# Thread: 3*2 Matrix with complex elements

1. ## 3*2 Matrix with complex elements

Folks,

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?

Are the following subsets subspaces?

1) The set of 3*2 with real entries

2) The set of matrices with first row (0,0)

3) The set of matrices with first row (1,1)

I understand the dimension of a vector space is the number of vectros in any basis for the space, ie the number of coordinates necessary to specify any vector. How do I find out what its dimension is?

Thanks

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

the "naive" way is to ask yourself: how many complex numbers do i need to specify to identify my "vector"? for 3x2 matrices, that number is 6 (one for each entry).

however, mathematics professors being what they are, will usually insist you display a basis (a linearly independent spanning set). can you think of a set of six matrices,

each of which captures the idea of "a single matrix coordinate"? once you have done this, show linear independence and spanning.

again, with any vector space V, there are 3 things you need to show for any subset W:

1) W is non-empty (preferrably by showing the 0-element is a member. if the 0-element (0-vector) is not in W, you will not obtain a vector space).
2) if u,v are in W then their vector sum u+v must also be.
3) if c is any scalar in your underlying field, and u is in W, then cu must also be in W. be careful with this one. if you are working with a complex vector space, it is not sufficient to check this property for real scalars only.

attempt to show these properties (or give a counter-example) for each of the sets in your post. that's how it's done.

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

Originally Posted by Deveno
the "naive" way is to ask yourself: how many complex numbers do i need to specify to identify my "vector"? for 3x2 matrices, that number is 6 (one for each entry).

can you think of a set of six matrices, each of which captures the idea of "a single matrix coordinate"? once you have done this, show linear independence and spanning.
Sorry, I am still struggling to get a start...I dont know how to approach this? If its a 3*2 matrix then would it have 3 solutions in $\displaystyle {\mathbb{R}^3}$?

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

If I write a sample 3*2 matrix as

$\displaystyle \begin{bmatrix}2 & 5\\ 1 & 4 \\ 6 & 8 \end{bmatrix}$

THis matrix contains 6 real entries, but I am unbale to write expression since I believe I need it to be a 3*3 or a 2*3 matrix. How would I proceed to show its closed under vector addition and scalar multiplication?

Thanks

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

you're not being asked to "solve" a system of linear equations, you're being asked to display a basis for a vector space.

here is a similar problem:

find the dimension of the vector space of all 2x2 complex matrices.

i claim:

$\displaystyle \left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin {bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\ 1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix} \right\}$

is a basis for this vector space. clearly this is a spaning set, since:

$\displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix} = a\begin{bmatrix}1&0\\0&0\end{bmatrix} + b\begin{bmatrix}0&1\\0&0\end{bmatrix} + c\begin{bmatrix}0&0\\1&0\end{bmatrix} + d\begin{bmatrix}0&0\\0&1\end{bmatrix}$

now, to show linear independence, suppose that:

$\displaystyle \begin{bmatrix}0&0\\0&0\end{bmatrix} = c_1\begin{bmatrix}1&0\\0&0\end{bmatrix} + c_2\begin{bmatrix}0&1\\0&0\end{bmatrix} + c_3\begin{bmatrix}0&0\\1&0\end{bmatrix} + c_4\begin{bmatrix}0&0\\0&1\end{bmatrix}$

then:

$\displaystyle \begin{bmatrix}0&0\\0&0\end{bmatrix} = \begin{bmatrix}c_1&c_2\\c_3&c_4\end{bmatrix}$

so $\displaystyle c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.

since our basis has 4 elements, the dimension of the vector space is 4.

