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Basis, dimension and vector spaces

Yankel

Active member
Jan 27, 2012
398
Hello all,

I have these two sets (I couldn't use the notation {} in latex, don't know how).

V is the set of matrices spanned by these 3 matrices written below. W is a set of 2x3 matrices applying the rule a+e=c+f

\[V=span(\begin{pmatrix} 1 &1 &1 \\ 1 &3 &7 \end{pmatrix},\begin{pmatrix} 0 &0 &0 \\ 1 &1 &1 \end{pmatrix},\begin{pmatrix} 0 &0 &0 \\ 0 &2 &3 \end{pmatrix})\]

\[W=(\begin{pmatrix} a &b &c \\ e &f &g \end{pmatrix}|a+e=c+f)\]

For both sets, I need to determine if it is a sub-vector space, I need to find the basis and dimension.

For W, I think it is fairly easy to prove it is a vector space (sub space - assuming R2x3 is a vector space, which I am allowed to do in this example). However I am not so sure about the dimension. Can I write a=c+f-e and say that dim(W)=5 ?

My bigger problem is V. I managed to show that a matrix in V is from the shape:

\[\begin{pmatrix} a &a &a \\ a+b &3a+b+2c &7a+b+3c \end{pmatrix}\]

but I am not sure how from here I show that it is a vector space (I am looking for the 3 rules: 0 matrix, + and scalar *). In addition, I don't know what the dimension is (my intuition say dim(V)=3)

Can you assist please ?

Thank you in advance !
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
...I couldn't use the notation {} in latex, don't know how...
Precede the curly braces with a backslash. For example:

\{A,B,C\} gives you \(\displaystyle \{A,B,C\}\)

and

\left\{\frac{1}{2},\frac{1}{3},\frac{1}{4} \right\} gives you \(\displaystyle \left\{\frac{1}{2},\frac{1}{3},\frac{1}{4} \right\}\)
 

Yankel

Active member
Jan 27, 2012
398
Thanks, I'll try to remember it next time.

Any ideas about my question ?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Some general things that apply:

For starters, the set $\text{Mat}_{m \times n}(F)$ of all $m \times n$ matrices with entries in a field $F$ is always a vector space of dimension $mn$.

A subset $U$ of a vector space $V$ is a subspace if the following 3 statements hold:

1) $u,u' \in U \implies u+u' \in U$ (closure under vector addition)
2) $u \in V, \alpha \in F \implies \alpha u \in U$ (closure under scalar multiplication)
3) $0 \in U$ (this implies $U$ is non-empty, and conversely, if $U$ is non-empty, then: (2) implies for any $u \in U$, that $-u = (-1)u \in U$, and then (1) implies $u + -u = 0 \in U$, so (3) is sometimes replaced with the condition $U \neq \emptyset$)

For your first set, the span of any set automatically guarantees (1) and (2) (that's what span MEANS). So the real question is: is this set linearly independent, and if not, what is a maximal linearly independent subset? It's easiest in this case to proceed backwards. The set consisting of just the 3rd matrix is clearly linearly independent, since it is a non-zero matrix.

The set consisting of the last TWO matrices is linearly independent if and only if the second matrix is not a scalar multiple of the third (or vice versa). Since the 2,1-entry of the second matrix is 1, and the 2,1-entry of the third matrix is 0, the only way the second matrix could be a scalar multiple of the third is if it was the third matrix multiplied by 0. But this would make the second matrix 0, and it is NOT a 0-matrix, so the last two matrices form a linearly independent set.

Now all three matrices form a linearly independent set if and only if the first matrix is NOT a linear combination of the last two.

But ANY linear combination of the last two would have a 0 row as its first row, and the first row of the first matrix is NOT 0. Hence it is a linearly independent set, and thus is a basis for the set it spans, that is: $\text{dim}_{\ F}(V) = 3$.

To check if $W$ forms a subspace of $\text{Mat}_{2 \times 3}(\Bbb R)$ (which has dimension 6), let:

$A = \begin{bmatrix}a&b&c\\e&f&g \end{bmatrix}$

$A' = \begin{bmatrix}a'&b'&c'\\e'&f'&g' \end{bmatrix}$

be any two matrices belonging to $W$.

Then:

$A + A' = \begin{bmatrix}a+a'&b+b'&c+c'\\e+e'&f+f'&g+g' \end{bmatrix}$

Now we are given that:

$a+e = c+f$ and $a'+e' = c'+f'$, and we must show that:

$(a+a') + (e+e') = (c+c') + (f+f')$.

If you can do that, then that proves (1). I think that is enough to show you how to proceed with (2) and (3).

So if it IS a subspace we know that $1 \leq \text{dim}_{\ \Bbb R}(W) \leq 6$ (since it has non-zero matrices in it, for example the matrix consisting of all 1's).

One way to show it has a certain dimension, is to exhibit a basis with that number of elements. I'll get you started. Two linearly independent matrices in $W$ are:

$E_{12} = \begin{bmatrix}0&1&0\\0&0&0 \end{bmatrix}$

and:

$E_{23} = \begin{bmatrix}0&0&0\\0&0&1 \end{bmatrix}$

A third linearly independent element of $W$ is:

$E_{11} + E_{22} = \begin{bmatrix}1&0&0\\0&1&0 \end{bmatrix}$

This means that $3 \leq \text{dim}_{\ \Bbb R}(W) \leq 6$.

