space", then you DO treat each complex number as a number, not as matrix. the fact that these complex numbers may themselves be expressible as pairs of real numbers isn't relevant, unless you are considering the 3x2 matrices as a "real vector space".

as a complex vector space, the dimension is indeed 6, and NOT 12. the number 1 (= 1 + 0i) is in fact a basis for $\displaystyle \mathbb{C}$ over the field $\displaystyle \mathbb{C}$. i do not wish to seem pedantic about this, but you are completely wrong, and liable to confuse the thread-starter. HallsofIvy is completely correct that the underlying field does affect the dimension, but post #7 does not assert that the original question is to be interpreted as the matrices in question being a "real vector space".

this is contextually clear from the subsequent question:

1) The set of 3*2 with real entries (.....is it a subspace?)

this is sort of a "trick question" because the subset is closed under addition, but it is NOT closed under scalar multiplication, for such a (non-zero) matrix A, the matrix iA is not in the set.

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

Originally Posted by Deveno
if the underlying field of a vector space is $\displaystyle \mathbb{C}$, which is what one normally means by a "complex vector space", then you DO treat each complex number as a number, not as matrix. the fact that these complex numbers may themselves be expressible as pairs of real numbers isn't relevant, unless you are considering the 3x2 matrices as a "real vector space".

as a complex vector space, the dimension is indeed 6, and NOT 12. the number 1 (= 1 + 0i) is in fact a basis for $\displaystyle \mathbb{C}$ over the field $\displaystyle \mathbb{C}$. i do not wish to seem pedantic about this, but you are completely wrong, and liable to confuse the thread-starter. HallsofIvy is completely correct that the underlying field does affect the dimension, but post #7 does not assert that the original question is to be interpreted as the matrices in question being a "real vector space".

this is contextually clear from the subsequent question:

1) The set of 3*2 with real entries (.....is it a subspace?)

this is sort of a "trick question" because the subset is closed under addition, but it is NOT closed under scalar multiplication, for such a (non-zero) matrix A, the matrix iA is not in the set.
Assume the usual definitions for matrix addition (add corresponding elements) and multiplication by a scalar (multiply each element by the scalar).

1) The set of mxn matrices with real elements, and real scalars is a vector space of dimension mxn.

2) The set of mxn matrices with complex elements and real scalars is a vector space of dimension 2xmxn:

Example: 1x2 matrix: [(a,b) (c,d)] with basis E1 = [(1,0) (0,0)], E2 = [(0,1) (0,0)], E3 = [(0,0) (1,0)], E4 = [(0,0) (0,1)].
[(a,b) (c,d)] = aE1 + bE2 + cE3 + dE4

3) The set of real mxn matrices with complex scalars is not a vector space because iA is not a real matrix.

4) The set of complex mxn matrices with complex scalars is easily defined by usual definitions of matrix addition , and scalar multiplication. This is also a vector space because addition and multiplication by a scalar are defined and closed:

Addition is obvious and aA is also a complex matrix
a(A+B) = aA + bB because Z1(Z2+Z3) = Z1xZ2 + Z1xZ3 for individual elements

A basis and dimension for this case has to be thought out.

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

Originally Posted by bugatti79
So based on your post is this the answer to the following original question?

Given that the set of all 3*2 matrices with complex elements is a complex vector space, with the usual definitions of addition and scalr multiplication of matrices, what is its dimension?

3*2 matrix over the complex numbers

$\displaystyle \left[\begin{array}{ccc} a+bi & c+di \\ e+fi & g+hi\\j+ki & l+mi\end{array}\right]=a(1,0)+b(i,0)+c(1,0)+d(i,0)........$
Now you are really getting lost. The linear combination on the right is of pairs of Complex numbers, $\displaystyle C^2$, while on the left you have a 3 by 2 matrix. They can't be equal.

For example, if your set were the set of 2 by 2 matrices with complex entries, you could still have it as a vector space over the real numbers or the complex numbers. The first would mean that while the entries can be complex, your scalars must be real numbers. The second would allow both entries and scalars to be complex numbers.

As a vector space over the complex numbers, we could write a general 2 by 2 matrix as
$\displaystyle \begin{bmatrix}a+bi & c+di \\ e+fi ^& g+hi\end{bmatrix}= (a+ bi)\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+ (c+ di)\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}+(e+ fi)\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}+(g+ hi)\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$
and so has dimension 4.