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

Originally Posted by Deveno
you're not being asked to "solve" a system of linear equations, you're being asked to display a basis for a vector space.

here is a similar problem:

find the dimension of the vector space of all 2x2 complex matrices.

i claim:

$\displaystyle \left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin {bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\ 1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix} \right\}$

is a basis for this vector space. clearly this is a spaning set, since:

$\displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix} = a\begin{bmatrix}1&0\\0&0\end{bmatrix} + b\begin{bmatrix}0&1\\0&0\end{bmatrix} + c\begin{bmatrix}0&0\\1&0\end{bmatrix} + d\begin{bmatrix}0&0\\0&1\end{bmatrix}$

now, to show linear independence, suppose that:

$\displaystyle \begin{bmatrix}0&0\\0&0\end{bmatrix} = c_1\begin{bmatrix}1&0\\0&0\end{bmatrix} + c_2\begin{bmatrix}0&1\\0&0\end{bmatrix} + c_3\begin{bmatrix}0&0\\1&0\end{bmatrix} + c_4\begin{bmatrix}0&0\\0&1\end{bmatrix}$

then:

$\displaystyle \begin{bmatrix}0&0\\0&0\end{bmatrix} = \begin{bmatrix}c_1&c_2\\c_3&c_4\end{bmatrix}$

so $\displaystyle c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.

since our basis has 4 elements, the dimension of the vector space is 4.
I can only write 4 of the required 6 matrices...

$\displaystyle \left[ \begin{array}{ccc} 1 & 0 \\ 0 & 1\\ 1 & 1\end{array},\begin{array}{ccc} 0 & 1\\ 1 & 0\\ 0 & 0\end{array},\begin{array}{ccc} 1 & 1 \\ 0 & 0\\ 0 & 1\end{array},\begin{array}{ccc} 0 & 0 \\ 1 &1\\ 1 & 0\end{array}, \right]$

There is a lack of symmetry for me to complete....

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

One of the things you need to clarify is whether you are thinking of this as a vector space over the real numbers or over the complex numbers. For example, with the set of pairs of complex numbers, with the usual addition and scalar multiplication, over the real numbers, we can take (a+ bi, c+ di)= a(1, 0)+ b(i, 0)+ c(0, 1)+ d(0, i) so it has dimension 4. The same set over the complex numbers has (a+ bi, c+di)= (a+ bi)(1, 0)+ (c+di)(0, 1) and so the dimension is 2.

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

Originally Posted by HallsofIvy
One of the things you need to clarify is whether you are thinking of this as a vector space over the real numbers or over the complex numbers. For example, with the set of pairs of complex numbers, with the usual addition and scalar multiplication, over the real numbers, we can take (a+ bi, c+ di)= a(1, 0)+ b(i, 0)+ c(0, 1)+ d(0, i) so it has dimension 4. The same set over the complex numbers has (a+ bi, c+di)= (a+ bi)(1, 0)+ (c+di)(0, 1) and so the dimension is 2.
1) What is the significance of saying 'vector space over the real numbers' or 'vector space over the complex numbers'? You have written to different ways (RHS) of expressing the same LHS. I dont understand the difference.

2) Is the above an example for a 2*2 matrix?

3) How does one do the above for a 3*2?

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

the "underlying field" of a vector space can make a difference when talking about dimension.

i don't understand why you came up with those particular 4 3x2 matrices. why not the 6 matrices who have one entry 1, the rest 0?

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

Originally Posted by Deveno
you're not being asked to "solve" a system of linear equations, you're being asked to display a basis for a vector space.

here is a similar problem:

find the dimension of the vector space of all 2x2 complex matrices.

i claim:

$\displaystyle \left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin {bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\ 1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix} \right\}$

is a basis for this vector space. clearly this is a spaning set, since:

$\displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix} = a\begin{bmatrix}1&0\\0&0\end{bmatrix} + b\begin{bmatrix}0&1\\0&0\end{bmatrix} + c\begin{bmatrix}0&0\\1&0\end{bmatrix} + d\begin{bmatrix}0&0\\0&1\end{bmatrix}$

now, to show linear independence, suppose that:

$\displaystyle \begin{bmatrix}0&0\\0&0\end{bmatrix} = c_1\begin{bmatrix}1&0\\0&0\end{bmatrix} + c_2\begin{bmatrix}0&1\\0&0\end{bmatrix} + c_3\begin{bmatrix}0&0\\1&0\end{bmatrix} + c_4\begin{bmatrix}0&0\\0&1\end{bmatrix}$

then:

