Vector space dimension question

P-Jay1

Prove that the space of 2×2 real matrices forms a vector space of dimension 4 over
R. [12]

Im unsure, anyone any idea?

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boaz

If I understood right, we have to find an isomorphism from $M^{R}_{2x2}$ to $R^4$.

In other word, we need to find a linear transformation, $T:M^{R}_{2x2} \rightarrow R^4$ that $ker(T)=0$.

HINT: look at T[a,b;c,d]=(a,b,c,d)

Fredrik

Staff Emeritus
Gold Member
What's your book's definition of "dimension 4". I use the definition that says that a vector space is 4-dimensional if it contains a linearly independent set with 4 members, but no linearly independent set with 5 members. So to solve your problem, I would find a set of 4 linearly independent matrices and prove that every matrix is a linear combination of those four.

P-Jay1

So what does ker(T) stand for? Im still really clueless. So do I find an isomorphism from T[a,b;c,d] to (a,b,c,d)?

Fredrik

Staff Emeritus
Gold Member
So what does ker(T) stand for? Im still really clueless. So do I find an isomorphism from T[a,b;c,d] to (a,b,c,d)?
ker T is the set of all x such that Tx=0. Finding that isomorphism is one way to do it, but not the only one. Edit: Wait, you said "from T[a,b;c,d]". I don't know what you mean by that. I meant that one way to solve the problem is to find an isomorphism from the set of 2×2 matrices to ℝ4.

boaz

ker T is the set of all x such that Tx=0. Finding that isomorphism is one way to do it, but not the only one. Edit: Wait, you said "from T[a,b;c,d]". I don't know what you mean by that. I meant that one way to solve the problem is to find an isomorphism from the set of 2×2 matrices to ℝ4.
I guess I wasn't much clear. What I meant is that T takes a matrix [a,b;c,d] and gives a vector (a,b,c,d). Actually, I think that this is what he has meant.

P-Jay1

A = (a b c d)

A = a (1 0 0 0) + b (0 1 0 0) + c (0 0 1 0) + d (0 0 0 1)

Please excuse my lack of ability to lay this out correctly but each of these in brackets have 2 rows and 2 columns i.e 2 x 2. not 1 x4

a,b... Eℝ is a vector space over ℝ

Is this the correct approach?

boaz

it seems like you know the answer but don't know how to do this.
so first of all, lets state that "in order to prove that the space of 2×2 real matrices forms a vector space of dimension 4 over R. we will define an isomorphism from 2x2 real matrices space to R4". Later, we will say. "lets take a look at the following transformation T:M2x2->R4, T([a,b;c,d])=(a,b,c,d)"
now, what is left to do is to show that it is linear transformation and that it is a isomorphism.

regarding the linear transformation, it's not a big deal. i am sure you can handle it by yourself.
to show that T is isomorphism we need to prove that T is a linear transformation and ker(T)=0 or im(T)=4.

to do that - we need to know first what im(T) is equal. if we take a base of one space and we transform it using T we will get the base of the other space we are transforming to. so lets take the standard base of M_2x2(R). { [1,0;0,0], [0,1;0,0], [0,0;1,0], [0,0;0,1]}. from what we have defined before:
T([1,0;0,0])=(1,0,0,0); T([0,1;0,0])=(0,1,0,0) ... T([0,0;0,1])=(0,0,0,1).
in other terms, imT(M_2x2(R) ) = sp{(1,0,0,0),...,(0,0,0,1)}, but the set {(1,0,0,0)...(0,0,0,1)} is the standard base of R4 thus dim(imT)=4. so in conclusion, T is an isomorphism, and we proved what we have asked for.

Last edited:

HallsofIvy

Homework Helper
Any 2 by 2 matrix is of the form
$$\begin{bmatrix}a & b \\ c & d \end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}+ b\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}+ c\begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}+ d\begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}$$
The most "basic" definition of "dimension" is that it is the number of vectors in any space. It is easy to see that these four matrices are a basis.

Or, if as Boaz does, you wish to find an isomorphism,
$$f\left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right)= (a, b, c, d)$$
works nicely.

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