As a vector over the real numbers, where we only allow real numbers for scalars (but still can have complex numbers as entries in the matrix, we can write
$\displaystyle \begin{bmatrix}a+bi & c+di \\ e+fi ^& g+hi\end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+b\begin{bmatrix} i & 0 \\ 0 & 0\end{bmatrix}+ c\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}+ d\begin{bmatrix}0 & ui \\ 0 & 0\end{bmatrix}$$\displaystyle +e\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}+ f\begin{bmatrix}0 & 0 \\ i & 0\end{bmatrix}+g\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}+ h\begin{bmatrix}0 & 0 \\ 0 & i\end{bmatrix}$
and so has dimension 8.

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

Originally Posted by Hartlw
Assume the usual definitions for matrix addition (add corresponding elements) and multiplication by a scalar (multiply each element by the scalar).

1) The set of mxn matrices with real elements, and real scalars is a vector space of dimension mxn.

2) The set of mxn matrices with complex elements and real scalars is a vector space of dimension 2xmxn:

Example: 1x2 matrix: [(a,b) (c,d)] with basis E1 = [(1,0) (0,0)], E2 = [(0,1) (0,0)], E3 = [(0,0) (1,0)], E4 = [(0,0) (0,1)].
[(a,b) (c,d)] = aE1 + bE2 + cE3 + dE4

3) The set of real mxn matrices with complex scalars is not a vector space because iA is not a real matrix.

4) The set of complex mxn matrices with complex scalars is easily defined by usual definitions of matrix addition , and scalar multiplication. This is also a vector space because addition and multiplication by a scalar are defined and closed:

Addition is obvious and aA is also a complex matrix
a(A+B) = aA + bB because Z1(Z2+Z3) = Z1xZ2 + Z1xZ3 for individual elements

A basis and dimension for this case has to be thought out.
Basis and dimension for 4) are the same as for 3).

It was shown in 3) that any mxn complex matrix A can be expressed as a linear combination of mxnx2 linearly independent complex matrices. Then so can aA if a is a complex scalar.

The ordered pair (a,b) is identically a+ib in my posts in this thread.

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

Originally Posted by HallsofIvy
Now you are really getting lost. The linear combination on the right is of pairs of Complex numbers, $\displaystyle C^2$, while on the left you have a 3 by 2 matrix. They can't be equal.
Is my left hand side $\displaystyle \in {\mathbb{C}}^3?$

Thank you guys, I appreciated the kind help. SO I can extend the concept to a 3*2 matrix ie

Over the real number, the dimension is 12

$\displaystyle \begin{bmatrix}a+bi & c+di \\ e+fi ^& g+hi\\ j+ki ^& l+mi \end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 0\\0&0\end{bmatrix}+b\begin{bmatrix} i & 0 \\ 0 & 0\\0&0\end{bmatrix}+ c\begin{bmatrix}0 & 1 \\ 0 & 0\\0&0\end{bmatrix}+ d\begin{bmatrix}0 & i \\ 0 & 0\\0&0\end{bmatrix}$

$\displaystyle +e\begin{bmatrix}0 & 0 \\ 1 & 0\\0&0\end{bmatrix}+ f\begin{bmatrix}0 & 0 \\ i & 0\\0&0\end{bmatrix}+g\begin{bmatrix}0 & 0 \\ 0 & 1\\0&0\end{bmatrix}+ h\begin{bmatrix}0 & 0 \\ 0 & i\\0&0\end{bmatrix}$

$\displaystyle + j\begin{bmatrix}0 & 0 \\ 0 & 0\\1&0\end{bmatrix}+ k\begin{bmatrix}0 & 0 \\ 0 &0\\0&i\end{bmatrix}+ l\begin{bmatrix}0 & 0 \\ 0 & 0\\1&0\end{bmatrix}+ m\begin{bmatrix}0 & 0 \\ 0 & 0\\0&i\end{bmatrix}$

Over the complex numbers, dimension is 6?
Thanks

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

You have expressed a complex matrix (an mxn matrix with complex components) as a linear combination of mxnx2 linearly independent complex matrices, so the basis dimension is 12. Period.

There is another way of looking at it which is a matter of convention.

If you assume that the matrix is an ordered mxn n-tuple (a vector), you can write it out as an mxn n-tuple (vector).

if aij is a typical component of A, then:

A $\displaystyle \equiv$ (a11,a12...,a1n, a21,a22,....,a2n,........,am1, am2,......,amn)

The ntuple has mxn components. If they are real, they represent the real matrix as an mxn dimensional vector space with mxn base vectors (1,0,0,0,0,0...), (0,1,000000), (0,0,1,0000) .......

If the matrix is complex, then the vector form of A has complex components and it takes two real numbers to specify each component, so now the dimension is 2xmxn.
A set of 2xmxn base vectors is now: [(0,1),(0,0),(0,0),...........], [(1,0),(0,0),(,0),.......], [(0,0),(1,0),(0,0,0.....], [(0,0),(0,1)....], .......

