Answer: (i) \(\dim V=3\).
(ii) The three given vectors are a basis of \(V\).
(iii) \(W\) is not a vector space, since it does not contain the zero vector.
(iv) If \(\begin{pmatrix}x\\y\\z\\t\end{pmatrix}\in W\), then \(x+y\ne z+t\).
Row-reducing \(\mathbf M\) gives
\(\begin{pmatrix}1&2&3&4\\1&-1&2&3\\1&-3&3&5\\1&4&2&2\end{pmatrix}\sim\begin{pmatrix}1&2&3&4\\0&3&1&1\\0&0&5&8\\0&0&0&0\end{pmatrix}\).
There are three non-zero rows, so \(\operatorname{rank}\mathbf M=3\). Hence \(\dim V=3\).
The three given vectors are the first three columns of \(\mathbf M\), so they belong to \(V\). To show that they are linearly independent, suppose
\(a\begin{pmatrix}1\\1\\1\\1\end{pmatrix}+b\begin{pmatrix}2\\-1\\-3\\4\end{pmatrix}+c\begin{pmatrix}3\\2\\3\\2\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}\).
Then
\(a+2b+3c=0\),
\(a-b+2c=0\),
\(a-3b+3c=0\).
Subtracting the third equation from the first gives \(5b=0\), so \(b=0\). Then the first two equations become \(a+3c=0\) and \(a+2c=0\), which give \(c=0\) and then \(a=0\).
Thus the three vectors are linearly independent. Since \(V\) has dimension 3, these three linearly independent vectors in \(V\) form a basis of \(V\).
The set \(W\) is not a vector space, because \(\mathbf0\in V\), so \(\mathbf0\notin W\). A vector space must contain the zero vector.
Finally, any vector in \(V\) can be written as
\(\alpha\begin{pmatrix}1\\1\\1\\1\end{pmatrix}+\beta\begin{pmatrix}2\\-1\\-3\\4\end{pmatrix}+\gamma\begin{pmatrix}3\\2\\3\\2\end{pmatrix}=\begin{pmatrix}x\\y\\z\\t\end{pmatrix}\).
So
\(x=\alpha+2\beta+3\gamma\), \(y=\alpha-\beta+2\gamma\),
\(z=\alpha-3\beta+3\gamma\), \(t=\alpha+4\beta+2\gamma\).
Therefore
\(x+y=2\alpha+\beta+5\gamma\), and \(z+t=2\alpha+\beta+5\gamma\).
Hence every vector in \(V\) satisfies \(x+y=z+t\). Since \(V\) is 3-dimensional and the equation \(x+y=z+t\) defines a 3-dimensional subspace, this condition describes exactly \(V\).
Therefore, if a vector belongs to \(W\), meaning it is not in \(V\), it must satisfy \(x+y\ne z+t\).