$\displaystyle \begin{bmatrix}0&0\\0&0\end{bmatrix} = \begin{bmatrix}c_1&c_2\\c_3&c_4\end{bmatrix}$

so $\displaystyle c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.

since our basis has 4 elements, the dimension of the vector space is 4.
$\displaystyle \left[\begin{array}{ccc} 0 & 0 \\ 0 & 0\\0 & 0\end{array}\right]=c_1\left[ \begin{array}{ccc} 1 & 0 \\ 0 & 0\\0 & 0\end{array}\right]+c_2\left[ \begin{array}{ccc} 0 & 1\\ 0 & 0\\ 0 & 0\end{array}\right]+c_3\left[ \begin{array}{ccc} 0 & 0 \\ 1 & 0\\ 0 & 0\end{array}\right]+c_4\left[ \begin{array}{ccc} 0 & 0 \\ 0 & 1\\ 0 & 0\end{array}\right]+c_5\left[ \begin{array}{ccc} 0 & 0 \\ 0 & 0\\ 1 & 0\end{array}\right]+c_6\left[ \begin{array}{ccc} 0 & 0 \\ 0 & 0\\ 0& 1\end{array}\right]$

implies $\displaystyle c_1+c_2+c_3+c_4+c_5+c_6=0$

This implies the set is linearly independant with a dimension of 6? What is next?

My attempt based on HallsofIvy post

3*2 matrix over the real numbers
$\displaystyle \left[\begin{array}{ccc} a+bi & c+di \\ e+fi & g+hi\\j+ki & l+mi\end{array}\right]=a(1,0)+b(i,0)+c(1,0)+d(i,0)........$ Hence 12 dimensions

3*2 over the complex numbers

$\displaystyle \left[\begin{array}{ccc} a+bi & c+di \\ e+fi & g+hi\\j+ki & l+mi\end{array}\right]=(a+bi)(1,0)+(c+di)(0,1)+........$ Hence 6 dimensions.........?

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

Every element of the matrix is an ordered pair. If the matrix has only one non-zero element, all matrices with only this one element are a linear combination of two matrices, one with (0,1) as the sole element, and the other with (1,0) as the sole element. So for each element in the matrix you need two matrices with (1,0) and (0,1) as the sole elements. You need a total of 12 matrices, two for each element. The dimension of the vector space is 12.

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

Originally Posted by Hartlw
Every element of the matrix is an ordered pair. If the matrix has only one non-zero element, all matrices with only this one element are a linear combination of two matrices, one with (0,1) as the sole element, and the other with (1,0) as the sole element. So for each element in the matrix you need two matrices with (1,0) and (0,1) as the sole elements. You need a total of 12 matrices, two for each element. The dimension of the vector space is 12.
For both the real case and the complex case? Is this not conflicting with post # 7

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

Originally Posted by bugatti79
For both the real case and the complex case? Is this not conflicting with post # 7
For the real case, every base matrix has a 1 in one element and every other element a 0, for a total of six, the dimension is six. For complex case, see my previous post.

EDIT: I think post seven is discussing a complex vector space consisting of complex vectors a + bi. Not the same thing as a complex vector space each element of which is a matrix. You can probably think of each matrix as the sum of a real (a,0) at each location and imaginary, (,0,b) at each location, component . In any event, any complex matrix can be expressed as a linear combination of the 12 base matrices I described in my previous post.

EDIT: The matrices themselves satisfy requirements of linear vector space with the usual definition of matrix addition and multiplication by a scalar.

The base matrices are:

(0,1) (0,0)
(0,0) (0,0)
(0,0) (0,0)

(1,0) (0,0)
(0,0) (0,0)
(0,0) (0,0)

(0,0) (1,0)
(0,0) (0,0)
(0,0) (0,0)

..... for 12 base matrices (dim 12).

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

Originally Posted by Hartlw
I think post seven is discussing a complex vector space consisting of complex vectors a + bi.
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

$\displaystyle \left[\begin{array}{ccc} a+bi & c+di \\ e+fi & g+hi\\j+ki & l+mi\end{array}\right]=a(1,0)+b(i,0)+c(1,0)+d(i,0)........$

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

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.

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