The above converts the question to:
What is the dimension of a vector (n-tuple) with complex components? Another way to look at (which I,ve never heard of) is write double the number of components with the understanding that the first component is the real part of a complex number a1 and the second component is the imaginary part of a complex number a1, the THIRD component is the real part of a2, the fourth componet is the imaginary component of a2, etc.

It would be interesting to examine vectors with complex components.

Personally, I think my previous post was very concise and clear.

(You are trying to say something like a complex number is an ordered pair over a real Field, but that's reduntant when you specify the aij as complex.)

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

Originally Posted by Hartlw

You have expressed a complex matrix (an mxn matrix with complex components) as a linear combination of mxnx2 linearly independent complex matrices, so the basis dimension is 12. Period.

There is another way of looking at it which is a matter of convention.

If you assume that the matrix is an ordered mxn n-tuple (a vector), you can write it out as an mxn n-tuple (vector).

if aij is a typical component of A, then:

A $\displaystyle \equiv$ (a11,a12...,a1n, a21,a22,....,a2n,........,am1, am2,......,amn)

The ntuple has mxn components. If they are real, they represent the real matrix as an mxn dimensional vector space with mxn base vectors (1,0,0,0,0,0...), (0,1,000000), (0,0,1,0000) .......

If the matrix is complex, then the vector form of A has complex components and it takes two real numbers to specify each component, so now the dimension is 2xmxn.
A set of 2xmxn base vectors is now: [(0,1),(0,0),(0,0),...........], [(1,0),(0,0),(,0),.......], [(0,0),(1,0),(0,0,0.....], [(0,0),(0,1)....], .......

The above converts the question to:
What is the dimension of a vector (n-tuple) with complex components? Another way to look at (which I,ve never heard of) is write double the number of components with the understanding that the first component is the real part of a complex number a1 and the second component is the imaginary part of a complex number a1, the THIRD component is the real part of a2, the fourth componet is the imaginary component of a2, etc.

It would be interesting to examine vectors with complex components.

Personally, I think my previous post was very concise and clear.

(You are trying to say something like a complex number is an ordered pair over a real Field, but that's reduntant when you specify the aij as complex.)
Is it ok to continue with what I wrote?

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

"It would be interesting to examine vectors with complex components."

No luck on the internet. Halmos calls an n-tuple (a1,a2,a3...,an) with a1,a2,a3,...complex, an n-dimensional complex vector space, but doesn't exhibit an n-dimensional basis, which is obviously impossible because you can't express a+bi as ze for e a constant and a+bi arbitrary. So I can understand the confusion.

I think my posts above deal with the whole question very niceley and explicitly: Any complex mxn matrix can be expressed as a linear combination of 2xmxn linearly independent complex matrices. The dimension is 2xmxn.

EDIT:
Original Post

ques 1: all matrices with 0,0 in first row are a subspace because addition of these matrices or multiplication by a complex scalar results in a matrix with zeroes in the first row.

ques 2: matrices with 1's in the first row are not a subspace because after addition or multiplication by a scalar they no longer have 1's in the first row.

I wish I knew how to refer to previous posts in edit mode. But referring to your previous post, you can't refer tio a complex matrix over real numbers.

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

OK. I think we are home. From Mirsky:

"A vector X of order n over a Field F is an ordered set (x1,x2,...xn) of n numbers which belong to F.

By rules of vector addition, and multiplication by a scalar, the vectors X constitute a vector space Vn.

Any vector in Vn can be expressed as a linear combination of e1=(1,0,0,..,0), e2=(0,1,0,..,), ..., en = (0,0,0....,1)"

So the dimension of the space of all mxn matrices with elements over (any) F
is mxn (considering the matrix as identical to an ordered mxn n-tuple (vector)).

The 3x2 matrix with real or complex terms has dimension 6.

Thanks for your persistence. Sorry I confused things.

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

If I have a 3 dimensional complex vector space, a basis e1,e2,e3 each of the form a+bi, and an arbitrary vector (x1,x2,x3) with x1,x2,x3 complex, can I solve,

(x1,x2,x3) = p1e1 +p2e2 +p3e3 for p1,p2,p3?

The answer is yes, as illustrated by the one dimensional case:

x = pe all complex. x arbitrary and e a basis vector. solve for p. In complex form:

xR+i(PI) = [pR+i(pI)][eR+i(eI) and equate real and complex parts:
xR = pR(eR) - pI(eI)
xI = pI(eR) + pR(eI)
which are two eqs in the two unknowns pR and pI.

