# Sub vector spaces

#### Yankel

##### Active member
Hello,

I am struggling with this question...

U is a set of all matrices of order 3X3, in which there is at least one row of 0's.

W is the set of matrices:

a b
c b+c-3a

where a,b,c are real numbers.

V is the set of vectors: (x,y,z,w), for which 5(y-1)=z-5

which two of these statements are correct ?

a. U is a vector subspace with dimension of 9
b. U is a vector subspace with dimension of 6
c. W is a vector subspace with dimension of 2
d. W is a vector subspace with dimension of 3
e. V is a vector subspace with dimension of 3
f. V is a vector subspace with dimension of 2
g. V and U are not sub-spaces

I think that U should be, but I don't know, the 0's confuse me, and I don't know about W and V.
thank you...

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
U is not a vector space because it is not closed under addition. W and V are vector spaces. For example, suppose (x, y, z, w) and (x', y', z', w') are in V, so 5(y - 1) = z - 5 and 5(y' - 1) = z' - 5. This is equivalent to 5y = z and 5y' = z', so 5(y + y') = z + z', or 5(y + y' - 1) = z + z' - 5. Therefore, (x + x', y + y', z + z', w + w') is also in V. Note that if the equation were 5(y - 1) = z - 6, then V would not be closed under addition. You need to check the remaining axioms of a vector space.

In the case of W, one can independently choose a, b and c; then the fourth element of the matrix is fixed. Therefore, the dimension of W is 3. In other words, $\begin{pmatrix}1 & 0\\0 & -3\end{pmatrix}$, $\begin{pmatrix}0 & 1\\0 & 1\end{pmatrix}$ and $\begin{pmatrix}0 & 0\\1 & 1\end{pmatrix}$ are the basis elements. Similarly, for V we have four numbers and one equation relating them, so the dimension again is 3.