Let $V$ be the solution space of the following homogeneous linear system: $$\begin{align} x_1 − x_2 − 2x_3 + 2x_4 − 3x_5 &= 0\\ x_1 − x_2 − x_3 + x_4 − 2x_5 &= 0.\end{align} $$ Find $\dim(V)$ and a subspace $W$ of $\mathbb R^5$ such that $W$ contains $V$ and $\dim(W) = 4$. Justify your answer.

Not sure how to go about doing this.

  • $\begingroup$ "$V$ is a basis of $S$." There are many bases. $\endgroup$ Mar 16, 2020 at 6:08
  • $\begingroup$ The zero vector makes any set of vectors linearly dependent. Also, it's an element of every subspace, so you wouldn’t be adding anything. $\endgroup$
    – amd
    Mar 16, 2020 at 6:38
  • $\begingroup$ Have updated the question accordingly, thank you for the input $\endgroup$
    – XDXDXD
    Mar 16, 2020 at 6:49

1 Answer 1


Let $$A=\begin{bmatrix}1&-1&-2&2&-3\\1&-1&-1&1&-2\end{bmatrix}$$ The rows of $A$ are linearly independent, so the rank of $A$ is 2. Since A has 5 columns, the dimension of $V$, which is the solution space of $A$, is 5-2=3. Let $B=[1,-1,-2,2,-3].$ Let $W$ be the solution space of $B.$ $B$ has rank 1, so the dimesion of $W$ is 5-1=4. Any solution of both of the given equaions is a soluion of the first equation, so $V$ is a subspace of $W.$


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