MHB Dimension and Basis of a Relation in R5

[Logan Land]

Find dimension and basis of the set of all points in R5 whose coordinates satisfy the relation x1 +x2 +x3 +x4 =0.

shouldn't there be 5 vectors to satisfy the basis since they are asking about R5? but the relation only has x1, x2, x3, and x4.

or would my matrix just look like this

1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0

so x1=x2=x3=x4=0?dimension = 5 since its asking R5?

 

[Euge]

Find dimension and basis of the set of all points in R5 whose coordinates satisfy the relation x1 +x2 +x3 +x4 =0.

shouldn't there be 5 vectors to satisfy the basis since they are asking about R5? but the relation only has x1, x2, x3, and x4.

or would my matrix just look like this

1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0

so x1=x2=x3=x4=0?dimension = 5 since its asking R5?

Hi LLand314,

Since $x_4 = -x_1 - x_2 - x_3$, we let $x_1$, $x_2$, $x_3$, and $x_5$ be our free variables. So your vector space, call it $W$, is four-dimensional. A basis can be found by setting a free variable equal to one and every other free variable equal to 0. Setting $x_1 = 1$ and all other free variables equal to $0$, we get $x_4 = -1$. So $(1, 0, 0, -1, 0)$ is a basis vector. Show that the procedures three other basis vectors: $(0, 1, 0, -1, 0)$, $(0, 0, 1, -1, 0)$, and $(0, 0, 0, 0, 1)$. So a basis for $W$ is

$$\{(1,0,0,-1,0), (0,1,0,-1,0), (0,0,1,-1,0), (0,0,0,0,1)\}$$

 

[Logan Land]

Hi LLand314,

Since $x_4 = -x_1 - x_2 - x_3$, we let $x_1$, $x_2$, $x_3$, and $x_5$ be our our free variables. So your vector space, call it $W$, is four-dimensional. A basis can be found by setting a free variable equal to one and every other free variable equal to 0. Setting $x_1 = 1$ and all other free variables equal to $0$, we get $x_4 = -1$. So $(1, 0, 0, -1, 0)$ is a basis vector. Show that the procedures three other basis vectors: $(0, 1, 0, -1, 0)$, $(0, 0, 1, -1, 0)$, and $(0, 0, 0, 0, 1)$. So a basis for $W$ is

$$\{(1,0,0,-1,0), (0,1,0,-1,0), (0,0,1,-1,0), (0,0,0,0,1)\}$$

so is my dimension 4 or 5? my book says dim(R^n)=n so that would be 5 correct? but then you stated four-dimensional

 

[Euge]

so is my dimension 4 or 5? my book says dim(R^n)=n so that would be 5 correct? but then you stated four-dimensional

You're not computing the dimension of $\Bbb R^5$, but the dimension of a subspace of $\Bbb R^5$, which need not be $5$-dimensional. Recall that the dimension of a finite-dimensional vector space $V$ is equal to the number of elements in a basis for $V$. Since $W$ has a basis consisting of four vectors (the list is already displayed), $\text{dim}\, W = 4$.

 

[Logan Land]

You're not computing the dimension of $\Bbb R^5$, but the dimension of a subspace of $\Bbb R^5$, which need not be $5$-dimensional. Recall that the dimension of a finite-dimensional vector space $V$ is equal to the number of elements in a basis for $V$. Since $W$ has a basis consisting of four vectors (the list is already displayed), $\text{dim}\, W = 4$.

I see, that makes much more sense, the R^n isn't the same as $\Bbb R^5$

Still a little confused on when you found the basis
{(1,0,0,−1,0),(0,1,0,−1,0),(0,0,1,−1,0),(0,0,0,0,1)}
how does that equal 0 when x1+x2+x3+x4

 

[Euge]

I see, that makes much more sense, the R^n isn't the same as $\Bbb R^5$

Still a little confused on when you found the basis
{(1,0,0,−1,0),(0,1,0,−1,0),(0,0,1,−1,0),(0,0,0,0,1)}
how does that equal 0 when x1+x2+x3+x4

Sorry, but the question does not make sense to me. Each basis vector lies in $W$. For example, $(1,0,0,-1,0)$ lies in $W$ since $1 + 0 + 0 - 1 = 0$.

 

[Logan Land]

Sorry, but the question does not make sense to me. Each basis vector lies in $W$. For example, $(1,0,0,-1,0)$ lies in $W$ since $1 + 0 + 0 - 1 = 0$.

OOOOOOOO I see now, I was thinking that {(1,0,0,−1,0),(0,1,0,−1,0),(0,0,1,−1,0),(0,0,0,0,1)} all together needed to equal 0
I see now what you mean. That was actually an easy question I just was thinking of it completely wrong and making it difficult. thanks!