Prove that $\{E_{12},E_{23},E_{11}+E_{22},E_{13}+E_{22}\}$ is likewise linearly independent.

Since we can find some 2x3 matrices NOT in $W$, this narrows the dimension down to either 4 or 5. Can you find a 5th linearly independent matrix?
 

Yankel

Active member
Jan 27, 2012
398
Thank you for your detailed answer, it is very helpful. I want to ask something about my view of things.

For the set W, can't I look at the condition a+e=c+f, and make it a=c+f-e, saying that only 1 out of 6 parameters can be expressed by the others, and thus the dim(W)=5 ?

while a basis is:

\[\begin{pmatrix} 0 &1 &0 \\ 0 &0 &0 \end{pmatrix},\begin{pmatrix} 1 &0 &1 \\ 0 &0 &0 \end{pmatrix},\begin{pmatrix} -1 &0 &0 \\ 1 &0 &0 \end{pmatrix},\begin{pmatrix} 0 &0 &0 \\ 0 &1 &1 \end{pmatrix},\begin{pmatrix} 1 &0 &0 \\ 0 &0 &1 \end{pmatrix}\]

As for V, if I write V as

\[\begin{pmatrix} a &a &a \\ a+b &3a+b+2c &7a+b+3c \end{pmatrix}\]

Can't I just count 3 parameters and assume dim(V)=3 ?
It is the same answer you got, but is this reasoning valid ?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Thank you for your detailed answer, it is very helpful. I want to ask something about my view of things.

For the set W, can't I look at the condition a+e=c+f, and make it a=c+f-e, saying that only 1 out of 6 parameters can be expressed by the others, and thus the dim(W)=5 ?

while a basis is:

\[\begin{pmatrix} 0 &1 &0 \\ 0 &0 &0 \end{pmatrix},\begin{pmatrix} 1 &0 &1 \\ 0 &0 &0 \end{pmatrix},\begin{pmatrix} -1 &0 &0 \\ 1 &0 &0 \end{pmatrix},\begin{pmatrix} 0 &0 &0 \\ 0 &1 &1 \end{pmatrix},\begin{pmatrix} 1 &0 &0 \\ 0 &0 &1 \end{pmatrix}\]

As for V, if I write V as

\[\begin{pmatrix} a &a &a \\ a+b &3a+b+2c &7a+b+3c \end{pmatrix}\]

Can't I just count 3 parameters and assume dim(V)=3 ?
It is the same answer you got, but is this reasoning valid ?
Re-examine your proposed basis: the last two elements are not even members of $W$!

Yes, you can rephrase linear independence in terms of "parameters" but you must verify that these parameters lead to linear independence!! To see what I mean let's look at $V$:

We have:

$\begin{bmatrix}a&a&a\\a+b&3a+b+2c&7a+b+3c \end{bmatrix} =$

$a\begin{bmatrix}1&1&1\\1&3&7 \end{bmatrix} + b\begin{bmatrix}0&0&0\\1&1&1 \end{bmatrix} + c\begin{bmatrix}0&0&0\\1&2&3 \end{bmatrix}$

Linear independence MEANS that if these 3 sum to the 0-matrix, we MUST have:

$a = b = c = 0$.

In this particular case, that leads to the four equations:

$a = 0$
$a+b = 0$
$3a+b+2c = 0$
$7a+b+3c = 0$

and it is easy to see from equation (1) that equation (2) becomes $b = 0$, and then equation (3) becomes $2c = 0$, so $c = 0$.

To see why we HAVE to check linear independence consider the following example:

$V = \text{span}\left\{\begin{bmatrix}1&1\\1&0 \end{bmatrix}, \begin{bmatrix}1&1\\0&1 \end{bmatrix} \begin{bmatrix}1&1\\2&-1 \end{bmatrix}\right\}$

Again, we can write a "typical" element of $V$ as:

$\begin{bmatrix}a+b+c&a+b+c\\a+2c&b-c \end{bmatrix}$

which has "3 parameters", but watch closely what happens when we check for linear independence:

We get the 3 equations:

$a+b+c = 0$
$a+2c = 0$
$b-c = 0$

Subtracting (3) from (1), we get:

$a + 2c = 0$, which leads to $c = \dfrac{-a}{2}$

Then equation (2) gives:

$b = \dfrac{-a}{2}$

So, choosing $a = 2$, we get:

$2\begin{bmatrix}1&1\\1&0 \end{bmatrix} - 1\begin{bmatrix}1&1\\0&1 \end{bmatrix} - 1\begin{bmatrix}1&1\\2&-1 \end{bmatrix} = \begin{bmatrix}0&0\\0&0 \end{bmatrix}$

Which means that the set is linearly DEPENDENT (we can express one of the matrices as a linear combination of the other two). SO you have to be CAREFUL when expressing things via parameters, because there may be hidden relationships that are not immediately apparent.

Yes, with $W$ it turns out that you may freely choose $a,b,c,d$ and $g$ and then $f$ is determined, but you should be able to express this fact in terms of linear independence. The easiest way to do this is choose ONE of $a,b,c,d,g$ to be 1, and all the others 0, and show that that results in a basis. But it can be done in MANY ways, because bases aren't unique.

The reason I stress this is as a sanity check: "shortcuts" can lead to mistakes. Definitions exist for a reason: not to make things HARD for you, but to keep you from making unwarranted assumptions that can lead you astray.