Proving Subspace of a Matrix in R2

[jj48]

Hi.

if anybody can help me

Let S be the matrix a b
c d

with a constrain a = -2d, b= 3c -d

Prove that S is a subspace of R2?

 

[jj48]

matrix

a b
c d

 

[Tac-Tics]

It looks like you have misread the problem.

A matrix cannot be a subspace. Even a set of matrices cannot be a subspace of R^2, because R^2 consists of points, not matrices.

 

[HallsofIvy]

Hi.

if anybody can help me

Let S be the matrix a b
c d

with a constrain a = -2d, b= 3c -d

Prove that S is a subspace of R2?

As Tac-Tics said, you must have misread the problem. You can't "prove that S is a subspace of R²", it isn't true. A matrix is NOT a subspace of R².

It is possible that you were asked to show that the set of all matrices of the form
\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}
is a subspace of the M(2,2), the vector space of all 2 by 2 matrices. You would do that by showing that the set is closed under addition and scalar multiplication. That is, if
M= \begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}
and
N= \begin{bmatrix}-2y & 3x- y \\ x & y\end{bmatrix}
are matrices in this set and "a" is a number
are
M+ N= \begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}+ \begin{bmatrix}-2y & 3x- y \\ x & y\end{bmatrix}
and
aM= a\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}
also in that set?

 

[trambolin]

Maybe there is a condition on matrix and it is asked to consider the set of (c,d) pairs. Then RxR makes sense seemingly...

 

[jj48]

Yes, you are right HallsofIvy. how i profe that they r close under multiplication and addition. do i have to add and them factorise to go back to the original?

 

[jj48]

thank you HallsofIvy you were right. I read wrong the question. it is if it is a subspace of the M(2,2), the vector space of all 2 by 2 matricess.

How to i proof that they are close under adittion and multiplication and if i consider the set B = |B1 B2|
B1 = 0 -6
-2 0

B2 = 2 4
1 -1

find if B is bases of the first question?

 

[HallsofIvy]

I thought I had responded to this but I don't find it now.

You have the set of all 2 by 2 matrices of the form
\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}
another such is
\begin{bmatrix}-2y & 3x- y \\ x & y\end{bmatrix}
Their sum is
\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}+\begin{bmatrix}-2y & 3x- y \\ x & y\end{bmatrix}
= \begin{bmatrix}-2d-2y & 3c- d+ 3x- y \\ c+ x & d+ y\end{bmatrix}= \begin{bmatrix}-2(d+y) & 3(c+x)- (c+y)\\ c+ x & d+y \end{bmatrix}
Do you see how that final matrix also satisifies the definition of this set of matrices?

If a is any number then
a\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}= \begin{bmatrix}-2ad & 3ac- ad \\ ac & ad\end{bmatrix}
Again, do you see how this matrix satisfies the condition to be in the set?

To show that
\begin{bmatrix}0 & -6 \\ -2 & 0\end{bmatrix}
and
\begin{bmatrix}2 & 4 \\ 1 & -1\end{bmatrix}
form a basis for the subspace, you must show that they satify the conditions for a basis: that they are independent and that they span the subspace.
If
A\begin{bmatrix}0 & -6 \\ -2 & 0\end{bmatrix}+ B\begin{bmatrix}2 & 4 \\ 1 & -1\end{bmatrix}= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}
Setting the matrices on the two sides of that equation equal gives you four equations to solve for A and B. If the only solution is A= B= 0, the matrices are independent

Then, given any numbers, c, d so that
\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}
is in the set, you must show that there exist numbers A and B such that
A\begin{bmatrix}0 & -6 \\ -2 & 0\end{bmatrix}+ B\begin{bmatrix}2 & 4 \\ 1 & -1\end{bmatrix}= \begin{bmatrix}-2d & 3c-d \\ c & d\end{bmatrix}
Again, you set the two matrices on either side of the equation equal so you get 4 equations for A and B. Here, you must only show that those equations have a solution no matter what c and d are.

 

[jj48]

thank you very much i got it