So this illustrates that a vector with n complex components can be expressed as a linear combination of n linearly independent, arbitrary, complex vectors, not just (1,0,0,...), (0,1,0...), (0,0,1,...), .....(0,0,0...,1)

It probably follows logically in the abstract sense from what was said by Mirsky above, but I had to see it for myself.

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

Ok, Im finally piecing things together....
Given that the set of all 3*2 matrices with complex elements is a complex vector space, with the usual definitions of addition and scalar multiplication of matrices
Are the following subsets subspaces?

1) The set of 3*2 with real entries?

$\displaystyle V=( \begin{bmatrix}a_{11}& a_{12}\\ a_{21}& a_{22}\\ a_{31}& a_{32}\end{bmatrix}: a_{ij} \in \mathbb{C})$ Complex Vector Space.

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

Take scalar i $\displaystyle \in \mathbb{C}$

Therefore $\displaystyle i*A=\begin{bmatrix}i&i\\ i&0\\0 & i\end{bmatrix} \not \in W$. Hence not a subset.

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

Therefore $\displaystyle \alpha*A=\begin{bmatrix}\alpha &\alpha \\ \alpha&0 \\ 0 & \alpha\end{bmatrix} \not \in W$. Hence not a subset.

Now, for closed under addition, is this correct.

Let $\displaystyle A=\begin{bmatrix}1&1\\ a_{21}& a_{22}\\ a_{31}& a_{32}\end{bmatrix}$ and $\displaystyle B=\begin{bmatrix}1&1\\ b_{21}& b_{22}\\ b_{31}& b_{32}\end{bmatrix}$

Adding these 2 results in it NOT being closed under addition because of addition of corresponding first row will not result in elements of 1?

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

Originally Posted by bugatti79
Ok, Im finally piecing things together....
Given that the set of all 3*2 matrices with complex elements is a complex vector space, with the usual definitions of addition and scalar multiplication of matrices
Are the following subsets subspaces?

1) The set of 3*2 with real entries?

$\displaystyle V=( \begin{bmatrix}a_{11}& a_{12}\\ a_{21}& a_{22}\\ a_{31}& a_{32}\end{bmatrix}: a_{ij} \in \mathbb{C})$ Complex Vector Space.

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

Take scalar i $\displaystyle \in \mathbb{C}$

Therefore $\displaystyle i*A=\begin{bmatrix}i&i\\ i&0\\0 & i\end{bmatrix} \not \in W$. Hence not a subset.

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

Therefore $\displaystyle \alpha*A=\begin{bmatrix}\alpha &\alpha \\ \alpha&0 \\ 0 & \alpha\end{bmatrix} \not \in W$. Hence not a subset.

Now, for closed under addition, is this correct.

Let $\displaystyle A=\begin{bmatrix}1&1\\ a_{21}& a_{22}\\ a_{31}& a_{32}\end{bmatrix}$ and $\displaystyle B=\begin{bmatrix}1&1\\ b_{21}& b_{22}\\ b_{31}& b_{32}\end{bmatrix}$

Adding these 2 results in it NOT being closed under addition because of addition of corresponding first row will not result in elements of 1?
The only way to deal with this is rigid adherence to definition:

Vector Space: Addition, and multiplication by a scalar, are defined and closed.
A complex vector has complex entries.

1) Complex matrices with real scalars are a vector space.
2) Complex matrices with complex scalars are a vector space.
3) Real matrices with real scalars are a vector space.
4) Real matrices with complex scalars are not a vector space.

1 is a member of the set of complex numbers if it =1+0i

In the Field of complex matrices, elements with no imaginary components are complex matrices. So in the set of complex matrices an entry of 1 is a complex number.

A 3x2 matrix with real entries is not a subset of the original set because the original set doesn't contain real entries. But this is a matter of semantics. If you mean entries with imaginary component = 0, then your "real" matrix is a subset under addition and multiplication with real scalars, but not with complex scalars because you can get an entry with an i component.

Matrices with 1 in first row don't create a subset, as you say.

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

Originally Posted by bugatti79
Take take scalar $\displaystyle \alpha \in \mathbb{R}$

Therefore $\displaystyle \alpha*A=\begin{bmatrix}\alpha &\alpha \\ \alpha&0 \\ 0 & \alpha\end{bmatrix} \not \in W$. Hence not a subset.
If the above is wrong then why is the answer to the following question NOT a subset?

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

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

$\displaystyle 2A \not \in W since 2A=\begin{bmatrix} 2&2\\A*a_{21}&A*a_{22}\\A*a_{31}&A*a_{32} \end{bmatrix}$

It's wrong.

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

Originally Posted by Hartlw
It's wrong.
Sorry, I was editing my post while you replied to me. See update. Thanks

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