Thread: Vector space.

Let $\displaystyle V = M_{22}$ the set of all 2 x 2 matrices. Let the operation of vector addition in V be the usual matrix addition but let scalar multiplication in V be defined by:

$\displaystyle c \cdot A = cA^{T}$

Show that V is not a vector space by demonstrating with an example that one of the properties in the definition of a vector space fails to hold.

My attempt:
property I believe fails to hold (let r be any real number):
$\displaystyle r (A+B) = r(A) + r(B)$

I'll define both A and B as the 2 x 2 matrices:
$\displaystyle A=\begin{bmatrix}a & b\\c & d \end{bmatrix}$ and $\displaystyle B=\begin{bmatrix}e & f\\g & h \end{bmatrix}$

$\displaystyle r\begin{bmatrix}a & b\\c & d \end{bmatrix}^{T} + r\begin{bmatrix}e & f\\g & h \end{bmatrix}^{T}$ => $\displaystyle r\begin{bmatrix}a & c\\b & d \end{bmatrix} + r\begin{bmatrix}e & g\\f & h \end{bmatrix}$=>$\displaystyle \begin{bmatrix}ra & rb\\rc & rd \end{bmatrix} + \begin{bmatrix}re & rf\\rg & rh \end{bmatrix}$=> $\displaystyle \begin{bmatrix}ra+re & rb+rf\\rc+rg & rd+rh \end{bmatrix} \notin M_{22}$

Hence, V is not a vector space since it does not hold for the distributivity of scalar multiplication with respect to vector addition. Did I go about this proof correctly?

no, the matrix you would up with certainly is still a 2x2 matrix.

and, in fact, it IS true that r.(A+B) = r(A+B)^t = r(A^t + B^t) = r.A + r.B.

how about checking if r.(s.A) = (rs).A....?

Originally Posted by Deveno

no, the matrix you would up with certainly is still a 2x2 matrix.

and, in fact, it IS true that r.(A+B) = r(A+B)^t = r(A^t + B^t) = r.A + r.B.

how about checking if r.(s.A) = (rs).A....?

OK:
$\displaystyle r \cdot (s \cdot A) = r \cdot (s \cdot \begin{bmatrix}a & b\\c & d \end{bmatrix}^{T}) => r \cdot (s \cdot \begin{bmatrix}a & c\\b & d \end{bmatrix}) => r \cdot \begin{bmatrix}sa & sc\\sb & sd \end{bmatrix}) =>\begin{bmatrix}rsa & rsc\\rsb & rsd \end{bmatrix}$

While: $\displaystyle (r \cdot s) \cdot A => (r \cdot s)\begin{bmatrix}a & b\\c & d \end{bmatrix}^{T} => (r \cdot s) \begin{bmatrix}a & c\\b & d \end{bmatrix} => \begin{bmatrix}ra & rc\\rb & rd \end{bmatrix} \cdot \begin{bmatrix}sa & sc\\sb & sd \end{bmatrix} => \begin{bmatrix}rsa^{2} & rsc^{2}\\rsb^{2} & rsd^{2} \end{bmatrix}$

Clearly, $\displaystyle r \cdot (s \cdot A) \neq (r \cdot s) A$ and therefore, V is not a vector space. Correct?

Also, a general question: How would you approach such a question in a circumstance where you are pressed for time? I mean, in an exam, it is quite time consuming to check through all eight properties of a vector space to determine whether a set is a vector space or not. Any general or intuitive tips that you could give? Thanks in advance.

again, this is incorrect.

$\displaystyle r \cdot(s \cdot A) = r \cdot (sA^T) = r(sA^T)^T = r(s(A^T)^T) = r(sA) = (rs)A$

whereas:

$\displaystyle (rs) \cdot A = (rs)A^T$

if A =

$\displaystyle \begin{bmatrix}0&1\\0&0\end{bmatrix}$, for example, these two matrices are NOT the same.

$\displaystyle (r \cdot s)A$ makes no sense at all, a scalar multiplication is a map $\displaystyle F \times V \to V$, not $\displaystyle F \times F \to F$